Chapter-8: Class 11 Binomial Theorem

Formula for Binomial Theorem:

[a + b]ⁿ = [ ⁿC₀ × aⁿ ] + [ ⁿC₁ × (a^{n – 1}) × b ] + [ ⁿC₂ × (a ^{n – 2}) × b² ] + [ ⁿC₃ × (a^{n – 3} )× b³ ] + . . . . . . . . . . . . . + [ ⁿC_{n – 1} × a × (b^{n – 1}) ] + [ ⁿC_n × bⁿ ]

\(\Rightarrow\) Binomial Coefficient:

The coefficients ⁿC₀, ⁿC_1, ⁿC₂ . . . . . . . . . ⁿC_n occurring in the Binomial Theorem are known as Binomial coefficients

 

Some conclusions from Binomial Theorem:

(i) [x + y]ⁿ = [ ⁿC₀ × (xⁿ) ] + [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] + [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . + [ ⁿC_n × yⁿ ]

(ii) [x – y]ⁿ = [ ⁿC₀ × (xⁿ) ] – [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] – [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . +(-1)ⁿ [ ⁿC_n × yⁿ ]

(iii) [1 – x]ⁿ = [ ⁿC₀ ] – [ ⁿC₁ . x ] + [ ⁿC₂ . x² ] – [ ⁿC₃ . x³ ] + . . . . . . . . . . . . . . . . . . + (-1)ⁿ [ ⁿC_n . xⁿ ]

(iv) (a + b)ⁿ = \(\sum_{r\;=\;0}^{n}\) ⁿC_r (a)^{n – r} × b^r

 

IMPORTANT POINTS:

(1.) ⁿC_r = \(\frac{n!}{r!(n-r)!}\) where, n is a non-negative integer and [0 ≤ r ≤ n]

(2.) ⁿC₀ = ⁿC_n = 1

(3.) There are total (n + 1) terms in the expansion of (a + b)ⁿ

 

Exercise 8.1

 

Q.1: Expand the Expression (1 – 3x)⁵

 

Answer. Based on formula given in Binomial Theorem

By using Binomial Theorem:

Since, (1 – x)ⁿ = [ ⁿC₀ ] + [ ⁿC₁ × (-x) ] + [ ⁿC₂ × (-x)² ] + [ ⁿC₃ × (-x³) ] + . . . . . . . . . . . . . . . . . . + [ ⁿC_n × (-x)ⁿ ]

Therefore we get, (1 – 3x)⁵ = [ ⁵C₀ ] – [ ⁵C₁ × (3x) ] + [ ⁵C₂ × (3x)² ] – [ ⁵C₃ × (3x)³ ] + [ ⁵C₄ × (3x)⁴ ] – [ ⁵C₅ × (3x)⁵ ]

\(\\\Rightarrow\) 1 – [ 5 × (3x) ] + [ 10 × (9x²) ] – [ 10 × (27x³) ] + [ 5 × (81x⁴) ] – [ 1 × (243x⁵) ]

\(\\\Rightarrow\) 1 – 15x + 90x² – 270x³ + 405x⁴ – 243x⁵

Therefore we get, (1 – 3x)⁵ = 1 – 15x + 90x² – 270x³ + 405x⁴ – 243x⁵

 

Q.2: Expand the Expression \(\left ( \frac{5}{x}-\frac{x}{5} \right )^{5}\)

 

Answer. Based on formula given in Binomial Theorem

By using Binomial Theorem:

Since, [x – y]ⁿ = [ ⁿC₀ × (xⁿ) ] – [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] – [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . +(-1)ⁿ [ ⁿC_n × yⁿ ]

Therefore we get, \(\left ( \frac{5}{x}-\frac{x}{5} \right )^{5}:\)

[⁵C₀ × \(\left ( \frac{5}{x} \right )^{5}\) ] – [⁵C₁ × \(\left ( \frac{5}{x} \right )^{4}\) × \(\left ( \frac{x}{5} \right )^{1}\) ] + [⁵C₂ × \(\left ( \frac{5}{x} \right )^{3}\) × \(\left ( \frac{x}{5} \right )^{2}\) ] – [ ⁵C₃ × \(\left ( \frac{5}{x} \right )^{2}\) × \(\left ( \frac{x}{5} \right )^{3}\) ] + [⁵C₄ × \(\left ( \frac{5}{x} \right )^{1}\) × \(\left ( \frac{x}{5} \right )^{4}\) ] – [⁵C₅ × \(\left ( \frac{x}{5} \right )^{5}]\\\)

\(\\\Rightarrow \left [ 1\times \frac{3125}{x^{5}}\right ]-\left [ 5\times \frac{625}{x^{4}}\times \frac{x}{5}\right ]+\left [ 10\times \frac{125}{x^{3}}\times \frac{x^{2}}{25}\right ]-\left [ 10\times \frac{25}{x^{2}}\times \frac{x^{3}}{125} \right ]+\left [ 5\times \frac{5}{x}\times \frac{x^{4}}{625} \right ]-\left [ 1\times \frac{x^{5}}{3125} \right ]\\\) \(\\\Rightarrow \left [\frac{3125}{x^{5}}\right ]-\left [\frac{625}{x^{3}}\right ]+\left [ \frac{50}{x}\right ]-2x+\left [ \frac{3x^{3}}{25} \right ]-\left [\frac{x^{5}}{3125} \right ]\\\)

Therefore we get, \(\\\left ( \frac{5}{x}-\frac{x}{5} \right )^{5}\):

\(\\\left [\frac{x^{5}}{3125} \right ] +\left [ \frac{3x^{3}}{25} \right ]-2x+\left [ \frac{50}{x}\right ]-\left [\frac{625}{x^{3}}\right ]+ \left [\frac{3125}{x^{5}}\right ]\)

 

Q.3: Expand the Expression (3x – 2)⁶

 

Answer. Based on formula given in Binomial Theorem

By using Binomial Theorem:

[x – y]ⁿ = [ ⁿC₀ × (xⁿ) ] – [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] – [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . +(-1)ⁿ [ ⁿC_n × yⁿ ]

Therefore we get, (3x – 2)⁶ = [ ⁶C₀ × (3x)⁶ ] – [ ⁶C₁ × (3x)⁵ × (2) ] + [ ⁶C₂ × (3x)⁴ × (2)² ] – [ ⁶C₃ × (3x)³ × (2)³ ] + [ ⁶C₄ × (3x)² × (2)⁴ ] – [ ⁶C₅ × (3x)¹ × (2)⁵ ] + [ ⁶C₆ × (2)⁶ ]

\(\\\Rightarrow\) [1 × (729x⁶)] – [6 × (243x⁵) × 2] + [15 × (81x⁴) × 4] – [20 × (27x³) × 8 ] + [15 × (9x²) × 16] – [6 × (3x) × 32] + [1 × 64]

\(\\\Rightarrow\) 729x⁶ – 2916x⁵ + 4860x⁴ – 4320x³ + 2160x² – 576x + 64

Therefore we get, (3x – 2)⁶ = 729x⁶ – 2916x⁵ + 4860x⁴ – 4320x³ + 2160x² – 576x + 64

 

Q.4: Expand the Expression \(\left ( \frac{4}{x}+\frac{x}{3} \right )^{5}\)

 

Answer. Based on formula given in Binomial Theorem

By using Binomial Theorem:

Since, [x + y]ⁿ = [ ⁿC₀ × (xⁿ) ] + [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] – [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . + [ ⁿC_n × yⁿ ]

Therefore we get, \(\left ( \frac{4}{x}+\frac{x}{3} \right )^{5}\):

[ ⁵C₀ × \(\left ( \frac{4}{x} \right )^{5}\) ] + [ ⁵C₁ × \(\left ( \frac{4}{x} \right )^{4}\) × \(\left ( \frac{x}{3} \right )^{1}\) ] + [ ⁵C₂ × \(\left ( \frac{4}{x} \right )^{3}\) × \(\left ( \frac{x}{3} \right )^{2}\) ] + [ ⁵C₃ × \(\left ( \frac{4}{x} \right )^{2}\) × \(\left ( \frac{x}{3} \right )^{3}\) ] + [ ⁵C₄ × \(\left ( \frac{4}{x} \right )^{1}\) × \(\left ( \frac{x}{3} \right )^{4}\) ] + [ ⁵C₅ × \(\left ( \frac{x}{3} \right )^{5}]\\\)

\(\\\Rightarrow \left [ 1\times \frac{1024}{x^{5}}\;\right ]+\left [ 5\times \frac{256}{x^{4}}\times \frac{x}{3}\;\right ]+\left [ 10\times \frac{64}{x^{3}}\times \frac{x^{2}}{9}\;\right ]+\left [ 10\times \frac{16}{x^{2}}\times \frac{x^{3}}{27} \right ]+\left [ 5\times \frac{4}{x}\times \frac{x^{4}}{81}\; \right ]+\left [ 1\times \frac{x^{5}}{243} \right ]\\\\\) \(\\\Rightarrow \left [\frac{1024}{x^{5}}\right ]+\left [\frac{1280}{3x^{3}}\right ]+\left [ \frac{640}{9x}\right ]+\left [\frac{160x}{27}\right ]+\left [ \frac{20x^{3}}{81} \right ]+\left [\frac{x^{5}}{243} \right ]\\\)

Therefore we get, \(\left ( \frac{4}{x}+\frac{x}{3} \right )^{5}:\)

\(\\\Rightarrow \left [\frac{x^{5}}{243} \right ]+\left [ \frac{20x^{3}}{81} \right ]+\left [\frac{160x}{27}\right ]+\left [ \frac{640}{9x}\right ]+\left [\frac{1280}{3x^{3}}\right ]+\left [\frac{1024}{x^{5}}\right ]\)

 

Q.5: Expand the Expression \(\left ( \frac{1}{x}-\frac{x}{2} \right )^{6}\)

 

Answer. Based on formula given in Binomial Theorem

By using Binomial Theorem:

Since, [x – y]ⁿ = [ ⁿC₀ × (xⁿ) ] – [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] – [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . +(-1)ⁿ [ ⁿC_n × yⁿ ]

Therefore we get, \(\left ( \frac{1}{x}-\frac{x}{2} \right )^{6}\):

[ ⁶C₀ × \(\left ( \frac{1}{x} \right )^{6}\) ] – [ ⁶C₁ × \(\left ( \frac{1}{x} \right )^{5}\) × \(\left ( \frac{x}{2} \right )^{1}\) ] + [ ⁶C₂ × \(\left ( \frac{1}{x} \right )^{4}\) × \(\left ( \frac{x}{2} \right )^{2}\) ] – [ ⁶C₃ × \(\left ( \frac{1}{x} \right )^{3}\) × \(\left ( \frac{x}{2} \right )^{3}\) ] + [ ⁶C₄ × \(\left ( \frac{1}{x} \right )^{2}\) × \(\left ( \frac{x}{2} \right )^{4}\) ] – [ ⁶C₅ × \(\left ( \frac{1}{x} \right )^{1}\) × \(\left ( \frac{x}{2} \right )^{5}\) ] + [ ⁶C₆ ×\(\left ( \frac{x}{2} \right )^{6}]\\\)

\(\\\Rightarrow \left [ 1\times \frac{1}{x^{6}}\;\right ]-\left [ 6\times \frac{1}{x^{5}}\times \frac{x}{2}\;\right ]+\left [ 15\times \frac{1}{x^{4}}\times \frac{x^{2}}{4}\;\right ]-\left [ 20\times \frac{1}{x^{3}}\times \frac{x^{3}}{8} \right ]+\left [ 15\times \frac{1}{x^{2}}\times \frac{x^{4}}{16}\; \right ]-\left [ 6\times \frac{1}{x}\times \frac{x^{5}}{32} \right ]+\left [ 1\times \frac{x^{6}}{64} \right ]\\\) \(\\\Rightarrow \left [ \frac{1}{x^{6}}\;\right ]-\left [ \frac{3}{x^{4}}\right ]+\left [ \frac{15}{4x^{2}}\right ]-\left [ \frac{5}{2} \right ]+\left [\frac{15x^{2}}{16}\; \right ]-\left [\frac{3x^{4}}{16} \right ]+\left [\frac{x^{6}}{64} \right ]\\\)

Therefore we get, \(\left ( \frac{1}{x}-\frac{x}{2} \right )^{6}:\)

\(\\\Rightarrow \left [\frac{x^{6}}{64} \right ]-\left [\frac{3x^{4}}{16} \right ]+\left [\frac{15x^{2}}{16} \right ]-\left [ \frac{5}{2} \right ]+\left [ \frac{15}{4x^{2}}\right ]-\left [ \frac{3}{x^{4}}\right ]+\left [ \frac{1}{x^{6}}\right ]\)

 

Q.6: By using Binomial Theorem, Evaluate (98)⁴

 

Answer. Based on formula given in Binomial Theorem

(98)⁴ = (100 – 2)⁴

Now we get, by using Binomial Theorem:

Since, [x – y]ⁿ = [ ⁿC₀ × (xⁿ) ] – [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] – [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . +(-1)ⁿ [ ⁿC_n × yⁿ ]

Therefore we get, (100 – 2)⁴ = [ ⁴C₀ × (100)⁴ ] – [ ⁴C₁ × (100)³ × (2) ] + [ ⁴C₂ × (100)² × (2)² ] – [ ⁴C₃ × 100 × (2)³ ] + [⁴C₄ × (2)⁴]

\(\\\Rightarrow\) (1 × 100000000) – (4 × 1000000 × 2) + (6 × 10000 × 4) – (4 × 100 × 8 ) + (1 × 16)

\(\\\Rightarrow\) 100000000 – 8000000 + 240000 – 3200 + 16

Therefore we get, (98)⁴ = (100 – 2)⁴ = 92236816 = 9.2236816 × 10⁷

 

Q.7: By using Binomial Theorem, Evaluate (105)⁵

 

Answer. Based on formula given in Binomial Theorem

(105)⁵ = (100 + 5)⁵

Now we get, by using Binomial Theorem:

Since, [x + y]ⁿ = [ ⁿC₀ × (xⁿ) ] + [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] + [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . + [ ⁿC_n × yⁿ ]

Therefore we get, (100 + 5)⁵ = [ ⁵C₀ × (100)⁵ ] + [ ⁵C₁ × (100)⁴ × (5) ] + [ ⁵C₂ × (100)³ × (5)² ] + [ ⁵C₃ × (100)² × (5)³ ] + [ ⁵C₄ × (100) × (5)⁴ ] + [ ⁵C₅ × (5)⁵]

\(\\\Rightarrow\) (1 × 10000000000) + (5 × 100000000 × 5) + (10 × 1000000 × 5² ) + (10 × 10000 × 5³ ) + (5 × 100 × 5⁴ ) + ( 1 × 5⁵ )

\(\\\Rightarrow\) 10000000000 + 2500000000 + 250000000 + 12500000 + 312500 + 3125

Therefore we get, (105)⁵ = (100 + 5)⁵ = 12762815630 = 1.276281563 × 10¹⁰

 

 

Q.8: By using Binomial Theorem, Evaluate (104)⁴

 

Answer. Based on formula given in Binomial Theorem

(104)⁴ = (100 + 4)⁴

Now we get, by using Binomial Theorem:

Since, [x + y]ⁿ = [ ⁿC₀ × (xⁿ) ] + [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] – [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . + [ ⁿC_n × yⁿ ]

Therefore we get, (100 + 4)⁴ = [⁴C₀ × (100)⁴ ] + [ ⁴C₁ × (100)³ × (4) ] + [⁴C₂ × (100)² × (4)²] + [⁴C₃ × 100 × (4)³] + [⁴C₄ × (4)⁴ ]

\(\\\Rightarrow\) (1 × 100000000) + (4 × 1000000 × 4) + (6 × 10000 × 16) + (4 × 100 × 64 ) + (1 × 256)

\(\\\Rightarrow\) 100000000 + 16000000 + 960000 + 25600 + 256

Therefore we get, (104)⁴ = (100 + 4)⁴ = 116985856 = 1.16985856 × 10⁸

 

Q.9: By using Binomial Theorem, Evaluate (95)⁵

 

Answer Based on formula given in Binomial Theorem.

(95)⁵ = (100 – 5)⁵

Now we get, by using Binomial Theorem:

Since, [x – y]ⁿ = [ ⁿC₀ × (xⁿ) ] – [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] – [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . + (-1)ⁿ [ ⁿC_n × yⁿ ]

Therefore we get, (100 – 5)⁵ = [ ⁵C₀ × (100)⁵ ] – [ ⁵C₁ × (100)⁴ × (5) ] + [ ⁵C₂ × (100)³ × (5)² ] – [ ⁵C₃ × (100)² × (5)³ ] + [ ⁵C₄ × (100) × (5)⁴ ] – [ ⁵C₅ × (5)⁵ ]

\(\\\Rightarrow\) (1 × 10000000000) – (5 × 100000000 × 5) + (10 × 1000000 × 5² ) – (10 × 10000 × 5³ ) + (5 × 100 × 5⁴ ) – ( 1 × 5⁵ )

\(\\\Rightarrow\) 10000000000 – 2500000000 + 250000000 – 12500000 + 312500 – 3125

Therefore we get, (95)⁵ = (100 – 5)⁵ = 7737809375 = 7.737809375 × 10⁹

 

Q.10: By using Binomial Theorem, determine which number is greater (1.1)¹⁰⁰⁰⁰ or 1000.

 

Answer. Based on formula given in Binomial Theorem

Now we get, (1.1)¹⁰⁰⁰⁰ = (1 + 0.1)¹⁰⁰⁰⁰

Now we get, on applying Binomial Theorem, the first few terms in the expansion of (1 + 0.1)¹⁰⁰⁰⁰ are:

Since, [x + y]ⁿ = [ ⁿC₀ × (xⁿ) ] + [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] – [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . + [ ⁿC_n × yⁿ ]

Therefore we get, (1 + 0.1)¹⁰⁰⁰⁰ = [ ¹⁰⁰⁰⁰C₀ × (1)¹⁰⁰⁰⁰ ] + [ ¹⁰⁰⁰⁰C₁ × (1)⁹⁹⁹⁹ × 0.1 ] + other positive terms

\(\\\Rightarrow\) [1 × 1] + [10000 × 1 × 0.1] + other positive terms

\(\\\Rightarrow\) 1 + 1000 + other positive terms > 1000

Therefore we get, (1.1)¹⁰⁰⁰⁰ is greater than 1000.

 

Q.11: Find (a + b)⁵ – (a – b)⁵ . Hence evaluate \(\left(\sqrt{5}+\sqrt{3} \right )^{5}-\left ( \sqrt{5}-\sqrt{3}\right)^{5}\)

 

Answer. Based on formula given in Binomial Theorem

Now we get, by using Binomial Theorem:

Since, [x + y]ⁿ = [ ⁿC₀ × (xⁿ) ] + [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] + [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . + [ ⁿC_n × yⁿ ]

And, [x – y]ⁿ = [ ⁿC₀ × (xⁿ) ] – [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] – [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . + (-1)ⁿ [ ⁿC_n × yⁿ ]

Therefore we get, (a + b)⁵= [ ⁵C₀ × (a⁵) ] + [ ⁵C₁ × (a⁴) × b ] + [ ⁵C₂ × (a³) × b² ] + [ ⁵C₃ × (a²) × b³ ] + [ ⁵C₄ × (a¹) × b⁴ ] + [ ⁵C₅ × b⁵ ]

And, [a – b]⁵ = [ ⁵C₀ × (a⁵) ] – [ ⁵C₁ × (a⁴) × b ] + [ ⁵C₂ × (a³) × b² ] – [ ⁵C₃ × (a²) × b³ ] + [ ⁵C₄ × (a¹) × b⁴ ] – [ ⁵C₅ × b⁵ ]

Therefore we get, (a + b)⁵ – (a – b)⁵ = [ ⁵C₀ a⁵ + ⁵C₁ a⁴ b + ⁵C₂ a³b² + ⁵C₃ a²b³ + ⁵C₄ ab⁴ + ⁵C₅ b⁵ ] – [ ⁵C₀ a⁵ – ⁵C₁ a⁴b + ⁵C₂ a³b² – ⁵C₃ a²b³ + ⁵C₄ ab⁴ – ⁵C₅ b⁵ ]

\(\\\Rightarrow\) [ ⁵C₀ a⁵ + ⁵C₁ a⁴ b + ⁵C₂ a³b² + ⁵C₃ a²b³ + ⁵C₄ ab⁴ + ⁵C₅ b⁵ – ⁵C₀ a⁵ + ⁵C₁ a⁴b – ⁵C₂ a³b² + ⁵C₃ a²b³ – ⁵C₄ ab⁴ + ⁵C₅ b⁵ ]

\(\\\Rightarrow\) 2[ ⁵C₁ a⁴b + ⁵C₃ a²b³ + ⁵C₅ b⁵ ] = 2[ 5a⁴b + 10a²b³ + b⁵ ]

Therefore we get, (a + b)⁵ – (a – b)⁵ = 2b[ 5a⁴ + 10a² b² + b⁴ ] . . . . . . (1)

Now we get, on substituting a = \(\sqrt{5}\) and b = \(\sqrt{3}\) in equation (1) we will get:

\(\\\left(\sqrt{5}+\sqrt{3} \right )^{5}-\left ( \sqrt{5}-\sqrt{3}\right)^{5} = 2\sqrt{3}\times \left [ 5( \sqrt{5})^{4} +10( \sqrt{5})^{2}\times ( \sqrt{3} )^{2}+(\sqrt{3})^{4}\right ]\\\)

\(\\\Rightarrow 2\sqrt{3}\;[125+150+9]=568\sqrt{3}\\\)

Therefore we get, \(\left(\sqrt{5}+\sqrt{3} \right )^{5}-\left ( \sqrt{5}-\sqrt{3}\right)^{5}\)=\(568\sqrt{3}\)

 

Q.12: Find (1 + x)⁵ – (1 – x)⁵ . Hence evaluate \(\left ( 1+\sqrt{7} \right )^{5}-\left ( 1-\sqrt{7} \right )^{5}\)

 

Answer. Based on formula given in Binomial Theorem

Now we get, by using Binomial Theorem:

Since, [x + y]ⁿ = [ ⁿC₀ × (xⁿ) ] + [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] + [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . + [ ⁿC_n × yⁿ ]

And, [x – y]ⁿ = [ ⁿC₀ × (xⁿ) ] – [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] – [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . + (-1)ⁿ [ ⁿC_n × yⁿ ]

Therefore we get, (1 + x)⁵= [ ⁵C₀ × (1⁵) ] + [ ⁵C₁ × (1⁴) × (x) ] + [ ⁵C₂ × (1³) × (x)² ] + [ ⁵C₃ × (1²) × (x)³ ] + [ ⁵C₄ × (1¹) × (x)⁴ ] + [ ⁵C₅ × (x)⁵ ]

And, [1 – x]⁵ = [ ⁵C₀ × (1⁵) ] – [ ⁵C₁ × (1⁴) × (x) ] + [ ⁵C₂ × (1³) × (x)² ] – [ ⁵C₃ × (1²) × (x)³ ] + [ ⁵C₄ × (1¹) × (x)⁴ ] – [ ⁵C₅ × (x)⁵ ]

Therefore we get, (1 + x)⁵ – (1 – x)⁵ = [ ⁵C₀ + ⁵C₁ x + ⁵C₂ x² + ⁵C₃ x³ + ⁵C₄ x⁴ + ⁵C₅ x⁵ ] – [ ⁵C₀ – ⁵C₁ x + ⁵C₂ x² – ⁵C₃ x³ + ⁵C₄ x⁴ – ⁵C₅ x⁵ ]

\(\\\Rightarrow\) [ ⁵C₀ + ⁵C₁ x + ⁵C₂ x² + ⁵C₃ x³ + ⁵C₄ x⁴ + ⁵C₅ x⁵ – ⁵C₀ + ⁵C₁ x – ⁵C₂ x² + ⁵C₃ x³ – ⁵C₄ x⁴ + ⁵C₅ x⁵ ]

\(\\\Rightarrow\) 2[ ⁵C₁x + ⁵C₃ x³ + ⁵C₅ x⁵ ] = 2[ 5x + 10x³ + x⁵ ]

Therefore we get, (1 + x)⁵ – (1 – x)⁵ = 2x[ 5 + 10x² + x⁴ ] . . . . . . (1)

Now we get, on substituting x = \(\sqrt{7}\) in equation (1) we will get:

\(\\\Rightarrow \left ( 1+\sqrt{7} \right )^{7}-\left ( 1-\sqrt{7} \right )^{5}=2\sqrt{7}\left [ 5+10\left ( \sqrt{7} \right )^{2} + \left ( \sqrt{7} \right )^{4}\right ]\\\) \(\\\Rightarrow 2\sqrt{7}\;[5+70+49]=248\sqrt{7}\\\)

Therefore we get, \(\left(1+\sqrt{7} \right )^{5}-\left ( 1-\sqrt{7}\right)^{5}\)=\(\;248\sqrt{7}\)

 

Q.13 Show that 7ⁿ⁺¹ – 6n – 7 is divisible by 36, where n is the positive integer.

 

Answer. Based on formula given in Binomial Theorem

Equation 7ⁿ⁺¹ – 6n – 7 will be divisible by 36 if, 7ⁿ⁺¹ – 6n – 7 = 36p [where p is any natural number]

Now we get, by Binomial Theorem:

Since, [1 + x]^m = [ ^mC₀ ] + [ ^mC₁ . x ] + [ ^mC₂ . x² ] + [ ^mC₃ . x³ ] + . . . . . . . . . . . . + [ ^mC_m . x^m ]

Therefore we get, for x = 6 and m = n + 1 the equation becomes:

[1 + 6]ⁿ⁺¹ = [ ⁿ⁺¹C₀ ] + [ ⁿ⁺¹C₁ × 6 ] + [ ⁿ⁺¹C₂ × (6)² ] + [ ⁿ⁺¹C₃ × (6)³ ] + . . . . . . . . . . . . . + [ ⁿ⁺¹C_n+1 × (6)ⁿ⁺¹ ]

\(\\\Rightarrow\) [7]^{n + 1} = 1 + [(n + 1) × 6] + 6² [ ⁿ⁺¹C₂ + ⁿ⁺¹C₃ × (6) + . . . . . . . . . + ⁿ⁺¹C_{n + 1} × (6)^n–1 ]

\(\\\Rightarrow\) [7]^{n + 1} = 1 + 6n + 6 + 36 [ ⁿ⁺¹C₂ + ⁿ⁺¹C₃ × (6) + . . . . . . . . . + ^{n + 1}C_{n + 1} × (6)^{n – 1} ]

\(\\\Rightarrow\) (7)^{n + 1} – 6 n – 7 = 36p

Where p is any natural number and p = [^{n + 1}C₂ + ^{n + 1}C₃ × (6) + . . . . . . . . . . . . . . . . . . + ^{n + 1}C_{n + 1} × (6)^{n – 1}]

Therefore we get, 7ⁿ⁺¹ – 6n – 7 is divisible by 36 [where n is a positive integer]

 

 

Q.14 Prove that \(\sum_{r\;=\;0}^{n}\) 2^r × ⁿC_r = 3ⁿ

 

Answer. Based on formula given in Binomial Theorem

By using Binomial Theorem:

Since, (a + b)ⁿ = \(\sum_{r\;=\;0}^{n}\) ⁿC_r (a)^{n – r} × (b)^r

Therefore we get, on substituting the value of a = 1 and b = 2 in the above equation we will get:

(1 + 2)ⁿ = \(\sum_{r\;=\;0}^{n}\) ⁿC_r (1)^{n – r} × (2)^r

\(\\\Rightarrow\) 3ⁿ = \(\sum_{r\;=\;0}^{n}\) ⁿC_r × (2)^r

Therefore we get, \(\\\sum_{r\;=\;0}^{n}\) 2^r × ⁿC_r = 3ⁿ

 

Q.15 Show that 5ⁿ – 4n always leaves remainder 5 when divided by 16, where n is the positive integer.

 

Answer. Based on formula given in Binomial Theorem

Equation 5ⁿ – 4n will leave remainder 5 when divided by 16 if, 5ⁿ – 4n = 16p + 5 [where p is any natural number]

Now we get, by Binomial Theorem:

Since, [1 + x]ⁿ = [ ⁿC₀ ] + [ ⁿC₁ . x ] + [ ⁿC₂ . x² ] + [ ⁿC₃ . x³ ] + . . . . . . . . . . . . . . . . . . + [ ⁿC_n . xⁿ ]

Therefore we get, for x = 4 the equation becomes:

[1 + 4]ⁿ = [ ⁿC₀ ] + [ ⁿC₁ × 4 ] + [ ⁿC₂ × (4)² ] + [ ⁿC₃ × (4)³ ] + . . . . . . . . . . . . . . . . . . + [ ⁿC_n × (4)ⁿ ]

\(\\\Rightarrow\) [5]ⁿ = 1 + [(n + 1) × 4] + 4² [ ⁿC₂ + ⁿC₃ × (4) + . . . . . . . . . . . . . . . . . . + ⁿC_n × (4)ⁿ ]

\(\\\Rightarrow\) [5]ⁿ = 1 + 4n + 4 + 16 [ ⁿC₂ + ⁿC₃ × (4) + . . . . . . . . . . . . . . . . . . + ⁿC_n × (4)^n–1 ]

\(\\\Rightarrow\) 5ⁿ – 4n – 5 = 16p

Where p is any natural number and p = [ ⁿC₂ + ⁿC₃ × (4) + . . . . . . . . . . . . . . . . . . + ⁿC_n × (4)^n–1]

i.e. 5ⁿ – 4n = 16 + 5

Therefore we get, 5ⁿ – 4n will leave remainder 5 when divided by 16, where n is any natural number.

 

 

Q.16: Find (a + b)⁶ – (a – b)⁶ . Hence evaluate: \(\left(\sqrt{2}+\sqrt{3} \right )^{6}-\left ( \sqrt{2}-\sqrt{3}\right)^{6}\)

 

Answer. Based on formula given in Binomial Theorem

Now we get, by using Binomial Theorem:

Since, [x + y]ⁿ = [ ⁿC₀ × (xⁿ) ] + [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] + [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . + [ ⁿC_n × yⁿ ]

And, [x – y]ⁿ = [ ⁿC₀ × (xⁿ) ] – [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] – [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . + (-1)ⁿ [ ⁿC_n × yⁿ ]

Therefore we get, [a + b]⁶= [ ⁶C₀ × (a⁶) ] + [ ⁶C₁ × (a⁵) × b ] + [ ⁶C₂ × (a⁴) × b² ] + [ ⁶C₃ × (a³) × b³ ] + [ ⁶C₄ × (a²) × b⁴ ] + [ ⁶C₅ × a × b⁵ ] + [⁶C₆ × b⁶]

And, [a – b]⁶ = [ ⁶C₀ × (a⁶) ] – [ ⁶C₁ × (a⁵) × b ] + [ ⁶C₂ × (a⁴) × b² ] – [ ⁶C₃ × (a³) × b³ ] + [ ⁶C₄ × (a²) × b⁴ ] – [ ⁶C₅ × a × b⁵ ] + [⁶C₆ × b⁶]

Therefore we get, (a + b)⁶ – (a – b)⁶ = [ ⁶C₀ a⁶ + ⁶C₁ a⁵b + ⁶C₂ a⁴b² + ⁶C₃ a³b³ + ⁶C₄ a²b⁴ + ⁶C₅ ab⁵ + ⁶C₆ b⁶] – [ ⁶C₀ a⁶ – ⁶C₁ a⁵b + ⁶C₂ a⁴b² – ⁶C₃ a³b³ + ⁶C₄ a²b⁴ – ⁶C₅ ab⁵ + ⁶C₆ b⁶ ]

\(\\\Rightarrow\) [ ⁶C₀ a⁶ + ⁶C₁ a⁵b + ⁶C₂ a⁴b² + ⁶C₃ a³b³ + ⁶C₄ a²b⁴ + ⁶C₅ ab⁵ + ⁶C₆ b⁶ – ⁶C₀ a⁶ + ⁶C₁ a⁵b – ⁶C₂ a⁴b² + ⁶C₃ a³b³ – ⁶C₄ a²b⁴ + ⁶C₅ ab⁵ – ⁶C₆ b⁶ ]

\(\\\Rightarrow\) 2[ ⁶C₁ a⁵b + ⁶C₃ a³b³ + ⁶C₅ ab⁵ ] = 2[ 6a⁵b + 20a³b³ + 6ab⁵ ]

Therefore we get, (a + b)⁶ – (a – b)⁶ = 4ab[ 3a⁴ + 10a² b² + 3b⁴ ] . . . (1)

Now we get, on substituting a = \(\sqrt{2}\) and b = \(\sqrt{3}\) in equation (1) we will get:

\(\\(\sqrt{2}+\sqrt{3} )^{6}-(\sqrt{2}-\sqrt{3})^{6}=4 (\sqrt{2}. \sqrt{3})\left [ 3( \sqrt{2})^{4} +10 ( \sqrt{2})^{2}\times( \sqrt{3} )^{2}+3(\sqrt{3})^{4} \right ]\\\) \(\\\Rightarrow 4\sqrt{6}\;[12+60+27]=396\sqrt{6}\\\)

Therefore we get, \(\left(\sqrt{2}+\sqrt{3} \right )^{6}-\left ( \sqrt{2}-\sqrt{3}\right)^{6}\)=\(\;396\sqrt{6}\)

 

Q.17: By using Binomial Theorem, Evaluate (91)⁴

 

Answer. Based on formula given in Binomial Theorem

(91)⁴ = (100 – 9)⁴

Now we get, by using Binomial Theorem:

Since, [x – y]ⁿ = [ ⁿC₀ × (xⁿ) ] – [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] – [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . +(-1)ⁿ [ ⁿC_n × yⁿ ]

Therefore we get, (100 – 9)⁴ = [ ⁴C₀ × (100)⁴ ] – [ ⁴C₁ × (100)³ × (9) ] + [ ⁴C₂ × (100)² × (9)² ] – [ ⁴C₃ × 100 × (9)³ ] + [⁴C₄ × (9)⁴]

\(\\\Rightarrow\) (1 × 100000000) – (4 × 1000000 × 9) + (6 × 10000 × 81) – (4 × 100 × 729) + (1 × 6561)

\(\\\Rightarrow\) 100000000 – 36000000 + 4860000 – 291600 + 6561

Therefore we get, (91)⁴ = (100 – 9)⁴ = 68574961 = 6.8574961 × 10⁷

 

Q.18: By using Binomial Theorem, Evaluate (107)⁵

 

Answer. Based on formula given in Binomial Theorem

(107)⁵ = (100 + 7)⁵

Now we get, by using Binomial Theorem:

Since, [x + y]ⁿ = [ ⁿC₀ × (xⁿ) ] + [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] + [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . + [ ⁿC_n × yⁿ ]

Therefore we get, (100 + 7)⁵ = [ ⁵C₀ × (100)⁵ ] + [ ⁵C₁ × (100)⁴ × (7) ] + [ ⁵C₂ × (100)³ × (7)² ] + [ ⁵C₃ × (100)² × (7)³ ] + [ ⁵C₄ × (100) × (7)⁴ ] + [ ⁵C₅ × (7)⁵]

\(\\\Rightarrow\) (1 × 10000000000) + (5 × 100000000 × 7) + (10 × 1000000 × 7² ) + (10 × 10000 × 7³ ) + (5 × 100 × 7⁴ ) + ( 1 × 7⁵ )

\(\\\Rightarrow\) 10000000000 + 3500000000 + 490000000 + 34300000 + 1200500 + 16807

Therefore we get, (107)⁵ = (100 + 7)⁵ = 1402551731 = 1.402551731 × 10¹⁰

 

Q.19: Expand the Expression \(\left ( \frac{3}{2x}-\frac{x}{5} \right )^{6}\)

 

Answer. Based on formula given in Binomial Theorem

By using Binomial Theorem:

Since, [x – y]ⁿ = [ ⁿC₀ × (xⁿ) ] – [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] – [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . +(-1)ⁿ [ ⁿC_n × yⁿ ]

Therefore we get, \(\\\left ( \frac{3}{2x}-\frac{x}{5} \right )^{6}:\\\)

[⁶C₀ × \(\left ( \frac{3}{2x} \right )^{6}\) ] – [⁶C₁ × \(\left ( \frac{3}{2x} \right )^{5}\) × \(\left ( \frac{x}{5} \right )^{1}\) ] + [⁶C₂ × \(\left ( \frac{3}{2x} \right )^{4}\) × \(\left ( \frac{x}{5} \right )^{2}\) ] – [⁶C₃ × \(\left ( \frac{3}{2x} \right )^{3}\) × \(\left ( \frac{x}{5} \right )^{3}\) ] + [⁶C₄ × \(\left ( \frac{3}{2x} \right )^{2}\) × \(\left ( \frac{x}{5} \right )^{4}\) ] – [⁶C₅ × \(\left ( \frac{3}{2x} \right )^{1}\) × \(\left ( \frac{x}{5} \right )^{5}\) ] + [⁶C₆ ×\(\left ( \frac{x}{5} \right )^{6}]\\\)

\(\\\Rightarrow \left [ 1\times \frac{729}{64\times x^{6}}\;\right ]-\left [ 6\times \frac{243}{32\times x^{5}}\times \frac{x}{5}\;\right ]+\left [ 15\times \frac{81}{16\times x^{4}}\times \frac{x^{2}}{25}\;\right ]-\left [ 20\times \frac{27}{8x^{3}}\times \frac{x^{3}}{125} \right ]+\left [ 15\times \frac{9}{4x^{2}}\times \frac{x^{4}}{625}\; \right ]-\left [ 6\times \frac{3}{2x}\times \frac{x^{5}}{3125} \right ]+\left [ 1\times \frac{x^{6}}{15625} \right ]\\\) \(\\\Rightarrow \left [\frac{729}{64\; x^{6}}\;\right ]-\left [\frac{729}{80\;x^{4}}\;\right ]+\left [\frac{243}{80\;x^{2}}\;\right ]-\left [\frac{27}{50}\right ]+\left [\frac{27x^{2}}{500}\;\right ]-\left [\frac{9x^{4}}{3125} \right ]+\left [\frac{x^{6}}{15625} \right ]\\\)

Therefore we get, \(\left ( \frac{3}{2x}-\frac{x}{5} \right )^{6}:\\\)

\(\Rightarrow \left [\frac{x^{6}}{15625} \right ]-\left [\frac{9x^{4}}{3125} \right ]+\left [\frac{27x^{2}}{500}\;\right ]-\left [\frac{27}{50}\right ]+\left [\frac{243}{80\;x^{2}}\;\right ]-\left [\frac{729}{80\;x^{4}}\;\right ]+\left [\frac{729}{64\; x^{6}}\;\right ]\)

 

 

Q.20: Expand the Expression (5x – 3)⁶

 

Answer. Based on formula given in Binomial Theorem

By using Binomial Theorem:

Since, [x – y]ⁿ = [ ⁿC₀ × (xⁿ) ] – [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] – [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . +(-1)ⁿ [ ⁿC_n × yⁿ ]

Therefore we get, (5x – 3)⁶ = [ ⁶C₀ × (5x)⁶ ] – [ ⁶C₁ × (5x)⁵ × (3) ] + [ ⁶C₂ × (5x)⁴ × (3)² ] – [ ⁶C₃ × (5x)³ × (3)³ ] + [ ⁶C₄ × (5x)² × (3)⁴ ] – [ ⁶C₅ × (5x)¹ × (3)⁵ ] + [ ⁶C₆ × (3)⁶ ]

\(\\\Rightarrow\) [1 × (15625 x⁶)] – [6 × (3125 x⁵) × 3] + [15 × (625 x⁴) × 9] – [20 × (125 x³) × 27 ] + [15 × (25 x²) × 81] – [6 × (5x) × 243] + [1 × 729]

\(\\\Rightarrow\) 15625 x⁶ – 56250 x⁵ + 84375 x⁴ – 67500 x³ + 30375 x² – 7290 x + 729

Therefore we get, (5x – 3)⁶ = 15625 x⁶ – 56250 x⁵ + 84375 x⁴ – 67500 x³ + 30375 x² – 7290 x + 729

 

 

Exercise 8.2

Formulas:

1. The general term in the expansion of (a + b)ⁿ :

T_{r + 1} = ⁿC_r × (a)^{n – r} × b^r

2. The middle term in the expansion of (a + b)ⁿ :

(a). If n is even:

The middle term = \(\left ( \frac{n}{2}+1 \right )^{th}term\)

(b). If n is odd:

The middle term = \(\left ( \frac{n+1}{2}\right )^{th}term\;\;and\;\;\left ( \frac{n+1}{2} +1 \right )^{th}term\)

 

Q.1: Find the Coefficient of x⁶ in the expansion of (x + 2)⁹

 

Answer. Based on formula given in Binomial Theorem

Since, The general term in the expansion of (a + b)ⁿ is given by: T_{r + 1} = ⁿC_r × (a)^{n – r} × b^r

Therefore we get, on substituting n = 9, a = x and b = 2 in the above expression we will get:

T_{r + 1} = ⁹C_r × (x)^{9 – r} × 2^r . . . . . . . (1)

Now we get, on comparing the coefficient of x in equation (1) with x⁶, we will get:

i.e. (x)^{9 – r} = x⁶

(or) 9 – r = 6

Therefore we get, r = 3

Now we get, on substituting the value of r in equation (1) we will get:

T₄ = ⁹C₃ × (x)^{9 – 3} × 2³

\(\\\\\Rightarrow T_{4}=\frac{9!}{3!\;6!}\times x^{6}\times 8\\\\\) \(\\\\\Rightarrow T_{4}=\frac{9\times 8\times 7\times 6!}{3\times 2\times 1\times 6!}\times 8\;x^{6}= 672\;x^{6}\\\\\)

Therefore we get, the Coefficient of x⁶ in the expansion of (x + 2)⁹ = 672

 

 

Q.2: Find the Coefficient of a⁶ b⁸ in the expansion of (2a – 3b)¹⁴

 

Answer. Based on formula given in Binomial Theorem

Since, The general term in the expansion of (x + y)ⁿ is given by: T_{r + 1} = ⁿC_r × (x)^{n – r} × y^r

Therefore we get, on substituting n = 14, x = 2a and y = (-3b) in the above expression we will get:

T_{r + 1} = ¹⁴C_r × (2a)^{14 – r} × (-3b)^r

i.e. T_{r + 1} = [¹⁴C_r × (2)^{14 – r} (-3)^r ] (a)^{14 – r} × (b)^r . . . . . . . . . . (1)

Now we get, on comparing the coefficients of a and b in equation (1) with a⁶ b⁸, we will get:

i.e. a⁶ b⁸ = a^{14 – r} b^r

Therefore we get, r = 8

Now we get, on substituting the value of ‘r’ in equation (1) we will get:

T₉ = [ ¹⁴C₈ × (2)^{14 – 8} (-3)⁸ ] (a)^{14 – 8} × (b)⁸

i.e. T₉ = [ ¹⁴C₈ × (2)⁶ (-3)⁸ ] a⁶ b⁸

\(\\\\\Rightarrow T_{9}=\frac{14!}{8!\;6!}\times 64\times 729\times a^{6}\;b^{8}\\\\\) \(\\\\\Rightarrow T_{9}=\frac{14\times 13\times 12\times 11\times 10\times 9\times 8! }{6\times 5\times 4\times 3\times 2\times 1\times 8!}\times 64\times 729\times a^{6}\;b^{8}\\\\\) \(\\\\\Rightarrow T_{9} = 3003\times 64\times 729\times a^{6}\;b^{8}=140107968\;\;a^{6}b^{8}\\\\\)

Therefore we get, the Coefficient of a⁶ b⁸ in the expansion of (2a – 3b)¹⁴ = 140107968

 

Q.3: Write the general term in the expansion of (x³ – y²)⁵

 

Answer. Based on formula given in Binomial Theorem

The general term in the expansion of (a + b)ⁿ is given by: T_{r + 1} = ⁿC_r × (a)^{n – r} × b^r

Now we get, on substituting n = 5, a = x³ and b = y² we will get:

T_r+1 = ⁵C_r × (x³)^{5 – r} × (y²)^r

Therefore we get, the general term in the expansion of (x³ – y²)⁵:

T_r+1 = ⁵C_r × (x)^{15 – 3 r} × (y)^{2 r}

 

Q.4: Write the general term in the expansion of (x⁴ – xy²)⁹

 

Answer.Based on formula given in Binomial Theorem

The general term in the expansion of (a + b)ⁿ is given by: T_{r + 1} = ⁿC_r × (a)^{n – r} × b^r

Now we get, on substituting n = 9, a = x⁴ and b = xy² we will get:

T_r+1 = ⁹C_r × (x⁴)^{9 – r} × (xy²)^r

\(\Rightarrow\) T_r+1 = ⁹C_r × (x)^{36 – 4 r} × (x)^{2 r} × (y)^{2 r}

\(\Rightarrow\) T_r+1 = ⁹C_r × (x)^{36 – 4 r + 2 r} × (y)^{2 r}

\(\Rightarrow\) T_r+1 = ⁹C_r × (x)^{36 – 2 r} × (y)^{2 r}

Therefore we get, the general term in the expansion of (x⁴ – xy²)⁹:

T_r+1 = ⁹C_r × (x)^{36 – 2 r} × (y)^2r

 

 

Q.5: Find the 4^th term in the expansion of (2x + 3y)¹⁰

 

Answer.Based on formula given in Binomial Theorem

Since, The general term in the expansion of (a + b)ⁿ is given by: T_{r + 1} = ⁿC_r × (a)^{n – r} × b^r

Therefore we get, from the above equation the value of r should be 3; for finding out the values of 4^th Term (T₄)

Therefore we get, T_{3 + 1} = ⁿC₃ × (a)^{n – 3} × b³

Now we get, on substituting n =10, a = 2x and b = 3y we will get:

\(\\\Rightarrow\) T_{3 + 1} = ¹⁰C₃ × (2x)^{10 – 3} × (3y)³

\(\\\Rightarrow\) T₄ = ¹⁰C₃ × (2x)⁷ × (3y)³

\(\\\\\Rightarrow T_{4}=\frac{10!}{3!\; 7!}\times 128\;x^{7}\times 27\;y^{3}\\\\\) \(\\\\\Rightarrow T_{4}=\frac{10\times 9\times 8\times 7!}{3\times 2\times 1\times 7!}\times 3456\;x^{7}y^{3}\\\\\) \(\\\\\Rightarrow T_{4} = 120\times 3456\;x^{7}y^{3}=414720\;x^{7}y^{3}\\\\\)

Therefore we get, 4^th term in the expansion of (2x + 3y)¹⁰ = 414720 x⁷y³

 

Q.6 Find the 11^th term in the expansion of \(\left ( 8x-\frac{1}{2\sqrt{x}} \right )^{15}\)

Answer. Based on formula given in Binomial Theorem

Since, The general term in the expansion of (a + b)ⁿ is given by: T_{r + 1} = ⁿC_r × (a)^{n – r} × b^r

Therefore we get, from the above equation the value of r should be 10 for finding out the values of 11^th Term (T₁₁)

Therefore we get, T_{10 + 1} = ⁿC₁₀ × (a)^{n – 10} × b¹⁰

Now we get, on substituting n =15, a = 8x and b = \(\left (-\frac{1}{2\sqrt{x}} \right )\) we will get:

\(\\\Rightarrow\) T_{10 + 1} = ¹⁵C₁₀ × (8x)^{15 – 10} × \(\left (-\frac{1}{2\sqrt{x}} \right )^{10}\\\)

\(\\\Rightarrow\) T₁₁ = ¹⁵C₁₀ × (8x)⁵ × \(\left (-\frac{1}{2\sqrt{x}} \right )^{10}\\\)

\(\\\\\Rightarrow T_{11}=\frac{15!}{10!\; 5!}\times 8\times 8\times 8\times 8\times 8\times x^{5}\times \frac{1}{2^{10}}\times \left ( \frac{1}{\sqrt{x}} \right )^{10}\\\\\) \(\\\\\\\Rightarrow T_{11}=\frac{15\times 14\times 13\times 12\times 11\times 10!}{5\times 4\times 3\times 2\times 1\times 10!}\times 32\times x^{5}\times \frac{1}{x^{5}}\\\\\) \(\\\\\Rightarrow T_{11} = 3003\times 32\boldsymbol{=96096}\\\\\)

Therefore we get, 11^th term in the expansion of \(\left ( 8x-\frac{1}{2\sqrt{x}} \right )^{15}\)= 96096

 

Q.7: In the expansion of \(\left ( 5-\frac{x^{4}}{10} \right )^{9}\), Find the middle terms.

 

Answer. Based on formula given in Binomial Theorem

Here, n = 9

When n is odd then the middle terms in the expansion of (a + b)ⁿ are given by:

middle terms = \(\left ( \frac{n+1}{2}\right )^{th}term\;\;and\;\;\left ( \frac{n+1}{2} +1 \right )^{th}term\)

Therefore we get, the middle terms in the expansion of \(\left ( 5-\frac{x^{4}}{10} \right )^{9}\)are:

\(\Rightarrow \left ( \frac{9+1}{2}\right )^{th}term\;\;and\;\;\left ( \frac{9+1}{2} +1 \right )^{th}term\)

\(\Rightarrow\) 5^th term and 6^th term

Now we get, 5^th and 6^th terms in the expansion of \(\left ( 5-\frac{x^{4}}{10} \right )^{9}\) are:

T₅ = T_{4 + 1} = ⁹C₄ × (5)^{9 – 4} × \(\left ( -\frac{x^{4}}{10} \right )^{4}\)

\(\\\\\Rightarrow\) T₅ = ⁹C₄ × (5)⁵ × \(\left ( -\frac{x^{4}}{10} \right )^{4}\\\\\)

\(\\\\\Rightarrow T_{5}=\frac{9!}{4!\; 5!}\times (5)^{5}\times \left ( -\frac{x^{4}}{10} \right )^{4}\\\\\) \(\\\\\Rightarrow T_{5}=\frac{9\times 8\times 7\times 6\times 5!}{4\times 3\times 2\times 1\times 5!}\times 5\times 5\times 5\times 5\times 5\times \frac{1}{10^{4}}\times x^{16}\\\\\) \(\\\\\Rightarrow T_{5} = 126\times \frac{5}{16}\;x^{16} \boldsymbol{=\frac{315}{8}\;x^{16}}\\\\\)

Therefore we get, 5^th term \(=\frac{315}{8}\;x^{16}\)

Now we get, T₆ = T_{5 + 1} = ⁹C₅ × (5)^{9 – 5} × \(\left ( -\frac{x^{4}}{10} \right )^{5}\)

\(\\\\\Rightarrow\) T₆ = ⁹C₅ × (5)⁴ × \(\left ( -\frac{x^{4}}{10} \right )^{4}\\\\\)

\(\\\\\Rightarrow T_{6}=\frac{9!}{5!\; 4!}\times (5)^{4}\times \left ( -\frac{x^{4}}{10} \right )^{5}\\\\\) \(\\\\\Rightarrow T_{6}=\frac{9\times 8\times 7\times 6\times 5!}{5!\times (4\times 3\times 2\times 1)}\times 5\times 5\times 5\times 5\times \frac{-1}{10^{5}}\times x^{20}\\\\\) \(\\\\\Rightarrow T_{6} = -126\times \frac{1}{320}\;x^{20}\boldsymbol{=\frac{-63}{160}\;x^{20}}\\\\\)

Therefore we get, 6^th term \(=\frac{63}{160}\;x^{20}\)

Therefore we get, 5^th and 6^th terms are the middle terms in the expansion of \(\left ( 5-\frac{x^{4}}{10} \right )^{9}\)

And also, 5^th term \(=\frac{315}{8}\;x^{16}\) and 6^th term \(=\frac{-63}{160}\;x^{20}\)

 

Q.8: In the expansion of \(\left ( \frac{x}{2}+8y \right )^{10}\), Find the middle term.

 

Answer. Based on formula given in Binomial Theorem

Here, n = 10

When n is even, the middle term in the expansion of (a + b)ⁿ is given by:

\(\left ( \frac{n}{2}+1 \right )^{th}term\)

Therefore we get, the middle term in the expansion of \(\left ( \frac{x}{2}+8y \right )^{10}\) is:

\(\Rightarrow \left ( \frac{10}{2}+1 \right )^{th}term\) = 6^th term

Now we get, 6^th term in the expansion of \(\left ( \frac{x}{2}+8y \right )^{10}\) is:

T₆ = T_{5 + 1} = ¹⁰C₅ × \(\left ( \frac{x}{2} \right )^{10-5}\) × (8y)⁵

\(\\\Rightarrow\) T₆ = ¹⁰C₅ × \(\left ( \frac{x}{2} \right )^{5}\) × (8y)⁵

\(\\\Rightarrow T_{6}=\frac{10!}{5!\; 5!}\times \left ( \frac{x}{2} \right )^{5}\times (8y)^{5}\\\) \(\\\Rightarrow T_{6}=\frac{10\times 9\times 8\times 7\times 6\times 5!}{5\times 4\times 3\times 2\times 1\times 5!}\times \frac{x^{5}}{32}\times 8\times 8\times 8\times 8\times 8\times y^{5}\\\) \(\\\Rightarrow T_{6} = 252\times 1024\;x^{5}y^{5}\boldsymbol{=258048\;x^{5}y^{5}}\\\)

Therefore we get, 6^th term = 258048 x⁵ y⁵

Hence, the middle term in the expansion of \(\left ( \frac{x}{2}+8y \right )^{10}\) = 258048 x⁵ y⁵

 

Q.9: Prove that the coefficients of p^a and p^b are equal in the expansion of (1 + p)^{a + b}

 

Answer. Based on formula given in Binomial Theorem

Since, The general term in the expansion of (x + y)ⁿ is given by: T_{r + 1} = ⁿC_r × (x)^{n – r} × (y)^r

Now we get, on substituting n = (a + b), x =1 and y = p, we will get:

T_{r + 1} = ^{(a + b)}C_r × (1)^{(a + b) – r} × (p)^r

\(\Rightarrow\) T_{r + 1} = ^{(a + b)}C_r × (p)^r . . . . . . . . . . . . (1)

Now we get, on comparing the coefficient of p in equation (1) with p^a, we will get r = a

Therefore we get, from equation (1):

T _{a + 1} = ^{(a + b)}C_a × (p)^a

\(\\\Rightarrow T_{a+1}=\frac{(a+b)!}{a!\;(a+b-a)!}\times p^{a}\\\) \(\\\Rightarrow T_{a+1}=\frac{(a+b)!}{a!\;b!}\times p^{a}\\\)

Therefore we get, the coefficient of (p)^a = \(\frac{(a+b)!}{a!\;b!}\) . . . . . . . . . . (2)

Now we get, on comparing the coefficient of p in equation (1) with p^b, we will get r = b

Therefore we get, from equation (1):

T _{b + 1} = ^{(a + b)}C_b × (p)^b

\(\\\\\Rightarrow T_{b+1}=\frac{(a+b)!}{b!\;(a+b-b)!}\times p^{b}\\\\\) \(\\\\\Rightarrow T_{b+1}=\frac{(a+b)!}{b!\;a!}\times p^{b}\\\\\)

Therefore we get, the coefficient of (p)^b = \(\frac{(a+b)!}{b!\;a!}\) . . . . . . . . . . (3)

Now we get, on comparing equation (2) and equation (3) we conclude that the coefficients of p^a and p^b are equal in the expansion of (1 + p)^{a + b}.

Hence, Proved

 

Q.10: The coefficients of the (k + 1)^th, k^th and (k – 1)^th terms in the expansion of (x + 1)ⁿ are in the ratio 5 : 3 : 1. Find the values of n and k.

 

Answer. Based on formula given in Binomial Theorem

Since, The general term in the expansion of (a + b)ⁿ is given by: T_{r + 1} = ⁿC_r × (a)^{n – r} × b^r

Now we get on substituting a = x and b = 1 in the above equation:

T_{r + 1} = ⁿC_r × (x)^{n – r} × (1)^r

Now we get, (k + 1)^th term in the expansion of (x + 1)ⁿ :

T_{(k) + 1} = ⁿC_k × (x)^{n – k} × 1^{(k )}

i.e. T_{(k + 1)} = ⁿC_k × (x)^{n – k}

\(\\\Rightarrow T_{k+1} = \frac{n!}{k!\;(n-k)!}\times (x)^{n-k}\\\)

Therefore we get the coefficient of T_{(k + 1)}^th term = \(\frac{n!}{k!\;(n-k)!}\\\\\)

Now we get, (k)^th term in the expansion of (x + 1)ⁿ :

T_{(k – 1) + 1} = ⁿC_{k – 1} × (x)^{n – (k – 1)} × 1^{(k – 1)}

i.e. T_k = ⁿC_{k – 1} × (x)^{n – k + 1}

\(\\\\\Rightarrow T_{k} = \frac{n!}{(k-1)!\;(n-k+1)!}\times (x)^{n-k+1}\\\\\)

Therefore we get the coefficient of T_(k)^th term = \(\frac{n!}{(k-1)!\;(n-k+1)!}\)

And, (k – 1)^th term in the expansion of (x + 1)ⁿ :

T_{(k – 2 )+ 1} = ⁿC_{k – 2} × (x)^{n – (k – 2)} × 1^{(k – 2)}

i.e. T_{k – 1} = ⁿC_{k – 2} × (x)^{n – k + 2}

\(\\\\\Rightarrow T_{k-1} = \frac{n!}{(k-2)!\;(n-k+2)!}\times (x)^{n-k+2}\\\\\)

Therefore we get the coefficient of T_{(k – 1)}^th term = \(\frac{n!}{(k-2)!\;(n-k+2)!}\)

Now we get, according to the given conditions: Coefficients of T_{(k + 1)}, T_k and T_{k – 1} are in the ratio of 5 : 3 : 1

Therefore we get, \(\frac{T_{k-1}}{T_{k}}=\frac{1}{3}\\\)

\(\\\\\Rightarrow \left [ \frac{n!}{(k-2)!\;(n-k+2)!} \right ]\div \left [ \frac{n!}{(k-1)!\;(n-k+1)!} \right ]=\frac{1}{3}\\\\\) \(\\\\\Rightarrow \frac{(k-1)!\;(n-k+1)!}{(k-2)!\;(n-k+2)!}=\frac{1}{3}\\\\\) \(\\\\\Rightarrow \frac{(k-1)(k-2)!\;(n-k+1)!}{(k-2)!\;(n-k+2)(n-k+1)!}=\frac{1}{3}\\\\\) \(\\\Rightarrow \frac{k-1}{n-k+2}=\frac{1}{3}\\\) \(\\\Rightarrow 3k-3=n-k+2\\\)

(or) n – 4k + 5 = 0 . . . . . . . . . . . . . . . . (1)

And, \(\frac{T_{k+1}}{T_{k}}=\frac{5}{3}\\\)

\(\\\Rightarrow \left [ \frac{n!}{k!\;(n-k)!} \right ]\div \left [ \frac{n!}{(k-1)!\;(n-k+1)!} \right ]=\frac{5}{3}\\\) \(\\\Rightarrow \frac{(k-1)!\;(n-k+1)!}{k!\;(n-k)!}=\frac{5}{3}\\\) \(\\\Rightarrow \frac{(k-1)!\;[(n-k+1)\;(n-k)!]}{[k\;(k-1)!]\;(n-k)!}=\frac{5}{3}\\\) \(\\\Rightarrow \frac{n-k+1}{k}=\frac{5}{3}\\\) \(\\\Rightarrow 3n-3k+3=5k\\\)

(or) 3n – 8k + 3 = 0 . . . . . . . . . . . . . . . . . . (2)

On multiplying equation (1) by 3 and subtracting it to equation (2) we will get:

(3n – 8k + 3) – 3(n – 4k + 5) = 0

\(\\\Rightarrow\) 3n – 8k + 3 – 3n + 12k – 15 = 0

\(\\\Rightarrow\) 4k = 12

Therefore we get, k = 3

Now we get on substituting the value of k in equation (2) we will get:

3n – 8(3) + 3 = 0

Therefore we get, n = 7

 

Q.11: Prove that the coefficient of a^p in the expansion of (1 + 3a)^{2p – 1} is half the coefficient of a^p in the expansion of (1 + 3a)^2p

 

Answer. Based on formula given in Binomial Theorem

Since, The general term in the expansion of (x + y)ⁿ is given by: T_{r + 1} = ⁿC_r × (x)^{n – r} × (y)^r

For (1 + 3a)^2p:

On putting n = 2p, x =1 and y = 3a, we will get:

T_{r + 1} = ^(2p)C_r × (1)^{(3p) – r} × (3a)^r

\(\\\\\Rightarrow\) T_{r + 1} = ^(2p)C_r × (3a)^r . . . . . . . . . . . . (1)

Now we get, on comparing the coefficient of a in equation (1) with a^p, we will get r = p

Therefore we get, from equation (1):

T _{p + 1} = ^(2p)C_p × (3a)^p

\(\\\Rightarrow T_{p+1}=\frac{(2p)!}{p!\;(2p-p)!}\times (3a)^{p}\\\\\)

\(\\\Rightarrow T_{p+1}=\frac{(2p)!}{p!\;p!}\times 3^{p}\times (a)^{p}\\\\\)

Therefore we get, the coefficient of a^p \(=\frac{(2p)!}{p!\;p!}\times 3^{p}\) . . . . . . . . . . . . . . . . (2)

Now we get, for (1 + 3a)^{2p – 1} :

On substituting n = (2p – 1), x =1 and y = 3a, we will get:

T_{r + 1} = ^{(2p – 1)}C_r × (1)^{(2p – 1) – r} × (3a)^r

\(\Rightarrow\) T_{r + 1} = ^{(2p – 1)}C_r × (3a)^r . . . . . . . . . . . . (1)

Now we get, on comparing the coefficient of ‘a’ in equation (1) with ‘a^p ’, we will get r = p

Therefore we get, from equation (1):

T _{p + 1} = ^{(2p – 1)}C_p × (3a)^p

\(\\\\\Rightarrow T_{p+1}=\frac{(2p-1)!}{p!\;(2p-1-p)!}\times (3a)^{p}\\\\\) \(\\\\\Rightarrow T_{p+1}= \frac{(2p-1)!}{p!\;(p-1)!}\times (3)^{p}\times (a)^{p}\\\\\) \(\\\\\Rightarrow T_{p+1}=\frac{\frac{(2p)!}{2p}}{p!\times \frac{p!}{p}}\times (3)^{p}.(a)^{p}=\frac{(2p)!}{2(p!)^{2}}\times 3^{p}.a^{p}\\\)

Therefore we get, the coefficient of a^p = \(=\frac{(2p)!}{2(p!)^{2}}\times 3^{p}\) . . . . . . . . . . (3)

Now we get, on comparing equation (2) and equation (3):

\(\\\\\Rightarrow \frac{(2p)!}{p!\;p!}\times 3^{p}=\frac{(2p)!}{p!\;p!}\times 3^{p}\\\\\)

\(\\\Rightarrow\) ^(2p)C_p × 3^p = \(\frac{1}{2}\) ^{(2p – 1)}C_p × 3^p

Therefore we get, the coefficient of a^p in the expansion of (1 + 3a)^{2p – 1} is half the coefficient of a^p in the expansion of (1 + 3a)^2p

 

Q.12: For what values of ‘m’ the coefficient of x² in the expansion of (1+ 2x)^k is 140.

 

Answer. Based on formula given in Binomial Theorem

Since, The general term in the expansion of (a + b)ⁿ is given by: T_{r + 1} = ⁿC_r × (a)^{n – r} × (b)^r

Now we get, on substituting n = m, a =1 and b = 2x we will get:

T_{r + 1} = ^mC_r × (1)^{m – r} × (2x)^r

\(\Rightarrow\) T_{r + 1} = ^mC_r × (2)^r × (x)^r . . . . . . . . . . . . (1)

Now we get, on comparing the coefficient of x in equation (1) with x², we will get r = 2

Therefore we get, from equation (1):

T _{2 + 1} = ^mC₂ × (2)² × (x)²

Since, the coefficient of x² = 140 [GIVEN]

Therefore we get, 140 = ^mC₂ × (2)²

\(\Rightarrow 84=\left [ \frac{m!}{2!\;\;(m-2)!} \right ]\times 4\\\\\) \(\Rightarrow 21=\frac{m\times (m-1)\;\;(m-2)!}{[2\times 1]\times (m-2)!}\\\\\) \(\Rightarrow 42 = m^{2}-m\)

Therefore we get, m² – m – 42 = 0

Now we get, by splitting of middle term method, the roots of this quadratic equation are:

\(\Rightarrow\)m² – (7 – 6)m – 42 = 0

\(\Rightarrow\)m² – 7m + 6m – 42 = 0

\(\Rightarrow\) m(m – 7) +6(m – 7) = 0

i.e. (m + 6) (m – 7) = 0

Therefore we get, m = 7 or m = -6

Hence, for the coefficient of x² in the expansion of (1+ 2x)^k to be 140, the value of ‘m’ should be 7.

 

Q.13: In the expansion of (2x+3y)⁹, Find the middle terms.

 

Answer. Based on formula given in Binomial Theorem

Here, n = 9

When n is odd then the middle terms in the expansion of (a + b)ⁿ are given by:

middle terms = \(\left ( \frac{n+1}{2}\right )^{th}term\;\;and\;\;\left ( \frac{n+1}{2} +1 \right )^{th}term\)

Therefore we get, the middle terms in the expansion of (2x+3y)⁹ are:

\(\Rightarrow \left ( \frac{9+1}{2}\right )^{th}term\;\;and\;\;\left ( \frac{9+1}{2} +1 \right )^{th}term\)

\(\Rightarrow\) 5^th term and 6^th term

Now we get, 5^th and 6^th terms in the expansion of (2x+3y)⁹ are:

T₅ = T_{4 + 1} = ⁹C₄ × (2x)^{9 – 4} × (3y)⁴

\(\Rightarrow\) T₅ = ⁹C₄ × (2x)⁵ × (3y)⁴

\(\Rightarrow T_{5}=\frac{9!}{4!\; 5!}\times (2x)^{5}\times (3y)^{4}\) \(\Rightarrow T_{5}=\frac{9\times 8\times 7\times 6\times 5!}{4\times 3\times 2\times 1\times 5!}\times 32x^{5}\times 81y^{4}\\\\\)

\(\Rightarrow T_{5} = 126\times 2592\;x^{5}y^{4}\boldsymbol{=414720\;x^{5}y^{4}}\\\)

Therefore we get, 5^th term = 414720 x⁵ y⁴

Now we get, T₆ = T_{5 + 1} = ⁹C₅ × (2x)^{9 – 5} × (3y)⁵

\(\Rightarrow\) T₆ = ⁹C₅ × (2x)⁴ × (3y)⁵

\(\Rightarrow T_{6}=\frac{9!}{5!\times 4!}\times (2x)^{4}\times (3y)^{5}\) \(\Rightarrow T_{6}=\frac{9\times 8\times 7\times 6\times 5!}{5!\times (4\times 3\times 2\times 1)}\times 16x^{4}\times 243y^{5}\\\\\) \(\Rightarrow T_{6} = 126\times 3888\;x^{4}y^{5}\boldsymbol{=489888\;x^{4}y^{5}}\\\)

Therefore we get, 6^th term = 489888 x⁴ y⁵

Hence, 5^th and 6^th terms are the middle terms in the expansion of (2x+3y)⁹

And also, 5^th term = 414720 x⁵ y⁴ and 6^th term = 489888 x⁴ y⁵

 

 

Q.14: Find the Coefficient of x⁶ in expansion of (x + 3)¹¹

 

Answer. Based on formula given in Binomial Theorem

Since, The general term in the expansion of (a + b)ⁿ is given by: T_{r + 1} = ⁿC_r × (a)^{n – r} × b^r

Therefore we get, on substituting n = 11, a = x and b = 3 in the above expression we will get:

T_{r + 1} = ¹¹C_r × (x)^{11 – r} × 3^r . . . . . . . (1)

Now we get, on comparing the coefficient of x in equation (1) with x⁷, we will get:

i.e. (x)^{11 – r} = x⁶

(or) 11 – r = 6

Therefore we get, r = 5

Now we get, on substituting r in equation (1) we will get:

T_{5 + 1} = ¹¹C₅ × (x)^{11 – 5} × 3⁵

\(\Rightarrow T_{6}=\frac{11!}{5!\times 6!}\times x^{6}\times 3^{5}\) \(\Rightarrow T_{6}=\frac{11\times 10\times 9\times 8\times 7\times 6!}{5\times 4\times 3\times 2\times 1\times 6!}\times 243\;x^{6}\\\\\)

\(\Rightarrow T_{6} = 462\times 243\;x^{6}\boldsymbol{=112266\;x^{6}}\)

Therefore we get, the Coefficient of x⁶ in the expansion of (x + 3)¹¹ = 112266

 

Q.15: Write the general term in the expansion of (x²y³ – x³ y²)⁷

 

Answer. Based on formula given in Binomial Theorem

The general term in the expansion of (a + b)ⁿ is given by: T_{r + 1} = ⁿC_r × (a)^{n – r} × b^r

Now we get, on substituting n = 7, a = x²y³ and b = x³ y² we will get:

T_r+1 = ⁷C_r × (x²y³)^{7 – r} × (x³ y²)^r

\( \Rightarrow\) T_{r + 1} = ⁷C_r × (x²)^{7 – r} × (y³)^{7 – r} × (x)^{3 r} × (y)^{2 r}

\( \Rightarrow\) T_{r + 1} = ⁷C_r × (x)^{14 – 2 r} × (x)^{3 r}× (y)^{2 r}× (y)^{21 – 3r}

\( \Rightarrow\) T_{r + 1} = ⁷C_r × (x)^{14 – 2 r + 3r} × (y)^{21 – 3r + 2 r}

i.e. T_{r + 1} = ⁷C_r × (x)^{14 + r} × (y)^{21 – r}

Therefore we get, the general term in the expansion of (x²y³ – x³ y²)⁷:

T_{r + 1} = ⁷C_r × (x)^{14 + r} × (y)^{21 – r}

 

 

Miscellaneous Exercise:

Q.1: If 1^st term, 2^nd term and 3^rd term in the expansion of (a + b)ⁿ are 729, 7290 and 30375 respectively then find the value of ‘a’, ‘b’ and ‘c’.

 

Answer. Based on formula given in Binomial Theorem

The general term in the expansion of (a + b)ⁿ is given by: T_{r + 1} = ⁿC_r × (a)^{n – r} × b^r

Now we get, according to the given conditions:

T_{(0 + 1)} = ⁿC₀ × (a)^{n – 0} × b⁰

T₁ = aⁿ

i.e. aⁿ = 729 . . . . . . . . . . . . . . . . . . (1)

Also, T_{(1 + 1)} = ⁿC₁ × (a)^{n – 1} × b¹

T₂ = ⁿC₁ × (a)^{n – 1} × b

i.e. 7290 = ⁿC₁ × (a)^{n – 1} × b

\(\\\Rightarrow 7290=\frac{n!}{1!\;\;(n-1)!}\times a^{n}\times a^{-1}\times b\\\\\) \(\\\\\Rightarrow 7290=\frac{n(n-1)!}{(n-1)!}\times a^{n}\times \frac{b}{a}\;\;\;\)

Therefore we get, 7290 = n × aⁿ × \(\frac{b}{a}\) . . . . . . . . . . . . . . . . (2)

And, T_{(2 + 1)} = ⁿC₂ × (a)^{n – 2} × b²

T₃ = ⁿC₂ × (a)^{n – 2} × b²

i.e. 30375 = ⁿC₂ × (a)^{n – 2} × b²

\(\\\Rightarrow 30375=\frac{n!}{2!\;\;(n-2)!}\times a^{n}\times a^{-2}\times b^{2}\\\) \(\\\Rightarrow 30375=\frac{n\;(n-1)\;(n-2)!}{(2\times 1)\;\;(n-2)!}\times a^{n}\times \left ( \frac{b}{a} \right )^{2}\\\)

Therefore we get, 60750 = n (n – 1) × aⁿ × \(\left ( \frac{b}{a} \right )^{2}\) . . . . . . . . . . . . . . . . (3)

Now we get, on substituting equation (1) in equation (2) we will get:

\(\\\Rightarrow 7290 = n\times (729)\times \frac{b}{a}\\\) \(\\\Rightarrow \frac{10a}{b}=n\\\)

Therefore we get, n = \(\frac{10a}{b}\) . . . . . . . . . . . . . . . . . (4)

Now we get, on dividing equation (2) and equation (3) we will get:

\(\\\Rightarrow \frac{n \times a^{n} \times \frac{b}{a}}{n\times (n-1)\times a^{n}\times \left ( \frac{b}{a} \right )^{2} }=\frac{7290}{60750}\\\) \(\\\Rightarrow \frac{3}{25}=\frac{1}{(n-1)\times \frac{b}{a}}\\\) \(\\\Rightarrow 3n-3=\frac{25a}{b}\\\)

Now we get, from equation (4):

\(\frac{a}{b}\) = \(\frac{n}{10}\)

\(\Rightarrow 3n-3=25\times \left ( \frac{n}{10} \right )\) \(\Rightarrow 30n-30=25n\)

i.e. 30n – 30 = 25n

Therefore we get, n = 6

Now we get, on substituting the value of ‘n’ in equation (1) we will get:

\(\Rightarrow\) a⁶ = 729

i.e. a⁶ = 3⁶

Therefore we get, a = 3

Now we get, from equation (4):

\(\\\Rightarrow\) 6 = \(\frac{30}{b}\\\)

Therefore we get, b = 5

 

Q.2: If the coefficients of z² and z³ are equal in the expansion of (3 + bz)⁹. Find the value of ‘b’

 

Answer. Based on formula given in Binomial Theorem

Since, The general term in the expansion of (x + y)ⁿ is given by: T_{r + 1} = ⁿC_r × (x)^{n – r} × y^r

Therefore we get, on substituting n = 9, x = 3 and y = bz in the above expression we will get:

T_{r + 1} = ⁹C_r × (3)^{9 – r} × (bz)^r

T_{r + 1} = ⁹C_r × (3)^{9 – r} × b^r × z^r . . . . . (1)

Now we get, on comparing the coefficient of z in equation (1) with z², we will get: r = 2

On substituting the value of r in equation (1) we will get:

T_{(2 + 1)} = ⁹C₂ × (3)^{9 – 2} × b² × z²

\(\\\Rightarrow T_{3}=\frac{9!}{2!\;7!}\times 3^{7}\times b^{2}\times z^{2}\\\) \(\\\Rightarrow T_{3}=\frac{9\times 8\times 7!}{2\times 1\times 7!}\times 3^{7}\times b^{2}\times z^{2}\\\)

\(\\ \Rightarrow T_{3} = 36\times 3^{7}\times b^{2}\times z^{2}\) . . . . . . . . . . . . (2)

Now we get, on comparing the coefficient of z in equation (1) with z³, we will get: r = 3

On substituting the value of ‘r’ in equation (1) we will get:

T_{(3 + 1)} = ⁹C₃ × (3)^{9 – 3} × b³ × z³

\(\\\Rightarrow T_{4}=\frac{9!}{3!\;6!}\times 3^{6}\times b^{3}\times z^{3}\\\) \(\\\Rightarrow T_{4}=\frac{9\times 8\times 7\times 6!}{3\times 2\times 1\times 6!}\times 3^{6}\times b^{3}\times z^{3}\\\)

\(\\\Rightarrow T_{4} = 84\times 3^{6}\times b^{3}\times z^{3} . . . . . . . . . (3)\)

Now we get, according to the given conditions:

Coefficient of z² = Coefficient of z³

Therefore we get from equation (2) and equation (3):

\(\\\Rightarrow 36\times 3^{7}\times b^{2}=84\times 3^{6}\times b^{3}\\\) \(\\\Rightarrow b = \frac{36\times 3^{7}}{84\times 3^{6}}\\\)

Therefore we get, the value of b = \(\frac{9}{7}\)

 

Q.3: Find the coefficient of x⁶ in the product of (1 + 3x)⁷ (1 – 2x)⁶ by using binomial theorem.

 

Answer. Based on formula given in Binomial Theorem

By using Binomial Theorem:

Since, (1 + x)ⁿ = [ ⁿC₀ ] + [ ⁿC₁ × (x) ] + [ ⁿC₂ × (x)² ] + [ ⁿC₃ × (x³) ] + . . . . . . . . . . . . . . . . . . + [ ⁿC_n × (x)ⁿ ]

Therefore we get, (1 + 3x)⁷ = [ ⁷C₀ ] + [ ⁷C₁ × (3x) ] + [ ⁷C₂ × (3x)² ] + [ ⁷C₃ × (3x)³ ] + [ ⁷C₄ × (3x)⁴ ] + [ ⁷C₅ × (3x⁵) ] + [ ⁷C₆ × (3x)⁶ ] + [ ⁷C₇ × (3x)⁷ ]

\(\\\Rightarrow\) 1 + [ 7 × (3x) ] + [ 21 × (9x²) ] + [ 35 × (27x³) ] + [ 35 × (81x⁴) ] + [ 21 × (243x⁵) ] + [ 7 × 729x⁶ ] + [ 1 × 2187x⁷ ]

\(\\\Rightarrow\) 1 + 21x + 189x² + 945x³ + 2835x⁴ + 5103x⁵ + 5103x⁶ + 2187x⁷

Therefore we get, (1 + 3x)⁷ = 1 + 21x + 189x² + 945x³ + 2835x⁴ + 5103x⁵ + 5103x⁶ + 2187x⁷

Now we get, (1 – 2x)⁶:

By using Binomial Theorem:

(1 – 2x)⁶ = [ ⁶C₀ ] – [ ⁶C₁ × (2x) ] + [ ⁶C₂ × (2x)² ] – [ ⁶C₃ × (2x)³ ] + [ ⁶C₄ × (2x)⁴ ] – [ ⁶C₅ × (2x⁵) ] + [ ⁶C₆ × (2x)⁶ ]

\(\\\Rightarrow\) 1 – [ 6 × (2x) ] + [ 15 × (4x²) ] – [ 20 × (8x³) ] + [ 15 × (16x⁴) ] – [ 6 × (32x⁵) ] + [ 1 × 64x⁶ ]

\(\\\Rightarrow\) 1 – 12x + 60x² – 160x³ + 240x⁴ – 192x⁵ + 64x⁶

Therefore we get, (1 – 2x)⁶ = 1 – 12x + 60x² – 160x³ + 240x⁴ – 192x⁵ + 64x⁶

Now we get, (1 + 3x)⁷ (1 – 2x)⁶:

=(1 + 21x + 189x² + 945x³ + 2835x⁴ + 5103x⁵ + 5103x⁶ + 2187x⁷) × (1 – 12x + 60x² – 160x³ + 240x⁴ – 192x⁵ + 64x⁶)

Now we get, it not required to multiply the above equation, only terms with x⁵ are required and that can be identified by analyzing the above equation.

\(\\\Rightarrow\) [{1 × (- 192x⁵)} + {(21x) × (240x⁴)} + {(189x²) × (- 160x³)} + {(945x³) × (60x²)} + {(2835x⁴) × (- 12x)} + {(5103x⁵) × (1)}]

\(\\\Rightarrow\) [ -192x⁵ + 5040x⁵ – 30240x⁵+ 56700x⁵ – 34020x⁵ + 5103x⁵ ]

\(\\\Rightarrow\) 2391 x⁵

Therefore we get, the coefficient of x⁵ = 2391

 

Q.4: Show that (a – b) is a factor of (aⁿ – bⁿ), where ‘n’ is a positive integer and ‘a’ and ‘b’ are distinct integers.

 

Answer. Based on formula given in Binomial Theorem

Now we get, for (a – b) to be factor of (aⁿ – bⁿ), (aⁿ – bⁿ) = p(a – b) [where p is any natural number]

We can write aⁿ = (a – b + b)ⁿ = [(a – b) + b]ⁿ

Now we get, by Binomial Theorem:

Since, [x + y]ⁿ = [ ⁿC₀ × (xⁿ) ] + [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] + [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . + [ ⁿC_n × yⁿ ]

Therefore we get, for x = (a – b) and y = b the equation becomes:

[(a – b) + b]ⁿ = [ ⁿC₀ × (a – b)ⁿ ] + [ ⁿC₁ × (a – b)^{n – 1} × b ] + [ ⁿC₂ × (a – b)^{n – 2} × b² ] + [ ⁿC₃ × (a – b)^{n – 3} × b³ ] + . . . . . . . . . . . . . + [ ⁿC_{n – 1} × (a – b) × b^{n – 1} ] + [ ⁿC_n × bⁿ ]

\(\\\Rightarrow\) [a]ⁿ = [(a – b)ⁿ] + [ ⁿC₁ × (a – b)^{n – 1} × b ] + [ ⁿC₂ × (a – b)^{n – 2} × b² ] + [ ⁿC₃ × (a – b)^{n – 3} × b³ ] + . . . . . . . . . . . . . + [ ⁿC_{n – 1} × (a – b) × b^{n – 1} ] + [ bⁿ ]

Now we get, aⁿ – bⁿ = [(a – b)ⁿ] + [ ⁿC₁ × (a – b)^{n – 1} × b ] + [ ⁿC₂ × (a – b)^{n – 2} × b² ] + [ ⁿC₃ × (a – b)^{n – 3} × b³ ] + . . . . . . . . . . . . . + [ ⁿC_{n – 1} × (a – b) × b^{n – 1} ] + [ bⁿ ] – [ bⁿ ]

\(\\\Rightarrow\) aⁿ – bⁿ = (a – b) × [ (a – b)^{n – 1} + ⁿC₁ (a – b)^{n – 2} b + ⁿC₂ (a – b)^{n – 3} b² + ⁿC₃ (a – b)^{n – 4} b³ + . . . . . . . . . . . . . + ⁿC_{n – 1} b^{n – 1} ]

\(\\\Rightarrow\) (aⁿ – bⁿ) = (a – b) × p

Where, p = [ (a – b)^{n – 1} + ⁿC₁ (a – b)^{n – 2} b + ⁿC₂ (a – b)^{n – 3} b² + ⁿC₃ (a – b)^{n – 4} b³ + . . . . . . . . . . . . . + ⁿC_{n – 1} b^{n – 1} ] is any natural number.

Therefore we get, (a – b) is a factor of (aⁿ – bⁿ), where ‘n’ is a positive integer.

 

Q.5: Evaluate \(\left(\sqrt{7}+\sqrt{5} \right )^{6}-\left ( \sqrt{7}-\sqrt{5}\right)^{6}\)

 

Answer. Based on formula given in Binomial Theorem

Find (a + b)⁶ – (a – b)⁶

Now we get, by using Binomial Theorem:

Since, [x + y]ⁿ = [ ⁿC₀ × (xⁿ) ] + [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] + [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . + [ ⁿC_n × yⁿ ]

And, [x – y]ⁿ = [ ⁿC₀ × (xⁿ) ] – [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] – [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . + (-1)ⁿ [ ⁿC_n × yⁿ ]

Therefore we get, [a + b]⁶= [ ⁶C₀ × (a⁶) ] + [ ⁶C₁ × (a⁵) × b ] + [ ⁶C₂ × (a⁴) × b² ] + [ ⁶C₃ × (a³) × b³ ] + [ ⁶C₄ × (a²) × b⁴ ] + [ ⁶C₅ × a × b⁵ ] + [⁶C₆ × b⁶]

And, [a – b]⁶ = [ ⁶C₀ × (a⁶) ] – [ ⁶C₁ × (a⁵) × b ] + [ ⁶C₂ × (a⁴) × b² ] – [ ⁶C₃ × (a³) × b³ ] + [ ⁶C₄ × (a²) × b⁴ ] – [ ⁶C₅ × a × b⁵ ] + [⁶C₆ × b⁶]

Therefore we get, (a + b)⁶ – (a – b)⁶ = [ ⁶C₀ a⁶ + ⁶C₁ a⁵b + ⁶C₂ a⁴b² + ⁶C₃ a³b³ + ⁶C₄ a²b⁴ + ⁶C₅ ab⁵ + ⁶C₆ b⁶] – [ ⁶C₀ a⁶ – ⁶C₁ a⁵b + ⁶C₂ a⁴b² – ⁶C₃ a³b³ + ⁶C₄ a²b⁴ – ⁶C₅ ab⁵ + ⁶C₆ b⁶ ]

\(\\\Rightarrow\) [ ⁶C₀ a⁶ + ⁶C₁ a⁵b + ⁶C₂ a⁴b² + ⁶C₃ a³b³ + ⁶C₄ a²b⁴ + ⁶C₅ ab⁵ + ⁶C₆ b⁶ – ⁶C₀ a⁶ + ⁶C₁ a⁵b – ⁶C₂ a⁴b² + ⁶C₃ a³b³ – ⁶C₄ a²b⁴ + ⁶C₅ ab⁵ – ⁶C₆ b⁶ ]

\(\\\Rightarrow\) 2[ ⁶C₁ a⁵b + ⁶C₃ a³b³ + ⁶C₅ ab⁵ ] = 2[ 6a⁵b + 20a³b³ + 6ab⁵ ]

Therefore we get, (a + b)⁶ – (a – b)⁶ = 4ab[ 3a⁴ + 10a² b² + 3b⁴ ] . . . . . . (1)

Now we get, on substituting a = \(\sqrt{7}\) and b = \(\sqrt{5}\) in equation (1) we will get:

\(\\\left(\sqrt{7}+\sqrt{5} \right )^{6}-\left ( \sqrt{7}-\sqrt{5}\right)^{6}:\\\)

\(\\=4( \sqrt{7}\times \sqrt{5} )\left [ (3 \sqrt{7} )^{4} +10\times (\sqrt{7} )^{2}\times ( \sqrt{5})^{2}+3(\sqrt{5})^{4} \right ]\\\)

\(\\\Rightarrow 4\sqrt{35}\;[\;147+350+75\;]\\\)

Therefore we get, \(\left(\sqrt{7}+\sqrt{5} \right )^{6}-\left ( \sqrt{7}-\sqrt{5}\right)^{6}\)= \(2288\sqrt{35}\)

 

Q.6: Find the expansion of \(\left ( a^{3} +\sqrt{a^{3}-2}\right )^{4}+ \left ( a^{3} -\sqrt{a^{3}-2}\right )^{4}\)

 

Answer. Based on formula given in Binomial Theorem

The above given expression can be assumed as: (x + y)⁴ + (x – y)⁴

Now we get, by using Binomial Theorem:

Since, [x + y]ⁿ = [ ⁿC₀ × (xⁿ) ] + [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] + [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . + [ ⁿC_n × yⁿ ]

And, [x – y]ⁿ = [ ⁿC₀ × (xⁿ) ] – [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] – [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . + (-1)ⁿ [ ⁿC_n × yⁿ ]

Therefore we get, [x + y]⁴ = [⁴C₀ × (x⁴)] + [⁴C₁ × (x³) × (y)] + [⁴C₂ × (x²) × (y)²] + [⁴C₃ × (x¹) × (y)³] + [⁴C₄ × (x⁰) × (y)⁴]

And, [x – y]⁴ = [⁴C₀ × (x⁴) ] – [⁴C₁ × (x³) × (y) ] + [⁴C₂ × (x²) × (y)² ] – [⁴C₃ × (x¹) × (y)³ ] + [⁴C₄ × (x⁰) × (y)⁴ ]

Therefore we get, (x + y)⁴ + (x – y)⁴ = [ ⁴C₀ x⁴ + ⁴C₁ x³ y + ⁴C₂ x² y² + ⁴C₃ x y³ + ⁴C₄ y⁴ ] + [ ⁴C₀ x⁴ – ⁴C₁ x³ y + ⁴C₂ x² y² – ⁴C₃ x y³ + ⁴C₄ y⁴ ]

\(\\\Rightarrow\) [ ⁴C₀ x⁴ + ⁴C₁ x³y + ⁴C₂ x²y² + ⁴C₃ xy³ + ⁴C₄ y⁴ + ⁴C₀ x⁴ – ⁴C₁ x³y + ⁴C₂ x²y² – ⁴C₃ xy³ + ⁴C₄ y⁴ ]

\(\\\Rightarrow\) 2[ ⁴C₀ x⁴ + ⁴C₂ x² y² + ⁴C₄ y⁴] = 2[ x⁴ + 6x² y² + y⁴ ]

Therefore we get, (x + y)⁴ + (x – y)⁴ = 2[ x⁴ + y⁴ + 6 x² y² ] . . . . . . . . . . . . . . . . . . . (1)

Now we get, on substituting x = a³ and y = \(\sqrt{a^{3}-2}\) in equation (1) we will get:

\(\\\Rightarrow \left ( a^{3} +\sqrt{a^{3}-2}\right )^{4}+ \left ( a^{3} -\sqrt{a^{3}-2}\right )^{4}:\\\)

\(\\=2\times \left [ (a^{3})^{4}+ (\sqrt{a^{3}-2})^{4}+[6\times (a^{3})^{2}\times (\sqrt{a^{3}-2})^{2}] \right ]\\\)

\(\\=2\left [ a^{12}+ (a^{3}-2)^{2}+[6\times a^{6}\times (a^{3}-2)] \right ]\\\)

\(\\=2\left [\; a^{12}+ a^{6}+4-4a^{3}+6a^{9}-12a^{6} \;\right ]\\\)

\(\\\Rightarrow 2a^{12}+12a^{9}-22a^{6}-8a^{3}+8\)

Therefore we get, the expansion of \(\left ( a^{3} +\sqrt{a^{3}-2}\right )^{4}+ \left ( a^{3} -\sqrt{a^{3}-2}\right )^{4}\):

= \(2a^{12}+12a^{9}-22a^{6}-8a^{3}+8\)

 

Q.7: Find the approximate value of (0.98)⁵ using the first four terms of its expansion.

 

Answer. Based on formula given in Binomial Theorem

(0.98)⁵ = (1 – 0.02)⁵

Now we get, by using Binomial Theorem:

Since, [ x – y ]ⁿ = [ ⁿC₀ × (x)ⁿ ] – [ ⁿC₁ × (x)^{n – 1} × y ] + [ ⁿC₂ × (x)^{n – 2} × y² ] – [ ⁿC₃ × (x)^{n – 3} × y³ ] + . . . . . . . . . . . . . . . . . . (-1)ⁿ [ ⁿC_n × yⁿ ]

Therefore we get, (1 – 0.02)⁵ = [ ⁵C₀ × (1)⁵ ] – [ ⁵C₁ × (1)⁴ × (0.02) ] + [ ⁵C₂ × (1)³ × (0.02)² ] – [ ⁵C₃ × (1)² × (0.02)³ ]

= 1 – [5 × 1 × 0.02] + [5 × 1 × 0.0004] – [5 × 1 × 0.000008]

= 1 – 0.10 + 0.0020 + 0.000040

= 0.90204

Therefore we get, (0.98)⁵ = 0.90204

 

Q.8: The coefficient of the 5^th term from end and the 5^th term from the beginning in the expansion of \(\left ( 2^{\frac{1}{4}}+\frac{1}{3^{\frac{1}{4}}} \right )^{n}\) are in the ratio \(1:\sqrt{6}\).Find the value of ‘n’.

 

Answer. Based on formula given in Binomial Theorem

Since, The general term in the expansion of (a + b)ⁿ is given by: T_{k + 1} = ⁿC_k × (a)^{n – k} × b^k

Now we get on substituting a = \(\left ( 2\right )^{{\frac{1}{4}}}\) and b = \(\frac{1}{(3)^{\frac{1}{4}}}\) in the above equation:

T_{k + 1} = ⁿC_k ×\(\left ( \sqrt[4]{2} \right )^{n-k} \times \left ( \frac{1}{\sqrt[4]{3}}\right )^{k}\\\)

Now we get, 5^th term from beginning in the expansion of \(\left ( 2^{\frac{1}{4}}+\frac{1}{3^{\frac{1}{4}}} \right )^{n}\):

T_{(4 + 1)} = ⁿC₄ × \(\left ( \sqrt[4]{2} \right )^{n-4} \times \left ( \frac{1}{\sqrt[4]{3}} \right )^{4}\\\)

i.e. T₅ = ⁿC₄ × \(\left ( 2 \right )^{\frac{n}{4}-1} \times \left ( \frac{1}{3} \right )\\\)

\(\\\Rightarrow T_{5} = \frac{n!}{4!\;(n-4)!}\times (2)^{\frac{n}{4}}\times (2)^{-1}\times \frac{1}{3}\\\)

Therefore we get the coefficient of 5^th term from beginning \(\boldsymbol{= \frac{n!}{4!\;(n-4)!}\times (2)^{\frac{n}{4}}\times \frac{1}{6}}\\\)

Now we get, 5^th term from end in the expansion of \(\left ( 2^{\frac{1}{4}}+\frac{1}{3^{\frac{1}{4}}} \right )^{n}\):

1^st term, 2^nd term, 3^rd term . . . . . . . . . . . (n – 5)^th term, (n – 4)^th term, (n – 3)^th term, (n – 2)^th term, (n – 1)^th term, n^th term.

Therefore we get, 5^th term from end will be (n – 4)^th term from beginning.

i.e. T_{(n – 4) + 1} = ⁿC_{n – 4} × \(\left ( \sqrt[4]{2} \right )^{n-(n-4)} \times \left ( \frac{1}{\sqrt[4]{3}} \right )^{n-4}\\\)

i.e. T_{(n – 3)} = ⁿC_{n – 4} × \(\left ( \sqrt[4]{2} \right )^{4} \times \left ( \frac{1}{\sqrt[4]{3}} \right )^{n}\times \left ( \frac{1}{\sqrt[4]{3}} \right )^{-4}\\\)

\(\\\Rightarrow T_{(n-3)} = \frac{n!}{(n-4)!\;[(n-(n-4)]!}\times 2 \times \left ( \frac{1}{3} \right )^{\frac{n}{4}}\times 3\\\)

Therefore we get the coefficient of 5^th term from end [ T_{(n – 4)}^th term ] \(= \frac{n!}{(n-4)!\;[(4)]!} \times \left ( \frac{1}{3} \right )^{\frac{n}{4}}\times 6\\\)

Now we get, according to the given conditions:

The ratio of coefficient of 5^th term from end and the coefficient of 5^th term from beginning is \(1:\sqrt{6}\\\)

Therefore we get, \(\\ \frac{T_{n-3}}{T_{5}}=\frac{1}{\sqrt{6}}\\\)

\(\\\Rightarrow \left ( \frac{n!}{(n-4)!\;[(4)]!} \times \left ( \frac{1}{3} \right )^{\frac{n}{4}}\times 6 \right )\div\left ( \frac{n!}{4!\;(n-4)!}\times (2)^{\frac{n}{4}}\times \frac{1}{6} \right )=\frac{1}{\sqrt{6}}\\\) \(\\\Rightarrow \frac{4!\;(n-4)!}{(n-4)!\;4!}\times \frac{1}{(2)^{\frac{n}{4}}}\times \frac{1}{(3)^{\frac{n}{4}}}\times 6\times 6=\frac{1}{\sqrt{6}}\\\) \(\\\Rightarrow \frac{1}{(6)^{\frac{n}{4}}}\times 36=\frac{1}{\sqrt{6}}\\\) \(\\\Rightarrow 36\sqrt{6}=(6)^{\frac{n}{4}}\\\) \(\\\Rightarrow (6)^{2}\times (6)^{\frac{1}{2}}=6^{\frac{n}{4}}\\\) \(\\\Rightarrow (6)^{\frac{5}{2}}=(6)^{\frac{n}{4}\\}\)

On comparing RHS and LHS:

\(\\\Rightarrow \frac{5}{2}=\frac{n}{4}\\\)

\(\\\Rightarrow n = 10\)

Therefore we get, the value of n = 10.

 

Q.9: By using the Binomial Theorem, find the expansion of \(\left ( 1+\frac{x}{3}-\frac{3}{x} \right )^{4}\); where ≠ 0

 

Answer. Based on formula given in Binomial Theorem

The given expression \(\left ( 1+\frac{x}{3}-\frac{3}{x} \right )^{4}\) can be expanded as \(\left [\left (1-\frac{3}{x} \right )+\frac{x}{3} \right ]^{4}\)

Now we get, by using Binomial Theorem:

Since, [x + y]ⁿ = [ ⁿC₀ × (xⁿ) ] + [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] + [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . + [ ⁿC_n × yⁿ ]

Therefore we get, \(\left [\left (1-\frac{3}{x} \right )+\frac{x}{3} \right ]^{4}\) = [⁴C₀ × \(\left (1-\frac{3}{x} \right )^{4}\)] + [⁴C₁ × \(\left (1-\frac{3}{x} \right )^{3}\) × \(\left ( \frac{x}{3} \right )\)] + [⁴C₂ × \(\left (1-\frac{3}{x} \right )^{2}\) × \(\left ( \frac{x}{3} \right )^{2}\)] + [⁴C₃ × \(\left (1-\frac{3}{x} \right )^{1}\) × \(\left ( \frac{x}{3} \right )^{3}\) + [⁴C₄ × \(\left ( \frac{x}{3} \right )^{4}\)]

\(\\=\left [ 1\times \left (1-\frac{3}{x} \right )^{4} \right ]+\left [ 4 \times \left (1-\frac{3}{x} \right )^{3}\times \left ( \frac{x}{3} \right ) \right ]+\left [ 6\times \left (1-\frac{3}{x} \right )^{2} \times \left ( \frac{x}{3} \right )^{2} \right ]+\left [ 4 \times \left (1-\frac{3}{x} \right )^{1} \times \left ( \frac{x}{3} \right )^{3} \right ]+\left [ 1 \times \left ( \frac{x}{3} \right )^{4} \right ]\) . . . . . . . . . . . . . . . . . . . (1)

Now we get, \(\left (1-\frac{3}{x} \right )^{4}\):

By using Binomial Theorem:

Since, [x – y]ⁿ = [ ⁿC₀ × (xⁿ) ] – [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] – [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . (-1)ⁿ [ ⁿC_n × yⁿ ]

Therefore we get, \(\left (1-\frac{3}{x} \right )^{4}\) = [⁴C₀] – [⁴C₁ × \(\left ( \frac{3}{x} \right )^{1}\)] + [⁴C₂ × \(\left ( \frac{3}{x} \right )^{2}\)] – [⁴C₃ × \(\left ( \frac{3}{x} \right )^{3}\)] + [⁴C₄ × \(\left ( \frac{3}{x} \right )^{4}\)]

\(\\= \left [ 1 \right ] – \left [ 4 \times \left ( \frac{3}{x} \right ) \right ] + \left [ 6 \times \left ( \frac{9}{x^{2}} \right ) \right ] – \left [ 4\times \left ( \frac{27}{x^{3}} \right ) \right ] + \left [ 1 \times \left ( \frac{81}{x^{4}} \right ) \right ]\\\)

\(\\=\left [ 1-\frac{12}{x}+\frac{54}{x^{2}}-\frac{108}{x^{3}}+\frac{81}{x^{4}} \right ]\) . . . . . . . . (2)

And, \(\left (1-\frac{3}{x} \right )^{3}\) = [ ³C₀ ] – [ ³C₁ × \(\left ( \frac{3}{x} \right )^{1}\)] + [³C₂ × \(\left ( \frac{3}{x} \right )^{2}\)] – [³C₃ × \(\left ( \frac{3}{x} \right )^{3}\)]

\(\\= [1]-\left [ 3\times \left ( \frac{3}{x} \right ) \right ] + \left [ 3 \times \left ( \frac{9}{x^{2}} \right ) \right ] – \left [ 1\times \left ( \frac{27}{x^{3}} \right ) \right ]\\\)

\(\\=\left [1-\frac{9}{x}+\frac{27}{x^{2}}-\frac{27}{x^{3}} \right ]\) . . . . . . . . . . . . (3)

Now we get, from equation (1), equation (2) and equation (3):

\(\\\Rightarrow \left [ 1-\frac{12}{x}+\frac{54}{x^{2}}-\frac{108}{x^{3}}+\frac{81}{x^{4}}\right ]+\left [ 4 \times \left ( 1-\frac{9}{x}+\frac{27}{x^{2}}-\frac{27}{x^{3}} \right )\times \frac{x}{3} \right ]+\left [ 6\times \left (1+\frac{9}{x^{2}}-\frac{6}{x} \right )\times \frac{x^{2}}{9}\right ] +\left [ \left ( 4-\frac{12}{x} \right )\times \frac{x^{3}}{27} \right ]+\left [\frac{x^{4}}{81}\right ]\\\) \(\\\Rightarrow \left [ 1-\frac{12}{x}+\frac{54}{x^{2}}-\frac{108}{x^{3}}+\frac{81}{x^{4}}\right ]+\left [ \frac{4x}{3}-12+\frac{36}{x}-\frac{36}{x^{2}}\right ]+\\\\\\\\+\left [ \frac{2x^{2}}{3}+6-4x\right ] +\left [ \frac{4x^{3}}{27}-\frac{4x^{2}}{9} \right ]+\left [\frac{x^{4}}{81}\right ]\\\) \(\\\boldsymbol{\Rightarrow \left [ \frac{x^{4}}{81}+\frac{4x^{3}}{27}+\frac{2x^{2}}{9}-\frac{8x}{3}-5+\frac{24}{x}+\frac{18}{x^{2}}-\frac{108}{x^{3}}+\frac{81}{x^{4}}\right ]}\\\)

Therefore we get, the expansion of \(\left ( 1+\frac{x}{3}-\frac{3}{x} \right )^{4}\):

= \(\boldsymbol{\left [ \frac{x^{4}}{81}+\frac{4x^{3}}{27}+\frac{2x^{2}}{9}-\frac{8x}{3}-5+\frac{24}{x}+\frac{18}{x^{2}}-\frac{108}{x^{3}}+\frac{81}{x^{4}}\right ]}\)

 

 

Q.10: By using the Binomial Theorem, find the expansion of (6x² – 4ax + 6a²)³.

 

Answer. Based on formula given in Binomial Theorem

The given expression (6x² – 4ax + 6a²)³ can be expanded as [(6x² – 4ax) + 6a²)³]

By using Binomial Theorem:

Since, [x + y]ⁿ = [ ⁿC₀ × (xⁿ) ] + [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] + [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . + [ ⁿC_n × yⁿ ]

Therefore we get, [(6x² – 4ax) + 6a²)³] = [³C₀ × (6x² – 4ax)³] + [³C₁ × (6x² – 4ax)² × 6a²] + [³C₂ × (6x² – 4ax)¹ × (6a²)²] + [³C₃ × (6x² – 4ax)⁰ × (6a²)³]

= [1× (6x² – 4ax)³] + [3 × (6x² – 4ax)² × 6a²] + [3 × (6x² – 4ax) × (6a²)²] + [1 × (6a²)³]

= [(6x² – 4ax)³] + [(6x² – 4ax)² × 18a²] + [(6x² – 4ax) × 108a⁴ ] + [216a⁶] . . . . . . . . . . . . (1)

Now we get, [(6x² – 4ax)³]:

By using Binomial Theorem:

Since, [x – y]ⁿ = [ ⁿC₀ × (xⁿ) ] – [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] – [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . (-1)ⁿ [ ⁿC_n × yⁿ ]

Therefore we get, [(6x² – 4ax)³] = [³C₀ × (6x²)³] – [³C₁ × (6x²)²) × 4ax] + [³C₂ × 6x² × (4ax)²] – [³C₃ × (4ax)³]

= [1 × (216x⁶)] – [3 × (36x⁴) × 4ax] + [3 × (6x²)× 16a²x²] – [1 × (64a³x³)]

\(\Rightarrow\) [(6x² – 4ax)³] = [216x⁶ – 432 ax⁵ + 288 a²x⁴ – 64a³x³] . . . . . . . . . . (2)

From equation (1) and equation (2)

=[216x⁶ – 432 ax⁵ + 288 a²x⁴ – 64a³x³] + [(36x⁴ + 16a²x² – 48ax³) × 18a²] + [(6x² – 4ax) × 108a⁴ ] + [ 216a⁶ ]

=[216x⁶ – 432 ax⁵ + 288 a²x⁴ – 64a³x³ + 648a²x⁴ + 288a⁴x² – 864a³x³ + 648x²a⁴ – 432xa⁵ + 216a⁶]

=8[27x⁶ – 54 ax⁵ + 36 a²x⁴ – 8a³x³ + 81a²x⁴ + 36a⁴x² – 108a³x³ + 81x²a⁴ – 54xa⁵ + 27a⁶]

=8[27x⁶ – 54 ax⁵ + 117a² x⁴ – 116a³x³ + 117a⁴x² – 54a⁵x + 27a⁶]

Therefore we get, the expansion of (6x² – 4ax + 6a²)³ :

= 8[27x⁶ – 54 ax⁵ + 117a² x⁴ – 116a³x³ + 117a⁴x² – 54a⁵x + 27a⁶]

 

 

Q.11: By using the Binomial Theorem, find the expansion of \(\left ( 1+\frac{x}{5}-\frac{5}{x} \right )^{4}\); where ≠ 0

 

Answer. Based on formula given in Binomial Theorem

The given expression \(\left ( 1+\frac{x}{5}-\frac{5}{x} \right )^{4}\) can be expanded as \(\left [\left (1-\frac{5}{x} \right )+\frac{x}{5} \right ]^{4}\)

Now we get, by using Binomial Theorem:

Since, [x + y]ⁿ = [ ⁿC₀ × (xⁿ) ] + [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] + [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . + [ ⁿC_n × yⁿ ]

Therefore we get, \(\left [\left (1-\frac{5}{x} \right )+\frac{x}{5} \right ]^{4}\)

= [⁴C₀ × \(\left (1-\frac{5}{x} \right )^{4}\)] + [⁴C₁ × \(\left (1-\frac{5}{x} \right )^{3}\) × \(\left ( \frac{x}{5} \right )\)] + [⁴C₂ × \(\left (1-\frac{5}{x} \right )^{2}\) × \(\left ( \frac{x}{5} \right )^{2}\)] + [⁴C₃ × \(\left (1-\frac{5}{x} \right )^{1}\) × \(\left ( \frac{x}{5} \right )^{3}\) + [⁴C₄ × \(\left ( \frac{x}{5} \right )^{4}\)]

\(\\=\left [ 1\times \left (1-\frac{5}{x} \right )^{4} \right ]+\left [ 4 \times \left (1-\frac{5}{x} \right )^{3}\times \left ( \frac{x}{5} \right ) \right ]+\left [ 6\times \left (1-\frac{5}{x} \right )^{2} \times \left ( \frac{x}{5} \right )^{2} \right ]+\left [ 4 \times \left (1-\frac{5}{x} \right )^{1} \times \left ( \frac{x}{5} \right )^{3} \right ]+\left [ 1 \times \left ( \frac{x}{5} \right )^{4} \right ]\) . . . . . . . . . . . . . . . . . . . (1)

Now we get, \(\left (1-\frac{5}{x} \right )^{4}:\\\)

By using Binomial Theorem:

Since, [x – y]ⁿ = [ ⁿC₀ × (xⁿ) ] – [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] – [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . (-1)ⁿ [ ⁿC_n × yⁿ ]

Therefore we get, \(\left (1-\frac{5}{x} \right )^{4}\) = [⁴C₀] – [⁴C₁ × \(\left ( \frac{5}{x} \right )^{1}\)] + [⁴C₂ × \(\left ( \frac{5}{x} \right )^{2}\)] – [⁴C₃ × \(\left ( \frac{5}{x} \right )^{3}\)] + [⁴C₄ × \(\left ( \frac{5}{x} \right )^{4}\)]

\(\\= \left [ 1 \right ] – \left [ 4 \times \left ( \frac{5}{x} \right ) \right ] + \left [ 6 \times \left ( \frac{25}{x^{2}} \right ) \right ] – \left [ 4\times \left ( \frac{125}{x^{3}} \right ) \right ] + \left [ 1 \times \left ( \frac{625}{x^{4}} \right ) \right ]\\\)

\(\\=\left [ 1-\frac{20}{x}+\frac{150}{x^{2}}-\frac{500}{x^{3}}+\frac{625}{x^{4}} \right ]\) . . . . . . . . (2)

And, \(\left (1-\frac{5}{x} \right )^{3}:\\\)

[ ³C₀ ] – [ ³C₁ × \(\left ( \frac{5}{x} \right )^{1}\)] + [³C₂ × \(\left ( \frac{5}{x} \right )^{2}\)] – [³C₃ × \(\left ( \frac{5}{x} \right )^{3}\)]

\(\\= [1]-\left [ 3\times \left ( \frac{5}{x} \right ) \right ] + \left [ 3 \times \left ( \frac{25}{x^{2}} \right ) \right ] – \left [ 1\times \left ( \frac{125}{x^{3}} \right ) \right ]\\\)

\(\\=\left [ 1-\frac{15}{x}+\frac{75}{x^{2}}-\frac{125}{x^{3}} \right ]\) . . . . . . . . . . . . . . . . . . . (3)

Now we get, from equation (1), equation (2) and equation (3):

\(\\\Rightarrow \left [ 1-\frac{20}{x}+\frac{150}{x^{2}}-\frac{500}{x^{3}}+\frac{625}{x^{4}} \right ]+\left [ 4 \times \left [ 1-\frac{15}{x}+\frac{75}{x^{2}}-\frac{125}{x^{3}} \right ] \times \left ( \frac{x}{5} \right ) \right ]+\\\\\\+\left [ 6\times \left (1+\frac{25}{x^{2}} -\frac{10}{x}\right ) \times \left ( \frac{x}{5} \right )^{2} \right ]+\left [ 4 \times \left (1-\frac{5}{x} \right )\times \left ( \frac{x}{5} \right )^{3} \right ]+\left [ \left ( \frac{x}{5} \right )^{4} \right ]\\\) \(\\\Rightarrow \left [ 1-\frac{20}{x}+\frac{150}{x^{2}}-\frac{500}{x^{3}}+\frac{625}{x^{4}} \right ]+\left [ \frac{4x}{5}-12+\frac{60}{x}-\frac{100}{x^{2}} \right ]+\left [ \frac{6x^{2}}{25}+6-\frac{12x}{5}\right ]+\left [ \frac{4x^{3}}{125}-\frac{x^{2}}{25} \right ]+\left [ \frac{x^{4}}{625} \right ]\\\) \(\\\\\Rightarrow \left [\frac{50}{x^{2}}-\frac{500}{x^{3}}+\frac{625}{x^{4}}-\frac{8x}{5}-5+\frac{40}{x}+ \frac{x^{2}}{5}+\frac{4x^{3}}{125}+\frac{x^{4}}{625} \right ]\\\\\)

Therefore we get, the expansion of \(\left ( 1+\frac{x}{5}-\frac{5}{x} \right )^{4}\):

= \(\left [\frac{x^{4}}{625}+\frac{4x^{3}}{125}+ \frac{x^{2}}{5}-\frac{8x}{5}-5+\frac{40}{x}+\frac{50}{x^{2}}-\frac{500}{x^{3}}+\frac{625}{x^{4}}\right ]\)

 

Q.12: By using the Binomial Theorem, find the expansion of (5x³ – 2ax + 3a³)³.

 

Answer. Based on formula given in Binomial Theorem

The given expression (5x³ – 2ax + 3a³)³ can be expanded as [(5x³ – 2ax) + 3a³)³]

By using Binomial Theorem:

Since, [x + y]ⁿ = [ ⁿC₀ × (xⁿ) ] + [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] + [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . + [ ⁿC_n × yⁿ ]

Therefore we get, [ (5x³ – 2ax) + 3a³)³ ] = [ ³C₀ × (5x³ – 2ax)³ ] + [ ³C₁ × (5x³ – 2ax)² × 3a³ ] + [ ³C₂ × (5x³ – 2ax)¹ × (3a³)² ] + [ ³C₃ × (5x³ – 2ax)⁰ × (3a³)³ ]

= [1× (5x³ – 2ax)³] + [3 × (5x³ – 2ax)² × 3a³] + [3 × (5x³ – 2ax) × (3a³)²] + [1 × (3a³)³]

= [(5x³ – 2ax)³] + [(5x³ – 2ax)² × 9a³] + [(5x³ – 2ax) × 27a⁶] + [27a⁹] . . . . . . . . . . . . (1)

Now we get, [(5x³ – 2ax)³]:

By using Binomial Theorem:

Since, [x – y]ⁿ = [ ⁿC₀ × (xⁿ) ] – [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] – [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . (-1)ⁿ [ ⁿC_n × yⁿ ]

Therefore we get, [(5x² – 2ax)³] = [³C₀ × (5x³)³] – [³C₁ × (5x³)²) × 2ax] + [³C₂ × 5x³ × (2ax)²] – [³C₃ × (2ax)³]

= [1 × 125x⁹] – [3 × (25x⁶)× 2ax] + [3 × (5x³) × 4a²x²] – [1 × 8a³x³]

[(5x³ – 2ax)³] = [125x⁶ – 150ax⁵ + 60a²x⁴ – 8a³x³] . . . . . . . . . . (2)

From equation (1) and equation (2):

=[125x⁶ – 150ax⁵ + 60a²x⁴ – 8a³x³] + [(25x⁶ + 4a²x² – 20ax⁴) × 9a³] + [(5x³ – 2ax) × 27a⁶ ] + [ 27a⁹ ]

=[125x⁶ – 150ax⁵ + 60a²x⁴ – 8a³x³] + [225a³x⁶ + 36a⁵x² – 180a⁴x⁴] + [135a⁶x³ – 54a⁷x ] + [ 27a⁹ ]

=[125x⁶ – 150ax⁵ + 60a²x⁴ – 8a³x³ + 225a³x⁶ + 36a⁵x² – 180a⁴x⁴ + 135a⁶x³ – 54a⁷x + 27a⁹ ]

=[125x⁶ + 225a³x⁶ – 150ax⁵ + 60a²x⁴ – 180a⁴x⁴ – 8a³x³ + 135a⁶x³ + 36a⁵x² – 54a⁷x + 27a⁹ ]

Therefore we get, the expansion of (5x³ – 2ax + 3a³)³ :

=[125x⁶ + 225a³x⁶ – 150ax⁵ + 60a²x⁴ – 180a⁴x⁴ – 8a³x³ + 135a⁶x³ + 36a⁵x² – 54a⁷x + 27a⁹]

 

 

Q.13: Find the coefficient of x⁷ in the product of (1 + x)⁷ (1 + 3x)⁶ by using binomial theorem.

 

Answer. Based on formula given in Binomial Theorem

By using Binomial Theorem:

Since, (1 + x)ⁿ = [ ⁿC₀ ] + [ ⁿC₁ × (x) ] + [ ⁿC₂ × (x)² ] + [ ⁿC₃ × (x³) ] + . . . . . . . . . . . . . . . . . . + [ ⁿC_n × (x)ⁿ ]

Therefore we get, (1 + x)⁷ = [ ⁷C₀ ] + [ ⁷C₁ × (x) ] + [ ⁷C₂ × (x)² ] + [ ⁷C₃ × (x)³ ] + [ ⁷C₄ × (x)⁴ ] + [ ⁷C₅ × (x⁵) ] + [ ⁷C₆ × (x)⁶ ] + [ ⁷C₇ × (x)⁷

\(\\\Rightarrow\) 1 + [ 7x ] + [ 21x² ] + [ 35x³ ] + [ 35x⁴ ] + [ 21x⁵ ] + [ 7x⁶ ] + [ x⁷ ]

Therefore we get, (1 + x)⁷ = 1+7x + 21x² + 35x³ + 35x⁴ + 21x⁵ + 7x⁶ + x⁷

Now we get, (1 + 3x)⁶:

By using Binomial Theorem:

(1 + 3x)⁶ = [ ⁶C₀ ] + [ ⁶C₁ × (3x) ] + [ ⁶C₂ × (3x)² ] + [ ⁶C₃ × (3x)³ ] + [ ⁶C₄ × (3x)⁴ ] + [ ⁶C₅ × (3x⁵) ] + [ ⁶C₆ × (3x)⁶ ]

\(\\\Rightarrow\) 1 + [ 6 × (3x) ] + [ 15 × (9x²) ] + [ 20 × (27x³) ] + [ 15 × (81x⁴) ] + [ 6 × (243x⁵) ] + [ 1 × 729x⁶ ]

\(\\\Rightarrow\) 1 + 18x + 135x² + 540x³ + 1215x⁴ + 1458x⁵ + 729x⁶

Therefore we get, (1 + 3x)⁶ = 1 + 18x + 135x² + 540x³ + 1215x⁴ + 1458x⁵ + 729x⁶

Now we get, (1 + x)⁷ (1 + 3x)⁶:

(1+7x + 21x² + 35x³ + 35x⁴ + 21x⁵ + 7x⁶ + x⁷) × (1 + 18x + 135x² + 540x³ + 1215x⁴ + 1458x⁵ + 729x⁶)

Now we get, it not required to multiply the above equation, only terms with x⁵ are required and that can be identified by analysing the above equation.

\(\\\Rightarrow\) [{(7x) × (729x⁶)} + {(21x²) × (1458x⁵)} + {(35x³) × (1215x⁴)} + {(35x⁴) × (540x³)} + {(21x⁵) × (135x²)} + {(7x⁶) × (18x)} + {(x⁷) × (1)}]

\(\\\Rightarrow\) [5103x⁵ + 30618x⁵ + 42525x⁵+ 18900x⁵ + 2835x⁷ + 126x⁷ + x⁷]

\(\\\Rightarrow\\\) 72551 x⁷

Therefore we get, the coefficient of x⁷ = 72551

 

Q.14: Evaluate \(\left(\sqrt{3}+\sqrt{7} \right )^{5}-\left ( \sqrt{3}-\sqrt{7}\right)^{5}\)

 

Answer. Based on formula given in Binomial Theorem

Find (a + b)⁵ – (a – b)⁵

Now we get, by using Binomial Theorem:

Since, [x + y]ⁿ = [ ⁿC₀ × (xⁿ) ] + [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] + [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . + [ ⁿC_n × yⁿ ]

And, [x – y]ⁿ = [ ⁿC₀ × (xⁿ) ] – [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] – [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . + (-1)ⁿ [ ⁿC_n × yⁿ ]

Therefore we get, [a + b]⁵ = [ ⁵C₀ × (a⁵) ] + [ ⁵C₁ × (a⁴) × b ] + [ ⁵C₂ × (a³) × b² ] + [ ⁵C₃ × (a²) × b³ ] + [ ⁵C₄ × (a¹) × b⁴ ] + [ ⁵C₅ × b⁵ ]

And, [a – b]⁵ = [ ⁵C₀ × (a⁵) ] – [ ⁵C₁ × (a⁴) × b ] + [ ⁵C₂ × (a³) × b² ] – [ ⁵C₃ × (a²) × b³ ] + [ ⁵C₄ × (a) × b⁴ ] – [ ⁵C₅ × b⁵ ]

Therefore we get, (a + b)⁵ – (a – b)⁵ = [⁵C₀ a⁵ + ⁵C₁ a⁴b + ⁵C₂ a³b² + ⁵C₃ a²b³ + ⁵C₄ ab⁴ + ⁵C₅ b⁵] – [⁵C₀ a⁵ – ⁵C₁ a⁴ b + ⁵C₂ a³b² – ⁵C₃ a²b³ + ⁵C₄ ab⁴ – ⁵C₅ ab⁵]

\(\\\Rightarrow\) [⁵C₀ a⁵ + ⁵C₁ a⁴b + ⁵C₂ a³b² + ⁵C₃ a²b³ + ⁵C₄ ab⁴ + ⁵C₅ b⁵ – ⁵C₀ a⁵ + ⁵C₁ a⁴ b – ⁵C₂ a³b² + ⁵C₃ a²b³ – ⁵C₄ ab⁴ + ⁵C₅ ab⁵]

\(\\\Rightarrow\) 2[ ⁵C₁ a⁴b + ⁵C₃ a²b³ + ⁵C₅ b⁵ ] = 2[ 5a⁴b + 10a²b³ + b⁵ ]

Therefore we get, (a + b)⁵ – (a – b)⁵ = 2b[ 5a⁴ + 10a² b² + b⁴ ] . . . . . . (1)

Now we get, on substituting a = \(\sqrt{3}\) and b = \(\sqrt{7}\) in equation (1) we will get:

\(\\\left(\sqrt{3}+\sqrt{7} \right )^{5}-\left ( \sqrt{3}-\sqrt{7}\right)^{5}:\\\)

\(\\=2(\sqrt{7})\times \left [ 5( \sqrt{3})^{4} +10\times ( \sqrt{3})^{2}\times (\sqrt{7})^{2}+(\sqrt{7})^{4} \right ]\\\)

\(\\\Rightarrow 2\sqrt{7}\left [\;45+210+49\; \right ]\\\)

Therefore we get, \(\left(\sqrt{3}+\sqrt{7} \right )^{5}-\left ( \sqrt{3}-\sqrt{7}\right)^{5}\)= \(608\sqrt{7}\)

 

Q.15: Find the expansion of \(\left ( a^{2} +\sqrt{a^{2}-5}\right )^{4}+ \left ( a^{2} -\sqrt{a^{2}-5}\right )^{4}\)

 

Answer. Based on formula given in Binomial Theorem

The above given expression can be assumed as: (x + y)⁴ + (x – y)⁴

Now we get, by using Binomial Theorem:

Since, [x + y]ⁿ = [ ⁿC₀ × (xⁿ) ] + [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] + [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . + [ ⁿC_n × yⁿ ]

And, [x – y]ⁿ = [ ⁿC₀ × (xⁿ) ] – [ ⁿC₁ × (x^{n – 1}) × y ] + [ ⁿC₂ × (x^{n – 2}) × y² ] – [ ⁿC₃ × (x^{n – 3}) × y³ ] + . . . . . . . . . . . . . . . . . . + (-1)ⁿ [ ⁿC_n × yⁿ ]

Therefore we get, [x + y]⁴ = [⁴C₀ × (x⁴)] + [⁴C₁ × (x³) × (y)] + [⁴C₂ × (x²) × (y)²] + [⁴C₃ × (x¹) × (y)³] + [⁴C₄ × (x⁰) × (y)⁴]

And, [x – y]⁴ = [⁴C₀ × (x⁴) ] – [⁴C₁ × (x³) × (y) ] + [⁴C₂ × (x²) × (y)² ] – [⁴C₃ × (x¹) × (y)³ ] + [⁴C₄ × (x⁰) × (y)⁴ ]

Therefore we get, (x + y)⁴ + (x – y)⁴ = [ ⁴C₀ x⁴ + ⁴C₁ x³ y + ⁴C₂ x² y² + ⁴C₃ x y³ + ⁴C₄ y⁴ ] + [ ⁴C₀ x⁴ – ⁴C₁ x³ y + ⁴C₂ x² y² – ⁴C₃ x y³ + ⁴C₄ y⁴ ]

\(\\\Rightarrow\) [ ⁴C₀ x⁴ + ⁴C₁ x³y + ⁴C₂ x²y² + ⁴C₃ xy³ + ⁴C₄ y⁴ + ⁴C₀ x⁴ – ⁴C₁ x³y + ⁴C₂ x²y² – ⁴C₃ xy³ + ⁴C₄ y⁴ ]

\(\\\Rightarrow\) 2[ ⁴C₀ x⁴ + ⁴C₂ x² y² + ⁴C₄ y⁴] = 2[ x⁴ + 6x² y² + y⁴ ]

Therefore we get, (x + y)⁴ + (x – y)⁴ = 2[ x⁴ + y⁴ + 6 x² y² ] . . . . . . . . . . . . . . . . . . . (1)

Now we get, on substituting x = a² and y = \(\sqrt{a^{2}-5}\) in equation (1) we will get:

\(\\\Rightarrow \left ( a^{2} +\sqrt{a^{2}-5}\right )^{4}+ \left ( a^{2} -\sqrt{a^{2}-5}\right )^{4}:\\\)

\(\\=2\times \left [ (a^{2})^{4}+ \left ( \sqrt{a^{2}-5} \right )^{4}+6\times (a^{2})^{2}\times \left ( \sqrt{a^{2}-5} \right )^{2} \right ]\\\) \(\\\Rightarrow 2\left [ a^{8}+ (a^{2}-5)^{2}+[6a^{4}\times (a^{2}-5)] \right ]\\\) \(\\\Rightarrow 2 \left [ a^{8}+a^{4}+ 25 + 10a^{2}+6a^{6}-30a^{4} \right ]\\\)

Therefore we get, the expansion of \(\left [ a^{2} +\sqrt{a^{2}-5}\right ]^{4}+ \left [ a^{2} -\sqrt{a^{2}-5}\right ]^{4}:\)

=\(2 \left [ a^{8}+6a^{6}-29a^{4}+10a^{2}+25 \right ]\)

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