#### NCERT Class 11 Chapter-4: Principles of Mathematical Induction

Practice Exercise 4.1

Prove the following through principle of mathematical induction for all values of n, where n is a natural number.

1: $$1 + 3 + 3^{2} + …. + 3^{n – 1} = \frac{\left ( 3^{n} – 1 \right )}{2}$$

Sol: Based on the formulae given in Principles of Mathematical Induction

The given statement is as follows:

P(n) : $$1 + 3 + 3^{2} + …. + 3^{n – 1} = \frac{\left ( 3^{n} – 1 \right )}{2}$$

Now, for n = 1

P(1) = $$\frac{\left ( 3^{1} – 1 \right )}{2}$$ = $$\frac{\left ( 3 – 1 \right )}{2}$$= $$\frac{2}{2}$$= 1

Thus, the P(n) is true for n=1

P(k) be true, where k is a positive integer.

$$1 + 3 + 3^{2} + ….+ 3^{k – 1} = \frac{\left ( 3^{k} – 1 \right )}{2}$$ . . . . . . . . . . (1)

We will prove that P(k+1) is also true as follows:

P(k + 1):

=$$\\1 + 3 + 3^{2} + ….+ 3^{k – 1} + 3^{\left (k + 1 \right ) – 1}\\$$

=$$\\\left (1 + 3 + 3^{2} + ….+ 3^{k – 1} \right )+ 3^{k}\\$$ [Using equation (1)]

=$$\\\frac{\left (3^{k} – 1 \right )}{2} + 3^{k}\\$$

=$$\\\frac{\left (3^{k} – 1 \right )+ 2.3^{k}}{2}\\$$

=$$\\\frac{(1 + 2)3^{k} – 1}{2}\\$$

=$$\\\frac{3.3^{k} – 1}{2}\\$$

=$$\\\frac{3^{k + 1} – 1}{2}$$

Thus, whenever P(k) proves to be true, P(k+1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

2: $$1^{3}\, +\, 2^{3}\, +\, 3^{3}\, +\, ……\, +\, n^{3}$$ = $$\left ( \frac{n\;(n+1)}{2} \right )^{2}$$

Sol: Based on the formulae given in Principles of Mathematical Induction

The given statement is as follows:

P(n): $$1^{3}\, +\, 2^{3}\, +\, 3^{3}\, +\, ……\, +\, n^{3}$$ = $$\left ( \frac{n\;(n+1)}{2} \right )^{2}$$

Now, for n = 1

P(1): $$1^{3} = 1 = \left ( \frac{1\left ( 1 + 1 \right )}{2} \right )^{2}$$= $$\left (\frac{1\times 2}{2} \right )^{2}$$= $$1^{2}$$ = 1

Thus, the P(n) is true for n = 1

Let, P(k) be true, where ‘k’ is a positive integer as follows.

$$1^{3}\, +\, 2^{3}\, +\, 3^{3}\, +\, ……\, +\, k^{3} = \left ( \frac{k(k+1)}{2} \right )^{2}$$ . . . . . . . . . . . . . . . . . (1)

Now, we will prove P(k + 1) is also true.

P(k + 1):

=$$\\1^{3}\, +\, 2^{3}\, +\, 3^{3}\, +\, ……\, +\, k^{3} + \left (k + 1 \right )^{3}\\$$

=$$\\\left (1^{3}\, +\, 2^{3}\, +\, 3^{3}\, +\, ……\, +\, k^{3} \right ) + \left (k + 1 \right )^{3}\\$$

=$$\\\left ( \frac{k\left ( k + 1 \right )}{2} \right )^{2} +\left ( k + 1 \right )^{3}\\$$ [From equation (1)]

=$$\\\frac{k^{2} \left ( k + 1 \right )^{2}}{4} + \left ( k + 1 \right )^{3}\\$$

=$$\\\frac{k^{2} \left ( k + 1 \right )^{2} + 4\left ( k + 1 \right )^{3}}{4}\\$$

=$$\\\frac{ \left ( k + 1 \right )^{2}\left \{ k^{2} + 4\left ( k + 1 \right ) \right \}}{4}\\$$

=$$\\\frac{ \left ( k + 1 \right )^{2}\left \{ k^{2} + 4k + 4 \right \}}{4}\\$$

=$$\\\frac{ \left ( k + 1 \right )^{2} \left ( k + 2 \right )^{2}}{4}\\$$

=$$\\\frac{ \left ( k + 1 \right )^{2} \left ( k + 1 + 1 \right )^{2}}{4}\\$$

=$$\\\left [\frac{ \left ( k + 1 \right ) \left ( k + 1 + 1 \right )}{2} \right ]^{2}\\$$

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

3: $$1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + …. + \frac{1}{1 + 2 + 3 +… + n } = \frac{2n}{n + 1}$$

Sol: Based on the formulae given in Principles of Mathematical Induction

The given statement is:

P(n): $$1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + …. + \frac{1}{1 + 2 + 3 +… + n } = \frac{2n}{n + 1}$$

Now, for n = 1

P(1): $$1 = \frac{2 \times 1}{1 + 1} = \frac{2}{2} = 1$$

Thus, the P(n) is true for n=1 as follows

Let, P(k) be true, where k is a positive integer.

$$1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + …. + \frac{1}{1 + 2 + 3 +… + k } = \frac{2k}{k + 1}$$ . . . . . . . . . . . . . . (1)

Now, we will prove P(k + 1) is also true.

=$$\\1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + …. + \frac{1}{1 + 2 + 3 +… + k } + \frac{1}{1 + 2 + 3 +… + k + \left ( k + 1 \right )}\\$$

=$$\\\left [1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + …. + \frac{1}{1 + 2 + 3 +… + k } \right ]+ \frac{1}{1 + 2 + 3 +… + k + \left ( k + 1 \right )}\\$$

=$$\\\frac{2k}{k + 1}+ \frac{1}{1 + 2 + 3 +… + k + \left ( k + 1 \right )}\\$$ [From equation (1)]

= $$\\\frac{2k}{k + 1}+ \frac{1}{\left (\frac{\left ( k + 1 \right )\left ( k + 1 + 1 \right )}{2} \right )}$$ ($$\;1 + 2 + 3 + … + n = \frac{n\left ( n + 1 \right )}{2}\;\\$$)

= $$\\\frac{2k}{k + 1}+ \frac{2}{{\left ( k + 1 \right )\left ( k + 2 \right )}}\\$$

= $$\\\frac{2}{\left (k + 1 \right )} \left ( k + \frac{1}{k + 2} \right )\\$$

= $$\\\frac{2}{\left (k + 1 \right )} \left ( \frac{k^{2} + 2k + 1 }{k + 2}\right )\\$$

= $$\\\frac{2}{\left (k + 1 \right )} \left ( \frac{\left (k + 1 \right ) ^{2}}{k + 2}\right )\\$$

= $$\\\frac{2\left ( k + 1 \right )}{\left ( k + 2 \right )}\\$$

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

4: $$1.2.3 + 2.3.4 + … + n\left ( n + 1 \right )\left ( n + 2 \right ) = \frac{n \left ( n + 1 \right ) \left ( n + 2 \right )\left ( n + 3 \right )}{4}$$

Sol: Based on the formulae given in Principles of Mathematical Induction

The given statement is as follows:

P(n): $$1.2.3 + 2.3.4 + … + n\left ( n + 1 \right )\left ( n + 2 \right ) = \frac{n \left ( n + 1 \right ) \left ( n + 2 \right )\left ( n + 3 \right )}{4}$$

Now, for n = 1

P(1): $$1.2.3 = 6$$= $$\frac{1 \left ( 1 + 1 \right ) \left ( 1 + 2 \right )\left ( 1 + 3 \right )}{4}$$= $$\frac{1.2.3.4}{4}$$= 6

Thus, the P(n) is true for n=1.

Let, P(k) be true, where k is a positive integer.

$$1.2.3 + 2.3.4 + … + k\left ( k + 1 \right )\left ( k + 2 \right ) = \frac{k \left ( k + 1 \right ) \left ( k + 2 \right )\left ( k + 3 \right )}{4}$$ . . . . . . . . . . (1)

Now, we will prove P(k + 1) is true.

=$$\\1.2.3 + 2.3.4 + … + k\left ( k + 1 \right )\left ( k + 2 \right ) + \left ( k + 1 \right )\left ( k + 2 \right )\left ( k + 3 \right )\\$$

=$$[1.2.3 + 2.3.4 + … + k( k + 1)( k + 2 ) ] +n (k + 1)( k + 2 )( k + 3 )\\$$ [By using equation (1)]

=$$\\\frac{k \left ( k + 1 \right ) \left ( k + 2 \right )\left ( k + 3 \right )}{4} + \left ( k + 1 \right )\left ( k + 2 \right )\left ( k + 3\right )\\$$

Now, by using equation (1) :

=$$\\\left ( k + 1 \right )\left ( k + 2 \right )\left ( k + 3\right )\left ( \frac{k}{4} + 1\right )\\$$

=$$\\\frac{\left ( k + 1 \right )\left ( k + 2 \right )\left ( k + 3\right )\left ( k + 4 \right )}{4}\\$$

=$$\\\frac{\left ( k + 1 \right )\left ( k + 1 + 1 \right )\left ( k + 1 + 2\right )\left ( k + 1 + 3 \right )}{4}\\$$

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

5: $$1.3 + 2. 3^{2}+ 3.3^{3} +…+ n. 3^{n} = \frac{(2n-1)3^{n+1}\, +\, 3}{4}$$

Sol: Based on the formulae given in Principles of Mathematical Induction

The given statement is:

P(n): $$1.3 + 2. 3^{2}+ 3.3^{3} +…+ n. 3^{n} = \frac{(2n-1)3^{n+1}\, +\, 3}{4}$$

Now, for n = 1:

=$$\\\frac{\left ( 2.1 – 1 \right )3^{1+1} + 3}{4}$$= $$\frac{3^{2} + 3}{4}$$= $$\frac{12}{4}$$= 3

Thus, the P(n) is true for n=1.

Let, P(k) be true, where k is a positive integer.

1.3 + 2. 3^{2}+ 3.3^{3} +…+ k. 3^{k} = \frac{(2k-1)3^{k+1}\, +\, 3}{4} . . . . . . . (1)

Now, we will prove P(k + 1) is also true:

=$$\\1.3 + 2.3^{2} + 3.3^{3} + … + k.3^{k} + \left ( k + 1 \right ).3^{k+1}\\$$

=$$\\\frac{\left ( 2k -1 \right )3^{k+1}+ 3}{4} + \left ( k + 1 \right )3^{k + 1}$$ [By using equation (1)]

=$$\\\frac{\left ( 2k -1 \right )3^{k+1}+ 3 + 4\left ( k + 1 \right )3^{k + 1}}{4}\\$$

= $$\\\frac{3^{k + 1}\left \{ 2k – 1 + 4\left ( k + 1 \right ) \right \} + 3}{4}\\$$

=$$\\\frac{3^{k + 1}\left \{ 6k + 3 \right \} + 3}{4}\\$$

=$$\\\frac{3^{k + 1}.3\left \{ 2k + 1 \right \} + 3}{4}\\$$

= $$\\\frac{3^{\left (k + 1 \right )+ 1}\left \{ 2k + 1 \right \} + 3}{4}\\$$

=$$\\\frac{\left \{ 2\left ( k + 1 \right )- 1 \right \}3^{\left ( k + 1 \right ) + 1} + 3}{4}\\$$

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

6: $$1.2 + 2.3 + 3.4 + … + n.\left ( n + 1 \right ) = \left [ \frac{n\left ( n + 1 \right )\left ( n + 2 \right )}{3} \right ]$$

Sol: Based on the formulae given in Principles of Mathematical Induction

The given statement is:

P(n): $$1.2 + 2.3 + 3.4 + … + n.\left ( n + 1 \right ) = \left [ \frac{n\left ( n + 1 \right )\left ( n + 2 \right )}{3} \right ]$$

Now, for n = 1:

=$$\frac{1\left ( 1 + 1 \right )\left ( 1 + 2 \right )}{3}$$= $$\frac{1.2.3}{3}$$= 2

Thus, the P(n) is true for n=1.

Let, P(k) be true, where k is a positive integer.

$$1.2 + 2.3 + 3.4 + … + k.\left ( k + 1 \right ) = \left [ \frac{k\left ( k + 1 \right )\left ( k + 2 \right )}{3} \right ]$$ . . . . . . . . . (1)

Now, we will prove P(k + 1) is also true:

=$$\\\left [1.2 + 2.3 + 3.4 + … + k.\left ( k + 1 \right ) \right ] + \left ( k + 1 \right )\left ( k + 2 \right )\\$$

=$$\\\frac{k\left ( k + 1 \right )\left ( k + 2 \right )}{3} + \left ( k + 1 \right )\left ( k + 2 \right )\\$$

Now, by using equation (1):

=$$\\\left ( k + 1 \right )\left ( k + 2 \right )\left ( \frac{k}{3} + 1 \right )\\$$

=$$\\\frac{\left ( k + 1 \right )\left ( k + 2 \right )\left ( k + 3 \right )}{3}\\$$

=$$\\\frac{\left ( k + 1 \right )\left ( k + 1 + 1 \right )\left ( k + 1 + 2 \right )}{3}\\$$

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

7: $$1.3 + 3.5 + 5.7 + … + \left ( 2n – 1 \right )\left ( 2n + 1 \right ) = \frac{n\left ( 4n^{2} + 6n – 1 \right )}{3}$$

Sol: Based on the formulae given in Principles of Mathematical Induction

The given statement is:

$$1.3 + 3.5 + 5.7 + … + \left ( 2n – 1 \right )\left ( 2n + 1 \right ) = \frac{n\left ( 4n^{2} + 6n – 1 \right )}{3}\\$$

Now, for n = 1:

$$\\\frac{1\left ( 4.1^{2} + 6.1 – 1 \right )}{3}$$= $$\frac{4 + 6 – 1}{3}$$= $$\frac{9}{3}$$ = 3

Thus, the P(n) is true for n = 1

Let, P(k) be true, where k is a positive integer.

$$1.3 + 3.5 + 5.7 + … + \left ( 2k – 1 \right )\left ( 2k + 1 \right ) = \frac{k\left ( 4k^{2} + 6k – 1 \right )}{3}$$ . . . . . . . . . (1)

Now we will prove P(k + 1) is also true:

=$$\\1.3 + 3.5 + 5.7 + … + \left ( 2k – 1 \right )\left ( 2k + 1 \right ) + \left \{ 2\left ( k + 1 \right ) – 1 \right \}\left \{ 2\left ( k + 1 \right ) + 1\right \}\\$$

=$$\\\frac{k\left ( 4k^{2} + 6k – 1\right )}{3} + \left ( 2k + 2 – 1 \right )\left ( 2k + 2 + 1 \right )\\$$

Now, by using equation (1):

=$$\\\frac{k\left ( 4k^{2} + 6k – 1\right )}{3} + \left ( 2k + 1 \right )\left ( 2k + 3 \right )\\$$

=$$\\\frac{k\left ( 4k^{2} + 6k – 1\right )}{3} + \left ( 4k^{2} + 8k + 3 \right )\\$$

=$$\\\frac{k\left ( 4k^{2} + 6k – 1\right )+ 3\left ( 4k^{2} + 8k + 3 \right )}{3}\\$$

=$$\\\frac{4k^{3} + 6k^{2} – k + 12k^{2} + 24k + 9}{3}\\$$

=$$\\\frac{4k^{3} + 18k^{2} + 23k + 9}{3}\\$$

=$$\\\frac{4k^{3} + 14k^{2} + 9k + 4k^{2} + 14k + 9}{3}\\$$

=$$\\\frac{k\left (4k^{2} + 14k + 9 \right )+ 1\left (4k^{2} + 14k + 9 \right )}{3}\\$$

=$$\\\frac{\left (k + 1 \right )\left (4k^{2} + 14k + 9 \right )}{3}\\$$

=$$\\\frac{\left (k + 1 \right )\left (4k^{2} + 8k + 4 + 6k + 6 – 1 \right )}{3}\\$$

=$$\\\frac{\left ( k + 1 \right )\left \{ 4\left ( k^{2} + 2k + 1\right ) + 6\left ( k + 1 \right ) – 1\right \}}{3}\\$$

=$$\\\frac{\left ( k + 1 \right )\left \{ 4\left ( k + 1\right )^{2} + 6\left ( k + 1 \right ) – 1\right \}}{3}\\$$

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

8: $$1.2 + 2.2^{2} + 3.2^{2} + … + n.2^{n} = \left ( n – 1 \right )2^{n + 1} + 2$$

Sol: Based on the formulae given in Principles of Mathematical Induction

The given statement is:

P(n): $$1.2 + 2.2^{2} + 3.2^{2} + … + n.2^{n} = \left ( n – 1 \right )2^{n + 1} + 2$$

Now, for n = 1:

=$$\left ( 1 – 1 \right )2^{1 + 1} + 2$$ = 0 + 2= 2

Thus, the P(n) is true for n=1.

Let P(k) be true, where k is a positive integer:

$$\\1.2 + 2.2^{2} + 3.2^{2} + … + k.2^{k} = \left ( k – 1 \right )2^{k + 1} + 2$$ . . . . . . . . (1)

Now we will prove P(k + 1) is also true:

=$$[ 1.2 + 2.2^{2} + 3.2^{2} + … + k.2^{k} ] + ( k + 1 ). 2^{k + 1}\\$$

=$$\\\left ( k – 1 \right )2^{k + 1} + 2 + \left ( k + 1 \right ).2^{k + 1}\\$$

=$$\\2^{k + 1}\left \{ \left ( k – 1 \right ) + \left ( k + 1 \right ) \right \} + 2\\$$

=$$\\2^{k + 1}.2k + 2\\$$

=$$\\k.2^{\left ( k + 1 \right ) + 1} + 2\\$$

=$$\\\left \{ \left ( k + 1 \right ) – 1 \right \}2^{\left ( k + 1 \right ) + 1} + 2\\$$

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

9: $$\frac{1}{2} + \frac{1}{4} +\frac{1}{8} + … + \frac{1}{2^{n}} = 1 – \frac{1}{2^{n}}$$

Sol: Based on the formulae given in Principles of Mathematical Induction

The given statement is:

$$\frac{1}{2} + \frac{1}{4} +\frac{1}{8} + … + \frac{1}{2^{n}} = 1 – \frac{1}{2^{n}}\\$$

Now, for n = 1:

=$$1 – \frac{1}{2^{1}}$$= $$\frac{1}{2}$$

Thus, the P (n) is true for n = 1.

Let, P (k) be true, where k is a positive integer:

$$\frac{1}{2} + \frac{1}{4} +\frac{1}{8} + … + \frac{1}{2^{k}} = 1 – \frac{1}{2^{k}}$$ . . . . . . (1)

Now, we will prove P (k + 1) is also true:

=$$\\\left (\frac{1}{2} + \frac{1}{4} +\frac{1}{8} + … + \frac{1}{2^{k}} \right ) + \frac{1}{2^{k + 1}}\\$$

=$$\\\left (1 – \frac{1}{2^{k}} \right ) + \frac{1}{2^{k + 1}}\\$$

Now, by using equation (1):

=$$\\1 – \frac{1}{2^{k}} + \frac{1}{2^{1}.2^{k}}\\$$

=$$\\1 – \frac{1}{2^{k}}\left (1 – \frac{1}{2} \right )\\$$

=$$\\1 – \frac{1}{2^{k}}\left ( \frac{1}{2} \right )\\$$

=$$\\1 – \frac{1}{2^{k + 1}}\\$$

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

10: Provide a proof for the following with the help of mathematical induction principle for all values of n, where it is a natural number.

$$\frac{1}{2\cdot 5} + \frac{1}{5\cdot 8} + \frac{1}{8\cdot 11} +…… + \frac{1}{(3n-1)(3n+2)} = \frac{n}{(6n+4)}$$

Solution: Based on the formulae given in Principles of Mathematical Induction

The given Statement is:
Q(n): $$\frac{1}{2\cdot 5} + \frac{1}{5\cdot 8} + \frac{1}{8\cdot 11} +…… + \frac{1}{(3n-1)(3n+2)} = \frac{n}{(6n+4)}$$

Now, for n = 1

$$\\Q (1) = \frac{1}{2\cdot 5} = \frac{1}{(6(1)+4)} = \frac{1}{2\cdot 5}\\$$

Thus, Q(1) proves to be true.

Let’s assume Q(p) is true, where p is a natural number.

Q(p):

= $$\frac{1}{2\cdot 5} + \frac{1}{5\cdot 8} + \frac{1}{8\cdot 11} +…… + \frac{1}{(3p-1)(3p+2)} = \frac{p}{(6p+4)}$$ . . . . . . . . . . . . . . (1)

Now, we have to prove that Q(p+1) is also true.

Since, Q (p) is true, we have:

Q (p+1):

= $$\\\frac{1}{2\cdot 5} + \frac{1}{5\cdot 8} + \frac{1}{8\cdot 11} +…… + \frac{1}{(3p-1)(3p+2)} + \frac{1}{[3(p + 1)-1][3(p + 1)+2]}\\$$

Now, by using equation (1):

=$$\\\frac{p}{(6p + 4)} + \frac{1}{[3p + 3 – 1]\;[3p + 3 + 2]}\\$$

=$$\\\frac{p}{(6p + 4)} + \frac{1}{(3p + 2)\;(3p + 3 + 2)}\\$$

=$$\\\frac{p}{2(3p + 2)} + \frac{1}{[3p + 3 – 1]\;[3p + 3 + 2]}\\$$

=$$\\\frac{p}{2(3p + 2)} + \frac{1}{[3p + 3 – 1]\;[3p + 3 + 2]}\\$$

=$$\\\frac{1}{2(3p + 2)}\left [ \frac{p}{2} +\frac{1}{(3p + 5)} \right ]\\$$

=$$\\\frac{1}{2(3p + 2)}\left [\frac{p(3p + 5) + 2}{(3p + 5)} \right ]\\$$

=$$\\\frac{1}{2(3p + 2)}\left [\frac{3p^{2} + 5p + 2}{(3p + 5)} \right ]\\$$

=$$\\\frac{1}{2(3p + 2)}\left [\frac{(3p + 2)(p + 1)}{(3p + 5)} \right ]\\$$

=$$\\\frac{p + 1}{6p + 10}\\$$

=$$\\\frac{p + 1}{6(p + 1) + 4}\\$$

Thus, whenever Q (p) proves to be true, Q (p+1) subsequently proves to be true.

Therefore, with the help of induction principle it can be proved that the statement Q(n) is true for all the natural numbers.

11: Provide a proof for the following with the help of mathematical induction principle for all values of n, where it is a natural number.

$$\frac{1}{1\cdot 2\cdot 3} + \frac{1}{2\cdot 3\cdot 4} + \frac{1}{3\cdot 4\cdot 5} + …….+ \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)}$$

Solution: Based on the formulae given in Principles of Mathematical Induction

The given statement is:

Q(n): $$\frac{1}{1\cdot 2\cdot 3} + \frac{1}{2\cdot 3\cdot 4} + \frac{1}{3\cdot 4\cdot 5} + …….+ \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)}$$

Now, for n = 1:

$$Q(1) = \frac{1}{1\cdot 2\cdot 3} = \frac{(1)((1)+3)}{4((1)+1)((1)+2)} = \frac{1}{1\cdot 2\cdot 3}\\$$

Thus, Q(1) proves to be true.

Let’s assume Q(p) is true, where p is a natural number:

Q (p):

=$$\frac{1}{1\cdot 2\cdot 3} + \frac{1}{2\cdot 3\cdot 4} + \frac{1}{3\cdot \cdot 4\cdot 5} + …….+ \frac{1}{p(p + 1)(p + 2)} = \frac{p(p + 3)}{4(p + 1)(p+2)}$$. . . . . . . . . . . (1)

Now, we have to prove that Q (p+1) is also true.

Since, Q (p) is true, we have:

Q (p + 1):

=$$\\\frac{p(p + 3)}{4(p + 1)(p + 2)} + \frac{1}{(p + 1)(p + 2)(p + 3)}\\$$

Now, using equation (1):

=$$\frac{1}{(p + 1)(p + 2)}\left [ \frac{p(p + 3)}{4} + \frac{1}{(p + 3)} \right ]\\$$

=$$\\\frac{1}{(p + 1)(p + 2)}\left [ \frac{p(p + 3)^{2} + 4}{4(p + 3)} \right ]\\$$

=$$\\\frac{1}{(p + 1)(p + 2)}\left [ \frac{p(p^{2} + 6p + 9) + 4}{4(p + 3)} \right ]\\$$

=$$\\\frac{1}{(p + 1)(p + 2)}\left [ \frac{p(p^{2} + 6p + 9) + 4}{4(p + 3)} \right ]\\$$

=$$\\\frac{1}{(p + 1)(p + 2)}\left [ \frac{p^{3} + 6p^{2} + 9p + 4}{4(p + 3)} \right ]\\$$

=$$\frac{1}{(p + 1)(p + 2)}\left [ \frac{p(p^{2} + 2p + 1) + 4(p^{2} + 2p + 1)}{4(p + 3)} \right ]\\$$

=$$\\\frac{1}{(p + 1)(p + 2)}\left [ \frac{p(p + 1)^{2} + 4(p + 1)^{2}}{4(p + 3)} \right ]\\$$

=$$\\\frac{1}{(p + 1)(p + 2)}\left [ \frac{(p + 1)^{2}(p + 4)}{4(p + 3)} \right ]\\$$

=$$\\\frac{(p + 1)[(p + 1) + 3]}{4[(p + 1) + 1][(p + 1)+ 2]}\\$$

Thus, whenever Q(p) proves to be true, Q(p+1) subsequently proves to be true.

Therefore, with the help of induction principle it can be proved that the statement Q(n) is true for all the natural numbers.

12: Provide a proof for the following with the help of mathematical induction principle for all values of n, where it is a natural number.

$$a + ar + ar^{2} + ……. + ar^{n-1} = \frac{a(r^{n}-1)}{r-1}$$

Solution: Based on the formulae given in Principles of Mathematical Induction

The given statement is:

Q(n): $$a + ar + ar^{2} + ……. + ar^{n-1} = \frac{a(r^{n}-1)}{r-1}$$

Now, for n = 1

$$Q(1) = a = \frac{a(r^{(1)}-1)}{r-1} = a\\$$

Thus, Q (1) proves to be true.

Let’s assume Q (p) is true, where p is a natural number.

Q (p) = $$a + ar + ar^{2} + ……. + ar^{p-1} = \frac{a(r^{p}-1)}{r-1}$$ . . . . . . . . (1)

Now, we have to prove that Q (p+1) is also true.

Since, Q (p) is true, we have:

Q(p+1) = $$a + ar + ar^{2} + ……. + ar^{p-1} + ar^{(p + 1)-1}$$

Now, using Equation (1):

=$$\\\frac{a(r^{p} – 1)}{r – 1} + ar^{p}\\$$

=$$\\\frac{a(r^{p} – 1) + ar^{p}(r – 1)}{r – 1}\\$$

=$$\\\frac{a(r^{p} – 1) + ar^{p}(r – 1) – ar^{p}}{r – 1}\\$$

=$$\\\frac{ar^{p} – a + ar^{p + 1} – ar^{p}}{r – 1}\\$$

=$$\\\frac{ar^{p + 1} – a}{r – 1}\\$$

=$$\\\frac{a(r^{p + 1} – 1)}{r – 1}\\$$

Thus, whenever Q(p) proves to be true, Q(p+1) subsequently proves to be true.

Therefore, with the help of induction principle it can be proved that the statement Q(n) is true for all the natural numbers.

13: Provide a proof for the following with the help of mathematical induction principle for all values of n, where it is a natural number.

$$\left ( 1\, +\, \frac{3}{1} \right )\left ( 1\, +\, \frac{5}{4} \right )\left ( 1\, +\, \frac{7}{9} \right )……. \left ( 1\, +\, \frac{(2n+1)}{ n^{2}} \right ) = \left ( n+1 \right )^{2}$$

Solution: Based on the formulae given in Principles of Mathematical Induction

The given statement is:

Q(n): $$\left ( 1\, +\, \frac{3}{1} \right )\left ( 1\, +\, \frac{5}{4} \right )\left ( 1\, +\, \frac{7}{9} \right )……. \left ( 1\, +\, \frac{(2n+1)}{ n^{2}} \right ) = \left ( n+1 \right )^{2}$$

Now, for n = 1:

$$Q(1) = \left ( 1\, +\, \frac{3}{1} \right ) = 4\\$$ $$\\\Rightarrow \left ( n+1 \right )^{2} = (1 + 1)^{2} = 4\\$$

Thus, Q (1) proves to be true.

Let’s assume Q (p) is true, where p is a natural number.

Q (p):

=$$\left ( 1\, +\, \frac{3}{1} \right )\left ( 1\, +\, \frac{5}{4} \right )\left ( 1\, +\, \frac{7}{9} \right )……. \left ( 1\, +\, \frac{(2p+1)}{ p^{2}} \right ) = \left ( p+1 \right )^{2}$$ . . . . . . . . . . . (1)

Now we have to prove that Q(p+1) is also true.

Since Q (p) is true, we have:

Q(p+1):

=$$\left ( 1\, +\, \frac{3}{1} \right )\left ( 1\, +\, \frac{5}{4} \right )\left ( 1\, +\, \frac{7}{9} \right )……. \left ( 1\, +\, \frac{(2p+1)}{ p^{2}} \right )\left ( 1\, +\, \frac{(2(p+1) + 1}{ (p + 1)^{2}} \right )\\$$

Now, using equation (1):

=$$\\(p + 1)^{2}\left ( 1\, +\, \frac{(2(p+1) + 1}{ (p + 1)^{2}} \right )\\$$

=$$\\(p + 1)^{2}\left ( \frac{(p+1)^{2} + 2(p+1) + 1}{ (p + 1)^{2}} \right )\\$$

$$\\\Rightarrow$$ $$(p+1)^{2} + 2(p+1) + 1= \left [ (p + 1) + 1 \right ]^{2}\\$$
Thus, whenever Q(p) proves to be true, Q(p+1) subsequently proves to be true.

Therefore, with the help of induction principle it can be proved that the statement Q(n) is true for all the natural numbers.

14: Provide a proof for the following through principle of mathematical induction for all values of n, where n is a natural number.

$$\left ( 1\, +\, \frac{1}{1} \right )\left ( 1\, +\, \frac{1}{2} \right )\left ( 1\, +\, \frac{1}{3} \right )…….\left ( 1\, +\, \frac{1}{n} \right ) = (n+1)$$

Solution: Based on the formulae given in Principles of Mathematical Induction

The given statement is:

Q(n): $$\left ( 1\, +\, \frac{1}{1} \right )\left ( 1\, +\, \frac{1}{2} \right )\left ( 1\, +\, \frac{1}{3} \right )…….\left ( 1\, +\, \frac{1}{n} \right ) = (n+1)$$

Now, for n = 1:

$$Q(1) = \left ( 1\, +\, \frac{1}{1} \right ) = 2 \\ \Rightarrow (n+1) = 1 + 1 = 2\\$$

Thus, Q(1) proves to be true.

Let’s assume Q(p) is true, where p is a natural number.

Q(p):

=$$\left ( 1\, +\, \frac{1}{1} \right )\left ( 1\, +\, \frac{1}{2} \right )\left ( 1\, +\, \frac{1}{3} \right )…….\left ( 1\, +\, \frac{1}{p} \right ) = (p+1)$$. . . . .(1)

Now, we have to prove that Q (p+1) is also true.

Since, Q (p) is true, we have:

Q (p+1):

=$$\\\left ( 1\, +\, \frac{1}{1} \right )\left ( 1\, +\, \frac{1}{2} \right )\left ( 1\, +\, \frac{1}{3} \right )…….\left ( 1\, +\, \frac{1}{p} \right )\left ( 1\, +\, \frac{1}{p + 1} \right )\\$$

Now, using Equation (1):

=$$\\(p + 1)\left ( 1\, +\, \frac{1}{p + 1} \right )$$$$= (p + 1)\left (\frac{(p + 1) + 1}{p + 1} \right )\\$$

= (p + 1) + 1

Thus, whenever Q(p) proves to be true, Q(p+1) subsequently proves to be true.

Therefore, with the help of induction principle it can be proved that the statement Q(n) is true for all the natural numbers.

15: Provide a proof for the following through principle of mathematical induction for all values of n, where n is a natural number.

$$1^{2} + 3^{2} + 5^{2} +. …..+ (2n-1)^{2} = \frac{n(2n-1)(2n+1)}{3}$$

Solution : Based on the formulae given in Principles of Mathematical Induction

The given statement is:

Q(n): $$1^{2} + 3^{2} + 5^{2} +. …..+ (2n-1)^{2} = \frac{n(2n-1)(2n+1)}{3}$$

Now, for n = 1:

$$Q(1) = 1^{2} = 1\;and\; \frac{(1)(2(1)-1)(2(1)+1)}{3} = \frac{3}{3} = 1\\$$

Thus, Q(1) proves to be true.

Let’s assume Q(p) is true, where p is a natural number.

Q (p):

=$$1^{2} + 3^{2} + 5^{2} +. …..+ (2p-1)^{2} = \frac{p(2p-1)(2p+1)}{3}$$ . . . . . (1)

Now, we have to prove that Q(p+1) is also true.

Since, Q(p) is true, we have:

Q (p+1):

=$$\\1^{2} + 2^{2} + 3^{2} +……+ (2k – 1)^{2}] + (2\;(p + 1) – 1)^{2}\\$$

=$$\\\frac{p(2p – 1)(2p + 1)}{3} + (2p + 2 -1)^{2}\\$$

=$$\\\frac{p(2p – 1)(2p + 1)}{3} + (2p + 1)^{2}\\$$

=$$\\\frac{p(2p – 1)(2p + 1) + 3(2p + 1)^{2}}{3}\\$$

=$$\\\frac{(2p + 1)(p(2p – 1) + 3(2p + 1))}{3}\\$$

=$$\\\frac{(2p + 1)(2p^{2} – p + 6p + 3)}{3}\\$$

=$$\\\frac{(2p + 1)(2p^{2} + 5p + 3)}{3}\\$$

=$$\\\frac{(2p + 1)(2p^{2} + 2p + 3p + 3)}{3}\\$$

=$$\\\frac{(2p + 1)(2p(p + 1) + 3(p + 1))}{3}\\$$

=$$\\\frac{(2p + 1)(p + 1)(2p + 3)}{3}\\$$

=$$\\\frac{(p + 1)[2(p + 1) – 1][2(p + 1) + 1]}{3}\\$$

Thus, whenever Q(p) proves to be true, Q(p+1) subsequently proves to be true.

Therefore, with the help of induction principle it can be proved that the statement Q(n) is true for all the natural numbers.

16: Provide a proof for the following through principle of mathematical induction for all values of n, where n is a natural number.

$$\frac{1}{1\cdot 4} + \frac{1}{4\cdot 7} + \frac{1}{7\cdot 10} + …… + \frac{1}{(3n-2)(3n+1} = \frac{n}{3n+1}$$

Solution: Based on the formulae given in Principles of Mathematical Induction

The given statement is:

Q(n): $$\frac{1}{1\cdot 4} + \frac{1}{4\cdot 7} + \frac{1}{7\cdot 10} + …… + \frac{1}{(3n-2)(3n+1} = \frac{n}{3n+1}$$

Now, for n = 1:

$$Q(1) = \frac{1}{1.4} = \frac{1}{(3(1)-2)(3(1)+1)} = \frac{1}{1.4}\\$$

Thus, Q(1) proves to be true.

Let’s assume Q(p) is true, where p is a natural number.

Q(p):

=$$\frac{1}{1\cdot 4} + \frac{1}{4\cdot 7} + \frac{1}{7\cdot 10} + …… + \frac{1}{(3p – 2)(3p + 1)} = \frac{p}{3p + 1}$$ . . . . . . . . . . . . (1)

Now, we have to prove that Q(p+1) is also true.

Since, Q (p) is true, we have:

Q (p+1):

=$$\frac{1}{1\cdot 4} + \frac{1}{4\cdot 7} + \frac{1}{7\cdot 10} +…+ \frac{1}{(3p – 2)(3p + 1)} + \frac{1}{[3(p + 1) – 2][3(p + 1)+ 1]}$$

Now, using Equation (1):

=$$\\\frac{1}{(3p + 1)} + \frac{1}{(3p + 1)(3p + 4)]}\\$$

=$$\\\frac{1}{(3p + 1)}\left [ p + \frac{1}{(3p + 4)} \right ]\\$$

=$$\\\frac{1}{(3p + 1)}\left [ \frac{p(3p + 4) + 1}{(3p + 4)} \right ]\\$$

=$$\\\frac{1}{(3p + 1)}\left [ \frac{3p^{2} + 4p + 1}{(3p + 4)} \right ]\\$$

=$$\\\frac{1}{(3p + 1)}\left [ \frac{3p^{2} + 3p + p + 1}{(3p + 4)} \right ]\\$$

=$$\\\frac{1}{(3p + 1)}\left [ \frac{(3p + 1)(p + 1)}{(3p + 4)} \right ]\\$$

=$$\\\frac{(p + 1)}{3(p + 1) + 1}\\$$

Thus, whenever Q(p) proves to be true, Q(p+1) subsequently proves to be true.

Therefore, with the help of induction principle it can be proved that the statement Q(n) is true for all the natural numbers.

17: Provide a proof for the following through principle of mathematical induction for all values of n, where n is a natural number.

$$\frac{1}{3\cdot 5} + \frac{1}{5\cdot 7} + \frac{1}{7\cdot 9} + ……. + \frac{1}{(2n+1)(2n+3)} = \frac{n}{3(2n+3)}$$

Solution: Based on the formulae given in Principles of Mathematical Induction

The given statement is:

Q(n): $$\frac{1}{3\cdot 5} + \frac{1}{5\cdot 7} + \frac{1}{7\cdot 9} + ……. + \frac{1}{(2n+1)(2n+3)} = \frac{n}{3(2n+3)}$$

Now, for n = 1:

$$Q(1) = \frac{1}{3\cdot 5} = \frac{1}{3(2(1)+3)} = \frac{1}{3\cdot 5}\\$$

Thus, Q(1) proves to be true:

Let us assume that Q(p) is true, where p is a natural number.

Therefore, Q(p):

=$$\frac{1}{3\cdot 5} + \frac{1}{5\cdot 7} + \frac{1}{7\cdot 9} + ……. + \frac{1}{(2p + 1)(2p + 3)} = \frac{p}{3(2p + 3)}$$ . . . . . . . (1)

Now, we have to prove that Q (p+1) is also true.

Since, Q(p) is true, we have:

Q(p+1):

=$$\\\frac{1}{3\cdot 5} + \frac{1}{5\cdot 7} + \frac{1}{7\cdot 9} + ……. + \frac{1}{(2p + 1)(2p + 3)} + \frac{1}{[2(p + 1) + 1][2(p + 1) + 3]}\\$$

Now, Using Equation (1):

=$$\\\frac{p}{3(2p + 3)} + \frac{1}{(2p + 3)(2p + 5)}\\$$

=$$\\\frac{1}{3(2p + 3)}\left [ \frac{p}{3} + \frac{1}{(2p + 5)} \right ]\\$$

=$$\\\frac{1}{3(2p + 3)}\left [\frac{p(2p + 5) + 3}{(2p + 5)} \right ]\\$$

=$$\\\frac{1}{3(2p + 3)}\left [\frac{2p^{2} + 5p + 3}{(2p + 5)} \right ]\\$$

=$$\\\frac{1}{3(2p + 3)}\left [\frac{2p^{2} + 2p + 3p + 3}{(2p + 5)} \right ]\\$$

=$$\\\frac{1}{3(2p + 3)}\left [\frac{2p(p + 1) + 3(p + 1)}{(2p + 5)} \right ]\\$$

=$$\\\frac{1}{3(2p + 3)}\left [\frac{(2p + 3)(p + 1)}{(2p + 5)} \right ]\\$$

=$$\\\frac{p + 1}{3[2(p + 1) + 3]}\\$$

Thus, whenever Q(p) proves to be true, Q(p+1) subsequently proves to be true.

Therefore, with the help of induction principle it can be proved that the statement Q(n) is true for all the natural numbers.

18: Provide a proof for the following through principle of mathematical induction for all values of n, where n is a natural number.

$$1 + 2 + 3 + …… + n < \frac{1}{8}(2n+1)^{2}$$

Solution: Based on the formulae given in Principles of Mathematical Induction

The given statement is:

Q(n): $$1 + 2 + 3 + …… + n < \frac{1}{8}(2n+1)^{2}$$

Now, for n = 1:

$$Q(1) = 1 = \frac{1}{8}(2(1)+1)^{2} = \frac{9}{8} \\ \Rightarrow 1 < \frac{9}{8}\\$$

Thus, Q(1) proves to be true.

Let’s assume Q(p) is true, where p is a natural number.

Q(p) = $$1 + 2 + 3 + …… + p < \frac{1}{8}(2p+1)^{2}$$ . . . . . . (1)

Now, we have to prove that Q(p+1) is also true.

Since, Q (p) is true, we have:

Now, Using Equation (1):

=$$(1 + 2 + 3 + …… + p) + (p + 1) < \frac{1}{8}(2p+1)^{2} + (p + 1)\\$$

<$$\\\frac{1}{8}[(2p + 1)^{2} + 8(p + 1)]\\$$

<$$\\\frac{1}{8}[4p^{2} + 4p + 1 + 8p + 8]\\$$

<$$\\\frac{1}{8}[4p^{2} + 12p + 9]\\$$

<$$\\\frac{1}{8}[(2p +3)^{2}]\\$$

<$$\\\frac{1}{8}[(2(p + 1) + 1)^{2}]\\$$

Thus, $$\\(1 + 2 + 3 + …… + p) + (p + 1) < \frac{1}{8}(2p+1)^{2} + (p + 1)$$

Thus, whenever Q (p) proves to be true, Q (p+1) subsequently proves to be true.

Therefore, with the help of induction principle it can be proved that the statement Q (n) is true for all the natural numbers.

19: n (n + 1)(n + 5) is a multiple of 3.

Sol:

The given statement is:

P(n): n (n + 1) (n + 5) is a multiple of 3

Now, for = 1:

= 1(1 + 1)(1 + 5) = 12

Thus, the P(n) is true for n = 1.

Let, P(k) be true, where k is a positive integer.

k(k + 1)(k + 5) is a multiple of 3

Therefore, k(k + 1)(k + 5) = 3m, where m $$\in$$ N . . . . . . . . (1)

Now, we will prove P(k + 1) is also true.

=$$\\\left ( k + 1 \right )\left \{ \left ( k + 1 \right ) + 1 \right \}\left \{ \left ( k + 1 \right ) + 5\right \}$$

= $$\\\left ( k + 1 \right )\left ( k + 2 \right )\left \{ \left ( k + 5 \right ) + 1\right \}$$

= $$\\\left ( k + 1 \right )\left ( k + 2 \right )\left ( k + 5 \right ) + \left ( k + 1 \right )\left ( k + 2 \right )$$

= $$\\\left \{ k \left ( k + 1 \right )\left ( k + 5 \right ) + 2\left ( k + 1 \right )\left ( k + 5 \right ) \right \} + (k + 1)(k + 2)$$

= $$\\3m + (k + 1)\left \{ 2(k + 5) + (k + 2) \right \}$$

= $$\\3m + (k + 1)\left \{ 2k + 10 + k + 2\right \}$$

= $$\\3m + (k + 1)\left \{ 3k + 12 \right \}$$

= $$\\3m + 3(k + 1)(k + 4)$$

= $$\\3\left \{ m + (k + 1)(k + 4) \right \}$$

= $$\\3 \times q$$, where q = $$\left \{ m + (k + 1)(k + 4) \right \}$$ $$\in$$ N.

Therefore, $$\left ( k + 1 \right )\left \{ \left ( k + 1 \right ) + 1 \right \}\left \{ \left ( k + 1 \right ) + 5\right \}$$ is a multiple of 3.

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

20: $$10^{2n – 1} + 1$$ is divisible by 11.

Sol: Based on the formulae given in Principles of Mathematical Induction

The given statement is:

$$10^{2n – 1} + 1$$ is divisible by 11

Now, for n = 1

=$$10^{2.1 – 1} + 1$$ = 11

Thus we get, the P(n) is true for n = 1.

Then, P(k) is also true, where k is a positive integer.

$$10^{2k – 1} + 1$$ is divisible by 11.

$$10^{2k – 1} + 1$$ = 11m, where m $$\in$$ N . . . . . . . . . (1)

Now, we will prove P(k + 1) is also true:

=$$\\10^{2\left ( k + 1 \right )- 1} + 1$$

=$$\\10^{2k + 2 – 1} + 1$$

=$$\\10^{2k + 1} + 1$$

=$$\\10^{2}\left ( 10^{2k – 1} + 1 – 1 \right ) + 1$$

=$$\\10^{2}\left ( 10^{2k – 1} + 1 \right ) – 10^{2} + 1$$

=$$\\10^{2}. 11m – 100 + 1$$ [using equation (1)]

=$$\\100 \times 11m – 99$$

=$$\\11\left ( 100m – 9 \right )$$

= 11 r, where r = $$\left (100m – 9 \right )$$ is some natural number.

Therefore, $$10^{2\left ( k + 1 \right )- 1} + 1$$ is divisible by 11.

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

21. $$x^{2n} – y^{2n}$$ is divisible by x + y.

Sol: Based on the formulae given in Principles of Mathematical Induction

The given statement is:

P(n): $$x^{2n} – y^{2n}$$ is divisible by x + y.

Now, for n = 1

= $$x^{2 \times 1} – y^{2 \times 1}$$= $$x^{2} – y^{2}$$= (x + y) (x – y)

Therefore we conclude, it is divisible by (x + y).

Thus, the P(n) is true for n=1.

Let, P(k) is also true, where k is a positive integer.

$$x^{2k} – y^{2k}$$ is divisible by (x + y).

Let $$x^{2k} – y^{2k}$$ = m (x + y), where m $$\in$$ N . . . . (1)

Now from data, we will prove P(k + 1) is also true:

=$$\\x^{2\left ( k + 1 \right ) } – y^{2\left ( k + 1 \right )}$$

=$$\\x^{2k}.x^{2} – y^{2k}.y^{2}$$

=$$\\x^{2}\left ( x^{2k} – y^{2k} + y^{2k}\right ) – y^{2k}. y^{2}$$

=$$\\x^{2}\left \{ m \left ( x + y \right ) + y^{2k}\right \} – y^{2k}.y^{2}$$ [using equation (1)]

=$$\\m \left ( x + y \right )x^{2} + y^{2k}.x^{2} – y^{2k}.y^{2}$$

=$$\\m \left ( x + y \right )x^{2} + y^{2k}\left (x^{2} – y^{2} \right )$$

=$$\\m \left ( x + y \right )x^{2} + y^{2k}\left (x + y \right )\left ( x – y \right )$$

=$$\\\left ( x + y \right )\left \{ mx^{2} + y^{2k}\left ( x – y \right ) \right \}$$, which is the factor of (x + y).

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

22: $$3^{2n + 2} – 8n – 9$$ is divisible by 8.

Sol: Based on the formulae given in Principles of Mathematical Induction

The given statement is:

P(n): $$3^{2n + 2} – 8n – 9$$ is divisible by 8.

Now, for n = 1:

=$$3^{2.1 + 2} – 8.1 – 9$$= $$3^{4} – 17$$= 64

Therefore, it is divisible by 8.

Thus, the P(n) is true for n = 1.

Let, P(k) be true, where k is a positive integer.

$$3^{2k + 2} – 8k – 9$$ is divisible by 8.

$$3^{2k + 2} – 8k – 9 = 8m$$, where m $$\in$$ N. . . . . . (1)

Now from data, we will prove P(k + 1) is also true:

=$$\\3^{2\left (k + 1 \right ) + 2} – 8\left ( k + 1 \right ) – 9$$

=$$\\3^{2k + 2}.3^{2} – 8k – 8 – 9$$

=$$\\3^{2} \left ( 3^{2k + 2} – 8k – 9 + 8k + 9 \right ) – 8k – 17$$

=$$\\3^{2} \left ( 3^{2k + 2} – 8k – 9 \right ) + 3^{2}\left ( 8k + 9 \right ) – 8k – 17$$

=$$\\9.8m + 9\left ( 8k + 9 \right ) – 8k – 17$$

=$$\\9.8m + 72k + 81 – 8k – 17$$

=$$\\9.8m + 64k + 64$$

=$$\\8\left (9m + 8k + 8 \right )$$

= 8r, where r = $$\left (9m + 8k + 8 \right )$$ is a - natural number.

Therefore, $$3^{2\left (k + 1 \right ) + 2} – 8\left ( k + 1 \right ) – 9$$ is divisible by 8.

Thus we have, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

23: $$41^{n} – 14^{n}$$ is a multiple of 27.

Sol: Based on the formulae given in Principles of Mathematical Induction

The given statement is:

P(n): $$41^{n} – 14^{n}$$ is a multiple of 27

Now, for n = 1:

=$$41^{1} – 14^{1}$$= 27, which is the multiple of 27

Thus we get, the P(n) is true for n=1.

Lets assume, P(k) be true, where k is a positive integer.

$$41^{k} – 14^{k}$$ is a multiple of 27.

$$41^{k} – 14^{k}$$ = 27m , where m $$\in$$ N . . . . . . . . (1)

Now, we will prove P(k + 1) also is true:

=$$\\41^{k + 1} – 14^{k + 1}$$

=$$\\41^{k}.41 – 14^{k}.14$$

=$$\\41 \left (41^{k} – 14^{k} + 14^{k} \right ) – 14^{k}.14$$

=$$\\41 \left (41^{k} – 14^{k} \right ) + 41.14^{k} – 14^{k}.14$$

=$$\\41.27m + 14^{k} \left ( 41 – 14 \right )$$

=$$\\41.27m + 27.14^{k}$$

=$$\\27 \left (41m + 14^{k} \right )$$

=$$\\27 \times r$$, where r = $$\left (41m + 14^{k} \right )$$ is a natural number

Therefore, $$41^{k + 1} – 14^{k + 1}$$ is a multiple of 27.

Thus we have that, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore we conclude, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

24: $$\left ( 2n + 7 \right ) < \left ( n + 3 \right )^{2}$$

Sol: Based on the formulae given in Principles of Mathematical Induction

The given statement is:

P(n): $$\left ( 2n + 7 \right ) < \left ( n + 3 \right )^{2}$$

Now we have, for n = 1:

=$$\left ( 2.1 + 7 \right ) < \left ( 1 + 3 \right )^{2}$$= $$\left ( 9 \right ) < \left ( 4 \right )^{2}$$= $$9 < 16$$

Thus, the P(n) is true for n=1.

Let, P(k) be true, where k is a positive integer.

$$\left (2k + 7 \right ) < \left ( k + 3 \right )^{2}$$ . . . . . . . . (1)

We will prove P(k + 1) is also true:

=$$[ 2( k + 1) + 7 ]=[(2k + 7)+ 2]\\$$

=$$\\\left [ 2 \left ( k + 1 \right ) + 7 \right ]$$ = $$\left (2k + 7 \right ) + 2$$ < $$\left ( k + 3 \right )^{2} + 2\\$$ [using equation (1)]

=$$\\2\left ( k + 1 \right ) + 7 < k^{2} + 6k + 9 + 2\\$$

=$$\\2\left ( k + 1 \right ) + 7 < k^{2} + 6k + 11\\$$

Now,$$\\k^{2} + 6k + 11 < k^{2} + 8k + 16\\$$

$$\\2\left ( k + 1 \right ) + 7 < \left ( k + 4 \right )^{4}\\$$

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.