Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 4 Principle of Mathematical Induction Ex 4.1 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Maths Chapter 4 Principle of Mathematical Induction Ex 4.1

Prove the following by using the principle of mathematical induction for all n ∈ N?

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

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

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

4. 1.2.3 + 2.3.4 + 3.4.5 + … + n(n + 1)(n + 2)

= \(\frac{n(n + 1)(n + 2)(n + 3)}{4}\).

5. 1.3 + 2.3^{2} + 3.3^{2} + ……………. + n.3^{n} = \(\frac{(2 n-1) \cdot 3^{n+1}+3}{4}\)

6. 1.2 + 2.3 + 3.4 + ……………… + n(n + 1) = \(\frac{n(n + 1)(n + 2)}{3}\).

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

8. 1.2 + 2.2^{2} + 3.2^{2} + …………….. + n.2^{n} = (n – 1).2^{n+1} + 2.

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

10. \(\frac{1}{2.5}\) + \(\frac{1}{5.8}\) + \(\frac{1}{8.11}\) + ………………. + \(\frac{1}{(3n – 1)(3n + 2)}\) = \(\frac{n}{6n + 4}\).

11. \(\frac{1}{1.2.3}\) + \(\frac{1}{2.3.4}\) + \(\frac{1}{3.4.5}\) + …………….. + \(\frac{1}{n(n + 1)(n + 2)}\) = \(\frac{n(n + 3)}{4(n + 1)(n + 2)}\).

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

13. (1 + \(\frac{3}{1}\)) (1 + \(\frac{5}{4}\)) (1 + \(\frac{7}{9}\)) ……………. (1 + \(\frac{2 n+1}{n^{2}}\)) = (n + 1)^{2}

14. (1 + \(\frac{1}{1}\)) (1 + \(\frac{1}{2}\)) (1 + \(\frac{1}{3}\)) …………….. (1 + \(\frac{1}{n}\)) = n + 1.

15. 1^{2} + 3^{2} + 5^{2} + ………….. + (2n – 1)^{2} = \(\frac{n(2n – 1)(2n + 1)}{3}\).

16. \(\frac{1}{1.4}\) + \(\frac{1}{4.7}\) + \(\frac{1}{7.10}\) + …………….. + \(\frac{1}{(3n – 2)(3n + 1)}\) = \(\frac{n}{3n+1}\).

17. \(\frac{1}{3.5}\) + \(\frac{1}{5.7}\) + \(\frac{1}{7.9}\) + …………… + \(\frac{1}{(2n + 1)(2n + 3)}\) = \(\frac{n}{3(2n + 3)}\)

18. 1 + 2 + 3 + …………….. + n < \(\frac{1}{8}\) (2n + 1)^{2}.

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

20. 10^{2n-1} + 1 is divisible by 11.

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

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

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

24. (2n + 7) < (n + 3)^{2}.

Solutions to Questions 1 – 24:

1. Let P(n) be the given statement

i.e. P(n) : 1 + 3 + 3^{2} + ……………… + 3^{n-1} = \(\frac{3^{n}-1}{2}\).

Putting n = 1, P(1) = \(\frac{3 – 1}{2}\) = 1.

∴ P(n) is true for n = 1

Assume that P(k) is true.

So, P(k) : 1 + 3 + 3^{2} + ……………. + 3^{k-1} = \(\frac{3^{k}-1}{2}\)

We shall prove that P(k + 1) is true whenever P(k) is true.

Adding 3^{k} to both sides, we get

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

∴ P(k + 1) is also true whenever P(k) is true.

Hence, by principal of mathematical induction P(n) is true for all n ∈ N.

2. Let P(n) : 1^{3} + 2^{3} + 3^{3} + ……………. + n^{3} = \(\frac{n^{2}(n+1)^{2}}{4}\).

For n = 1, L.H.S. = 1^{3} = 1

and R.H.S. = \(\frac{1^{2}(1+1)^{2}}{4}\) = \(\frac{1×4}{4}\) = 1.

∴ L.H.S. = R.H.S., i.e; P(1) is true.

Let us suppose that P(k) is true,

∴ Putting n = k in (1), we have:

P(k) : 1^{3} + 2^{3} + 3^{3} + ……………… + k^{3} = \(\frac{k^{2}(k+1)^{2}}{4}\) …………….. (2)

Adding (k + 1)^{3} to both sides, we get

Thus, P(n) is true for n = k + 1, i.e., P(k + 1) is true.

∴ By Principle of Mathematical Induction, P(n) is true for all natural numbers n.

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

Putting n = 1, L.H.S. = 1, R.H.S. = \(\frac{2.1}{1+1}\) = \(\frac{2}{2}\) = 1.

L.H.S. = R.H.S. ∴ P(1) is true.

Let P(k) be true.

∴ Putting n = k, we get

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

Now we shall prove that P(k + 1) is true, whenever P(k) is true.

Adding \(\frac{1}{1 + 2 + 3 + ………….. + (k + 1)}\) to both sides, we get

∴ P(k + 1) is true, whenever P(k) is true.

Hence, by principle of mathematical induction, P(n) is true for all n ∈ N.

4. Let P(n) : 1.2.3 + 2.3.4 + 3.4.5 + … + n(n + 1)(n + 2)

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

For n = 1, L.H.S. = 1.23 = 6

and R.H.S. = \(\frac{1(1 + 1)(1 + 2)(1 + 3)}{4}\) = \(\frac{1×2×3×4}{4}\) = 6.

∴ L.H.S. = R.H.S., i.e., P(1) is true.

∴ Putting n = k in (1), we have:

P(k) : 1.2.3 + 2.3.4 + 3.4.5 + …………… + k(k + 1)(k + 2) = \(\frac{k(k + 1)(k + 2)(k + 3)}{4}\) ……………. (2)

We assume that P(k) is true.

Adding (k + 1)(k + 2)(k + 3) to both sides of (2), we have:

1.2.3 + 2.3.4 + 3.4.5 + ………….. + k(k + 1)(k + 2) + (k + 1)(k + 2)(k + 3)

∴ P(n) is true for n = k + 1, i.e., P(k + 1) is true.

∴ By Principle of Mathematical Induction, P(n) is true for all natural numbers n.

5. Let P(n) : 1.3 + 2.3^{2} + 3.3^{3} + … + n.3^{n} = \(\frac{(2 n-1) \cdot 3^{n+1}+3}{4}\)

Putting n = 1, L.H.S.= 1.3 = 3

and R.H.S. = \(\frac{(2-1) \cdot 3^{2}+3}{4}\) = \(\frac{12}{4}\) = 3.

∴ L.H.S. = R.H.S.

This shows that P(n) is true for n = 1.

Let P(n) be true for n = k.

∴ P(k) : 1.3 + 2.3^{2} + 3.3^{2} + ………… + k.3^{k} = \(\frac{(12 k-1) \cdot 3^{k+1}+3}{4}\) is true. ………………. (1)

Adding (k + 1).3^{k+1} to both sides of (1), we get

L.H.S. = 1.3 + 2.3^{2} + k.3^{2} + … + k.3^{k} + (k + 1).3^{k+1}

This shows P(n) is true for n = k + 1.

i.e., P(k + 1) is true, whenever P(k) is true.

Hence, by principle of mathematical induction P(n) is true for all values of n ∈ N.

6. Let P(n) : 1.2 + 2.3 + 3.4 + ………….. + n(n + 1) = \(\frac{n(n+1)(n+2)}{3}\)

For n = 1, L.H.S. = 1.2 = 2

and R.H.S. = \(\frac{n(n + 1)(n + 2)}{3}\) = \(\frac{1.2.3}{3}\) = 1.2 = 2.

i.e., L.H.S. = R.H.S.

So, P(1) is true.

We assume that P(n) is true for n = k.

i.e., 1.2 + 2.3 + 3.4 + … + k(k + 1) = \(\frac{k(k + 1)(k + 2)}{3}\)

Last term in L.H.S. is k(k + 1)

Replacing k by k + 1, we get (k + 1)(k + 2)

Adding it to both sides, we get

L.H.S. = 1.2 + 2.3 + 3.4 + … + k(k + 1) + (k + 1)(k + 2)

Thus, P(k + 1) is true, whenever P(k) is true.

Hence, by principle of mathematical induction, P(n) is true for all values of n ∈ N.

7. Let P(n) be the given statement.

i.e., P(n) = 1.3 + 3.5 + 5.7 + … + (2n – 1)(2n + 1)

= \(\frac{n\left(4 n^{2}+6 n-1\right)}{3}\)

Putting n = 1,

L.H.S. = 1.3 = 3

and R.H.S. = \(\frac{1 .\left(4.1^{2}+6.1-1\right)}{3}\) = \(\frac{4+6-1}{3}\) = \(\frac{9}{3}\) = 3.

L.H.S. = R.H.S.

∴ P(n) is true for n = 1.

Assume that P(n) is true for n = k, i.e., P(k) is true.

i.e., 1.3 + 3.5 + 3.7 + … + (2k – 1)(2k + 1) =\(\frac{k\left(4 k^{2}+6 k-1\right)}{3}\)

Last term in L.H.S. = (2k – 1)(2k + 1)

Replacing k by k + 1, we get

[2(k + 1) – 1][2(k + 1) + 1] = (2k + 1)(2k + 3)

Adding (2k + 1)(2k + 3) to both sides, we get

∴ L.H.S. = 1.3 + 3.5 + 5.7 + … + (2k – 1)(2k + 1) + (2k + 1)(2k + 3)

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

∴ P(k + 1) is true, whenever P(k) is true.

Hence, by principle of mathematical induction, P(n) is true for all n ∈ N.

8. Let P(n) be the given statement.

i.e., P(n) : 1.2 + 2.2^{2} + 3.2^{3} + … + n.2^{n} = (n – 1)2^{n+1} + 2

Putting n = 1, L.H.S. = 1.2 = 2

and R.H.S. = 0 + 2 = 2.

∴ P(n) is true for n = 1.

Assume that P(n) is true for n = k, i.e., P(k) is true, i.e.,

1.2 + 2.3 + 3.4 + … + k.2^{k} = (k – 1).2^{k+1} + 2

Last term in L.H.S. = k.2^{k}.

Replacing k by k + 1, we get the next term = (k + 1).2^{k+1}

Adding it to both sides, we get

L.H.S. = 1.2 + 2.2^{2} + 3.2^{3} + … + k.2^{k+1} + 1 + (k + 1).2^{k+1}

and R.H.S. = (k – 1).2^{k+1} + 2 + (k + 1).2^{k+1}

= 2^{k+1}.[k – 1 + k + 1] + 2 = 2k.2^{k+1} + 2

= k.2^{k+2} + 2.

This proves P(n) is true for n = k + 1.

Thus, P(k + 1) is true, whenever P(k) is true.

∴ By principle of mathematical induction, P(k) is true for all n ∈ N.

9. Let P(n) be the given statement.

i.e., P(n): \(\frac{1}{2}\) + \(\frac{1}{4}\) + \(\frac{1}{8}\) + ………….. + \(\frac{1}{2^{n}}\) = 1 – \(\frac{1}{2^{n}}\).

Putting n = 1, L.H.S. = \(\frac{1}{2}\)

and R.H.S. = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\).

∴ P(n) is true for n = 1.

Suppose P(n) is true for n = k.

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

Last term in L.H.S. = \(\frac{1}{2^{k}}\).

Replacing k by k + 1,

last term becomes \(\frac{1}{2^{k+1}}\).

Adding \(\frac{1}{2^{k+1}}\) to both sides, we get

L.H.S. = \(\frac{1}{2}\) + \(\frac{1}{4}\) + \(\frac{1}{8}\) + ……………… + \(\frac{1}{2^{k}}\) + \(\frac{1}{2^{k+1}}\)

and R.H.S. = 1 – \(\frac{1}{2^{k}}\) + \(\frac{1}{2^{k+1}}\) = 1 – \(\frac{1}{2^{k}}\) (1 – \(\frac{1}{2}\))

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

This shows P(n) is true for n = k + 1.

Thus, P(k + 1) is true, whenever P(k) is true.

Thus, by principle of mathematical induction, P(n) is true for all n ∈N.

10. Let the given statement be P(n) i.e.,

P(n) : \(\frac{1}{2.5}\) + \(\frac{1}{5.8}\) + \(\frac{1}{8.11}\) + ………….. + \(\frac{1}{(3n – 1)(3n + 2)}\) = \(\frac{n}{6n + 4}\).

Putting n = 1, L.H.S. = \(\frac{1}{2.5}\) = \(\frac{1}{10}\)

and R.H.S. = \(\frac{1}{6 + 4}\) = \(\frac{1}{10}\).

∴ P(n) is true for n = 1.

Assuming P(n) is true for n = k, i.e., P(k) is true, i.e.,

\(\frac{1}{2.5}\) + \(\frac{1}{3.8}\) + \(\frac{1}{8.11}\) + ………… + \(\frac{1}{(3k – 1)(3k

+ 1)}\) = \(\frac{k}{6k + 4}\).

Now, k^{th} term = \(\frac{1}{(3k – 1)(3k + 2)}\).

∴ (k + 1)^{th} term = \(\frac{1}{[3(k + 1) – 1][3(k + 1) + 1]}\) = \(\frac{1}{(3k + 2)(3k + 5)}\).

Adding this term to both sides, we get

L.H.S. = \(\frac{1}{2.5}\) + \(\frac{1}{3.8}\) + \(\frac{1}{8.11}\) + ………………. + \(\frac{1}{(3k – 1)(3k + 2)}\) + \(\frac{1}{(3k + 2)(3k + 5)}\)

and R.H.S.

This shows that P(n) is true for n = k + 1.

∴ P(k + 1) is true, whenever P(k) is true. So, by principle of mathematical induction, P(n) is true for all n ∈ N.

11. Let P(n) be the given statement, i.e.,

P(n) : \(\frac{1}{1.2.3}\) + \(\frac{1}{2.3.4}\) + \(\frac{1}{3.4.5}\) + ………….. + \(\frac{1}{n(n + 1)

(n + 2)}\) = \(\frac{n(n + 3)}{4(n + 1)(n + 2)}\).

Putting n = 1, L.H.S. = \(\frac{1}{1.2.3}\) = \(\frac{1}{6}\)

and R.H.S = \(\frac{1(1 + 3)}{4(1 + 1)(1 + 2)}\) = \(\frac{4}{4.2.3}\) = \(\frac{1}{6}\).

∴ P(n) is true for n = 1.

Assuming P(n) is true for n = k, i.e., P(k) is supposed to be true

Adding this term to both sides, we get

This shows P(n) is true for n = k + 1, i.e., P(k + 1) is true whenever P(k) is true.

Hence, by principle of mathematical induction, P(n) is true for all n ∈ N.

12. Let P(n) : a + ar + ar^{2} + … + ar^{n-1} = \(\frac{a\left(1-r^{n}\right)}{1-r}\), r ≠ 1 ………………. (1)

For n = 1, L.H.S. = a

and R.H.S. = \(\frac{a(1-r)}{1-r}\) = a.

∴ L.H.S. = R.H.S., i.e., P(1) is true.

Let us suppose that P(k) is true.

Putting n = k in (1), we have:

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

Adding ar^{k} to both sides of (2), we have:

a + ar + ar^{2} + ……………… + ar^{k-1} + ar^{k}

∴ P(n) is true for n = k + 1, i.e., P(k + 1) is true.

Thus, P(k + 1) is true, whenever P(&) is true.

∴ By principle of mathematical induction, P(n) is true for all natural numbers n.

13. Let the given statement be denoted by P(n), i.e.,

L.H.S. = R.H.S.

∴ P(n) is true for n = 1.

Suppose P(k) is true.

∴ P(n) is true for n = k + 1, i.e., P(k + 1) is true, whenever P(k) is true.

Hence, by principle of mathematical induction, P(n) is true for all n ∈ N.

14. Let the given statement be P(n), i.e.,

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

For n = 1, L.H.S. = 1 + \(\frac{1}{1}\) = 2

and R.H.S. = 1 + 1 = 2.

∴ P(n) is true for n = 1.

Let P(&) be true, i.e.,

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

Multiplying both sides by (1 + \(\frac{1}{k+1}\)), we get

L.H.S. = (1 + \(\frac{1}{1}\))(1 + \(\frac{1}{2}\))(1 + \(\frac{1}{3}\)) ………………. (1 + \(\frac{1}{k}\))(1 + \(\frac{1}{k+1)}\)

and R.H.S = (k + 1) (1 + \(\frac{1}{k + 1}\) = (k + 1)(\(\frac{k+1+1)}{k+1}\) = (k + 2)

= (\(\sqrt{k+1}\) + 1)

Therefore P(k + 1) is also true, whenever P(k) is true. Hence, by principle of mathematical induction, P(n) is true for all n ∈ N.

15. Let the given statement be P(n), i.e.,

P(n) : 1^{2} + 3^{2} + 5^{2} + … + (2n – 1)^{2} = \(\frac{n(2n – 1)(2n + 1)}{3}\).

For n = 1, L.H.S. = 1^{2} = 1

and R.H.S. = \(\frac{1.(2 – 1)(2 + 1)}{3}\) = \(\frac{1.1.3}{3}\) = 1.

∴ P(n) is true for n – 1.

Suppose P(&) is true for n = k i.e.,

1^{2} + 3^{2} + 5^{2} + …………… + (2k – 1)^{2} = \(\frac{k(2k – 1)(2k + 1)}{3}\).

Adding (2k + 1)^{2} to both sides, we get

L.H.S. = 1^{2} + 3^{2} + 5^{2} + ……………. + (2k – 1)^{2} + (2k + 1)^{2}

Thus, P(k + 1) is true.

∴ P(k + 1) is true, whenever P(k) is true.

∴ By principle of mathematical induction, Pin) is true for all values of n ∈ N.

16. Let the given statement be P(n), i.e.,

P(n) : \(\frac{1}{1.4}\) + \(\frac{1}{4.7}\) + \(\frac{1}{7.10}\) + …………….. + \(\frac{1}{(3n – 2)(3n + 1)}\) = \(\frac{n}{3n + 1}\).

For n = 1, L.H.S. = \(\frac{1}{1.4}\) = \(\frac{1}{4}\)

and R.H.S. = \(\frac{1}{3.1+1}\) = \(\frac{1}{4}\).

∴ P(n) is true for n = 1.

Assuming P(k) is true for some value of n = k.

∴ \(\frac{1}{1.4}\) + \(\frac{1}{4.7}\) + \(\frac{1}{7.10}\) + ………………. + \(\frac{1}{(3k – 2)(3k + 1)}\) = \(\frac{k}{3k + 1}\).

Adding \(\frac{1}{(3k + 1)(3k + 2)}\) both sides, we get

Therefore, P(k + 1) is true for n = k + 1, i.e., P(k + 1) is true, whenever P(k) is true. Hence, by principle of mathematical induction, P(n) is true for all n ∈ N.

17. Let P(n) be the given statement, i.e.,

P(n) : \(\frac{1}{3.5}\) + \(\frac{1}{5.7}\) + \(\frac{1}{7.9}\) + …………….. + \(\frac{1}{(2n + 1)(2n + 3)}\) = \(\frac{n}{3(2n + 3)}\).

For n = 1, L.H.S. = \(\frac{1}{3.5}\) = \(\frac{1}{15}\)

and R.H.S. = \(\frac{1}{3.(2 + 3)}\) = \(\frac{1}{3.5}\) = \(\frac{1}{15}\).

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

Suppose P(k) be true for n = k, i.e.,

\(\frac{1}{3.5}\) + \(\frac{1}{5.7}\) + \(\frac{1}{7.9}\) + ………….. + \(\frac{1}{(2k + 1)(2k + 3)}\) = \(\frac{k}{3(2k + 3)}\).

Adding \(\frac{1}{2(k + 3)(2k + 5)}\) to both sides, we get

L.H.S. = \(\frac{1}{3.5}\) + \(\frac{1}{5.7}\) + \(\frac{1}{7.9}\) + ……………. + \(\frac{1}{(2k + 1)(2k + 3)}\) + \(\frac{1}{(2k + 3)(2k + 5)}\)

Hence, P(k + 1) is true for n = k + 1, i.e., P(k + 1) is true, whenever P(k) is true.

Hence, by principle of mathematical induction, P(n) is true for all n ∈ N.

18. Let P(n) : 1 + 2 + 3 + …………. + n < \(\frac{1}{8}\) (2n + 1)^{2} ………………. (1)

For n = 1, (1) becomes 1 < \(\frac{1}{8}\) (2 + 1)^{2} ⇒ 1 < \(\frac{9}{8}\) ⇒ 1 < 1 \(\frac{1}{8}\)

which is true, i.e., P(1) is true.

Let us suppose that P(k) is true.

∴ Putting n = k in (1), we have:

1 + 2 + 3 + ………….. + k < \(\frac{1}{8}\) (2k + 1)^{2}

Adding (k + 1) to both sides of (2), we have:

⇒ 1 + 2 + 3 + …………… + k + (k + 1) < \(\frac{1}{8}\) (4k^{2} + 12k + 9)

⇒ 1 + 2 + 3 + ……………… + k + (k + 1) < \(\frac{1}{8}\) (2k + 3)^{2}

⇒ 1 + 2 + 3 + ……………. + k + (k + 1) < \(\frac{1}{8}\) [2(k + 1) + 1]^{2} ……………… (3)

∴ P(k + 1) is true.

∴ By principle of mathematical induction, P(n) is true for all natural numbers n.

19. Let the statement be denoted by P(n), i.e.,

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

For n = 1, n(n + 1 )(n + 5) = 1.2.6 = 12 = 3.4.

P(n) is true for n = 1.

Suppose P(k) is true for n = k i.e.

k(k + 1)(k + 5) = 3m (say)

or k^{3} + 6k^{2} + 5k = 3m …………………… (1)

Replacing k by k + 1 in L.H.S. of (1), we get

(k + 1)(k + 2)(k + 6) = k(k^{2} + 8k + 12) + (k^{2} + 8k + 12)

= k^{3} + 9k^{2} + 20k + 12

= (k^{3} + 6k^{2} + 5k) + (3k^{2} – 15k + 12)

= 3m + 3k^{2} + 15k + 12 [From (1)]

= 3(m + k^{2} + 5k + 4).

i.e., (k + 1)(k + 2)(k + 6) is a multiple of 3, i.e.,

P(k + 1) is a multiple of 3, if P(k) is a multiple of 3.

i.e., P(k + 1) is true, whenever P(k) is true.

Hence, by principle of mathematical induction, P(n) is true for all n ∈ N.

20. Let P(n) : 10^{(2n – 1)} + 1 is divisible by 11 for every natural number n.

For n = 1, P(1) = 10^{2-1} + 1 = 10 + 1 = 11,

which is divisible by 11

∴ P(1) is true.

Put n = k in (1), we have:

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

∴ 10^{2k-1} + 1 = 11m, for some integer m.

⇒ 10^{2k-1} = 11m – 1 …………………. (1)

Putting n = k + 1(k + 1) in 10^{2n-1} + 1, it becomes

10^{2(k+1)-1} + 1 = 10^{2k+1} + 1 = 10^{2k-1} = 10^{2} + 1

= 100.10^{2k-1} + 1 = 100(11m – 1) + 1 [Using (1)]

= 100 × 11m – 100 + 1

= 100 × 11m – 99

= 11(100m – 9), which is divisible by 11. [∵ 11 is a factor of R.H.S]

∴ 10^{2(k+1)-1} + 1 is divisible by 11.

∴ P(n) is true for n = k + 1, i.e; P(k + 1) is true.

∴ By principle of mathematical induction, P(n) is true for all natural numbers n.

21. Let the statement be P(n), i.e.,

P (n) : x^{2n} – y^{2n} is divisible by x + y. ………………. (1)

Putting n = 1, x^{2n} – y^{2n} = x^{2} – y^{2} = (x + y)(x – y),

which is divisible by x + y.

⇒ P(n) is true for n = 1.

Let P(k) be true, i.e,

x^{2k} – y^{2k} is divisible by x + y

or x^{2k} – y^{2k} = m(x + y).

or x^{2k} = m(x + y) + y^{2k} ………………… (2)

Replace k by k + 1 in x^{2k} from (2), we get

x^{2(k + 1)} – y^{2(k + 1)} = x^{2k+2} – y^{2k+2}

= x^{2}.x^{2k} – y^{2k+2}.

Putting the value of x^{2k} from (2), we get

x^{2(k + 1)} – y^{2(k + 1)} = x^{2}[m (x + y) + y^{2k}] – y^{2k} + 2

= m(x + y)x^{2} + x^{2}y^{2k} – y^{2k} + 2

= m(x + y)^{2} + y^{2k})(x^{2} – y^{2})

= m(x + y)x^{2} + (x + y)(x – y)y^{2k}

Therefore x^{2(k+1)} – y^{2(k + 1)} is divisible by (x + y).

i.e., P(k + 1) is true whenever P(k) is true.

By principle of mathematical induction, P(n) is true for all n ∈ N.

22. Let the statement be denoted by P(n), i.e;

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

For n = 1, 3^{2n+2} – 8n – 9 = 3^{2+2} – 8.1 – 9

= 3^{4} – 8 – 9 = 81 – 17 = 64,

which is divisible by 8.

Let P(k) be the true statement for some value of n = k.

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

or 3^{2k+2} – 8k – 9 = 3^{2(k + 1) + 2} – 8(k + 1) – 9

= 3^{2k + 2 + 2} – 8k – 8 – 9

= 3^{2}. 3^{2k+2} – 8k – 17

= 9.3^{2k+2} – 8k – 17.

Putting the value of 3^{2k+2} from (1), we get

2^{2k+4} – 8k – 17 = 9(8m + 8k + 9) – 8k – 17

= 72m + 72k + 81 – 8k – 17

= 72m + 64k + 64

= 8(9m + 8k + 8)

Hence, 3^{2k+4} – 8(k + 1) – 9 is divisible by 8.

i.e; P(k + 1) is true, whenever P(k) is true.

Hence, by principle of mathematical induction, P(n) is true for all n ∈ N.

23. Let P(n) be the given statement, i.e;

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

For n = 1, 41^{n} – 14^{n} = 41 – 14 = 27.

∴ P(n) is true for n = 1.

Let P(k) be true for some value of n = k.

P(k) : 41^{k} – 14^{k} = 27m, where m is a positive integer.

41^{k} = 27 + 14^{k} ……………… (1)

Replacing k by k + 1 in L.H.S. of (1), we get

41^{k + 1} – 14^{k + 1} = 41.41^{k} – 14^{k + 1}

= 41. (27m + 14^{k}) – 14^{k + 1} [From (1)]

= 27.41 m + 14^{k}(41 – 14)

= 27. 41m + 14^{k}.27

= 27[41m + 14^{k}]

This shows 41^{k + 1} – 14^{k + 1} is a multiple of 27.

or P(k + 1) is true, whenever P(k) is true.

Hence, by principal of mathematical induction, P(n) is true for all n ∈ N.

24. Let P(n) : (2n + 7) < (n + 3)^{2} ………………. (1)

For n = 1, (1) becomes 2 × 1 + 7 < (1 + 3)^{2} ⇒ 9 < 16, which is true.

∴ P(1) is true

Let us suppose that P(k) is true.

∴ Putting n = k in (1), we have:

2(k + 7) < (k + 3)^{2} ………………….. (2)

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

i.e; 2(k + 1) + 7 < (k + 4)^{2}.

We know that 2k + 7 < (k + 3)^{2}.

⇒ 2k + 7 + 2 < (k + 3)^{2} + 2 [Adding 2 t0 both sides]

⇒ 2(k + 1) + 7 < (k + 3)^{2} + 2 ……………… (3)

⇒ 2 (k + 1) + 7 < k^{2} + 6k + 11

Adding 2k + 5 to R.H.S., we get

⇒ 2(k + 1) + 7 < x^{2} + 7k + 16

or 2(k + 1) + 7 < (k + 4)^{2}.

Thus, P(k + 1) is true. Hence, P(n) is true for n = k + 1.

∴ By principal of mathematical induction, P(n) is true for all natural numbers n.