GSEB Solutions Class 11 Maths Chapter 4 Principle of Mathematical Induction Ex 4.1

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 + 3² + … + 3^n-1 = $\frac{3^{n}-1}{2}$.
2. 1³ + 2³ + 3³ + ………………. + n³ = ($\frac{n(n+1)}{2}$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 + 2.3² + 3.3² + ……………. + n.3ⁿ = $\frac{(2 n-1) \cdot 3^{n+1}+3}{4}$
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 + 2.2² + 3.2² + …………….. + n.2ⁿ = (n – 1).2ⁿ⁺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 + ar² + …………… + 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. 1² + 3² + 5² + ………….. + (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. x²ⁿ – y²ⁿ is divisible by x + y.
22. 3²ⁿ⁺² – 8n – 9 is divisible by 8.
23. 41ⁿ – 14ⁿ 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 + 3² + ……………… + 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 + 3² + ……………. + 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 + 3² + ……………….. + 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) : 1³ + 2³ + 3³ + ……………. + n³ = $\frac{n^{2}(n+1)^{2}}{4}$.
For n = 1, L.H.S. = 1³ = 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) : 1³ + 2³ + 3³ + ……………… + k³ = $\frac{k^{2}(k+1)^{2}}{4}$ …………….. (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 + 2.3² + 3.3³ + … + n.3ⁿ = $\frac{(2 n-1) \cdot 3^{n+1}+3}{4}$
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 + 2.3² + 3.3² + ………… + 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 + 2.3² + k.3² + … + 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 + 2.2² + 3.2³ + … + n.2ⁿ = (n – 1)2ⁿ⁺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 + 2.2² + 3.2³ + … + 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 + ar² + … + 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 + ar² + … + 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 + ar² + ……………… + 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) : 1² + 3² + 5² + … + (2n – 1)² = $\frac{n(2n – 1)(2n + 1)}{3}$.
For n = 1, L.H.S. = 1² = 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.,
1² + 3² + 5² + …………… + (2k – 1)² = $\frac{k(2k – 1)(2k + 1)}{3}$.
Adding (2k + 1)² to both sides, we get
L.H.S. = 1² + 3² + 5² + ……………. + (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) < $\frac{1}{8}$ (4k² + 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 k³ + 6k² + 5k = 3m …………………… (1)
Replacing k by k + 1 in L.H.S. of (1), we get
(k + 1)(k + 2)(k + 6) = k(k² + 8k + 12) + (k² + 8k + 12)
= k³ + 9k² + 20k + 12
= (k³ + 6k² + 5k) + (3k² – 15k + 12)
= 3m + 3k² + 15k + 12 [From (1)]
= 3(m + k² + 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 = 10² + 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) : x²ⁿ – y²ⁿ is divisible by x + y. ………………. (1)
Putting n = 1, x²ⁿ – y²ⁿ = x² – y² = (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
= x².x^2k – y^2k+2.
Putting the value of x^2k from (2), we get
x^{2(k + 1)} – y^{2(k + 1)} = x²[m (x + y) + y^2k] – y^2k + 2
= m(x + y)x² + x²y^2k – y^2k + 2
= m(x + y)² + y^2k)(x² – y²)
= m(x + y)x² + (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) : 3²ⁿ⁺² – 8n – 9 is divisible by 8.
For n = 1, 3²ⁿ⁺² – 8n – 9 = 3²⁺² – 8.1 – 9
= 3⁴ – 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
= 3². 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) : 41ⁿ – 14ⁿ is a multiple of 27.
For n = 1, 41ⁿ – 14ⁿ = 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 < k² + 6k + 11
Adding 2k + 5 to R.H.S., we get
⇒ 2(k + 1) + 7 < x² + 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.