Students get through Maths 1A Important Questions Inter 1st Year Maths 1A Mathematical Induction Important Questions which are most likely to be asked in the exam.

## Intermediate 1st Year Maths 1A Mathematical Induction Important Questions

Question 1.
Use mathematical induction to prove the statement 13 + 23 + 33 + … + n3 = $$\frac{n^{2}(n+1)^{2}}{4}$$, ∀ n ∈ N
Solution:
Let p(n) be the statement:
13 + 23 + 33 + … + n3 = $$\frac{n^{2}(n+1)^{2}}{4}$$ and let S(n) be the sum on the L.H.S.
Since S(1) = 13 = 13 = $$\frac{1^{2}(1+1)^{2}}{4}$$ = 1 = 13
∴ The formula is true for n = 1.
Assume that the statement p(n) is true for n = k
(i.e.,) S(k) = 13 + 23 + 33 + … + k3 = $$\frac{k^{2}(k+1)^{2}}{4}$$
We show that the formula is true for n = k + 1
(i.e.,) We show that S(k + 1) = $$\frac{(k+1)^{2}(k+2)^{2}}{4}$$
We observe that
S(k + 1) = 13 + 23 + 33 + … + k3 + (k + 1)3
= S(k) + (k + 1)3
= $$\frac{k^{2}(k+1)^{2}}{4}$$ + (k + 1)3

∴ The formula holdes for n = k + 1
∴ By the principle of mathematical induction p(n) is true for all n ∈ N
(i.e.,) the formula
13 + 23 + 33 + … + k3 = $$\frac{n^{2}(n+1)^{2}}{4}$$ is true for all n ∈ N.

Question 2.
Use mathematical induction to prove the statement
$$\sum_{k=1}^{n}(2 k-1)^{2}$$ = $$\frac{n(2 n-1)(2 n+1)}{3}$$ for all n ∈ N.
Solution:
Let p(n) be the statement:
12 + 32 + 52 + … + (2n – 1)2 = $$\frac{n(2 n-1)(2 n+1)}{3}$$ for all n ∈ N
Let S(n) be the sum on the L.H.S.
∴ s(1) = 12 = $$\frac{1(2-1)(2+1)}{3}$$ = 1
∴ The formula is true for n = 1
Assume that the statement p(n) is true for n = k
(i.e.,) S(k) = 12 + 32 + 52 + … + (2k – 1)2 = $$\frac{k(2 k-1)(2 k+1)}{3}$$
We show that the formula is true for n = k + 1 (i.e.,) we show that S(k + 1) = $$\frac{(k+1)(2 k+1)(2 k+3)}{3}$$
We observe that
S(k + 1) = 12 + 32 + 52 + ….. + (2k – 1)2 + (2k + 1)2
= S(k) + (2k + 1)2

∴ The formula holds for n = k + 1
∴ By the principle of mathematical induction
p(n) is true ∀ n ∈ N.
i.e., $$\sum_{k=1}^{n}(2 k-1)^{2}$$ = $$\frac{n(2 n-1)(2 n+1)}{3}$$ for all n ∈ N.

Question 3.
Use mathematical induction to prove the statement 2 + 3.2 + 4.22 + … upto n terms = n.2n, ∀ n ∈ N. (May ’07)
Solution:
Let p(n) be the statement:
2 + 3.2 + 4.22 + … + (n + 1) . 2n-1 = n.2n and let S(n) be the sum on the L.H.S.
∵ S(1) = 2 = (1) . 21
∵ The statement is true for n = 1
Assume that the statement p(n) is true for n = k.
(i.e.,) S(k) = 2 + 3.2 + 4.22 + … + (k + 1).2k-1 = k . 2k
We show that the formula is true for n = k + 1 (i.e.,) we show that S(k + 1) = (k + 1). 2k + 1
We observe that
S(k + 1)= 2 + 3.2 + 4.22 + …(k + 1)2k + 1 + (k + 2) . 2k
= S(k) + (k + 2) . 2k
= k. 2k + (k + 2) . 2k
= (k + k + 2) 2k
= 2(k + 1) . 2k = 2k + 1 (k + 1)
∴ The formula holds for n = k + 1
∴ By the principle of mathematical induction, p(n) is true for all n ∈ N
(i.e.,) the formula
2 + 3.2 + 4.22 + … + (n + 1)2n – 1
= n.2n ∀ n ∈ N (i.e.,) the formula

Question 4.
Show that ∀ n ∈ N, $$\frac{1}{1.4}$$ + $$\frac{1}{4.7}$$ + $$\frac{1}{7.10}$$ + …….. upto n terms = $$\frac{n}{3 n+1}$$ (Mar. ’05)
Solution:
1, 4, 7, … are in A.R, whose nth term is
1 + (n – 1)3 = 3n – 2
4, 7, 10, … are in A.R, whose nth term is
4 + (n – 1)3 = 3n + 1
∴ The nth term of the given sum is
$$\frac{1}{(3 n-2)(3 n+1)}$$
Let p(n) be the statement:
$$\frac{1}{1.4}$$ + $$\frac{1}{4.7}$$ + $$\frac{1}{7.10}$$ + ….. + $$\frac{1}{(3 n-2)(3 n+1)}$$ = $$\frac{n}{3 n+1}$$
and S(n) be the sum on the L.H.S
Since S(1) = $$\frac{1}{1.4}$$ = $$\frac{1}{3(1)+1}$$ = $$\frac{1}{4}$$
∴ p(1) is true.
Assume the statement p(n) is true for n = k
(i.e)S(k) = $$\frac{1}{1.4}$$ + $$\frac{1}{4.7}$$ + $$\frac{1}{7.10}$$ + …….. + $$\frac{1}{(3 k-2)(3 k+1)}$$ = $$\frac{k}{3 k+1}$$
We show that the statement s p(n) is true for n = k + 1
(i.e) we show that S(k + 1) = $$\frac{k+1}{3 k+4}$$
We observe that

∴ The statement holds for n = k + 1.
∴ By the principle of mathematical induction, p(n) is true ∀ n ∈ N.
(ie.) $$\frac{1}{1.4}$$ + $$\frac{1}{4.7}$$ + $$\frac{1}{7.10}$$ + ……. + $$\frac{1}{(3 n-2)(3 n+1)}$$ = $$\frac{n}{3 n+1}$$ for all n Σ N.

Question 5.
Use mathematical induction to prove that 2n – 3 ≤ 22n – 2 for all n ≥ 5, n ∈ N.
Solution:
Let P(n) be the statement:
2n – 3 ≤ 2n – 2, ∀ n ≥ 5, n ∈ N.
Here we note that the basis of induction is 5.
Since 2.5 – 3 ≤ 25 – 2, the statement is true for n = 5.
Assume the statement is true for n = k, k ≥ 5.
i.e., 2K – 3 ≤ 2k – 2, for k ≥ 5.
We show that the statement is true for
n = k + 1, k ≥ 5
i.e., [2(k + 1) – 3] ≤ 2(k + 1) – 2, for k ≥ 5.
We observe that [2(k + 1) – 3]
= (2k – 3) + 2
≤ 2, (By inductive hypothesis)
≤ 2k – 2 + 2k – 2 for k ≥ 5
= 2.2k – 2
= 2(k + 1) – 2
∴ The statement P(n) is true for
n = k + 1, k ≥ 5.
∴ By the principle of mathematical induction, the statement is true for all n ≥ 5, n ∈ N.

Question 6.
Use Mathematical induction to prove that (1 + x)n > 1 + nx for n ≥ 2, x > -1, x ≠ 0.
Solution:
Let the statement P (n) be : (1 + x)n > 1 + nx
Here we note that the basis of induction is 2 and that x ≠ 0, x > -1
⇒ 1 + x > 0.
Since (1 + x)2 = t + 2x + x2 > 1 + 2x, the statement is true for n = 2.
Assume that the statement is true for n = k, k ≥ 2.
i.e., (1 + x)k > 1 + k x for k ≥ 2.
We show that the statement is true for n = k + 1,
i.e., (1 + x)k + 1 > + (k + 1)x.
We observe that (1 + x)k + 1
= (1 + x)k . (1 + x)
> (1 + kx) . (1 + x), (By inductive hypothesis)
= 1 + (k + 1)x + kx2
> 1 + (k + 1)x, (since kx2 > 0)
∴ The statement is true for n = k + 1.
∴ By the principle of mathematical induction, the statement P(n) is true for all n ≥ 2.
i.e.,(1 + x)n > 1 + nx, ∀ n ≥ 2, x > -1, x ≠ 0.

Question 7.
If x and y are natural numbers and x ≠ y, using mathematical induction, show that xn – yn is divisible by x – y for all n ∈ N. (June ’04)
Solution:
Let p(n) be the statement:
xn – yn is divisible by (x – y)
Since x1 – y1 = x – y is divisible by (x – y)
∴ The statement is true for n = 1
Assume that the statement is true for n = k
(i.e) xk – yk is divisible by x – y
Then xk – yk = (x – y)p, for some integer p  (1)
We show that the statement is true for n = k + 1. (i.e) we show that xk + 1 – yk + 1 is divisible by x – y from (1), we have
xk – yk = (x – y) p
⇒ xk = (x – y)p + yk
∴ xk + 1 = (x – y)p x + yk . x
⇒ xk + 1 – yk + 1 = (x – y) p x + yk x – yk + 1
= (x – y)p x + yk(x – y)
= (x – y)[px + yk],
(where px + yk is an integer)
∴ xk + 1 – yk + 1 is divisible by (x – y)
∴ The statement p(n) is true for n = k + 1
∴ By mathematical induction, p (n) is true for all n ∈ n
(i.e) xk – yk is divisible by (x – y) for all n Σ N.

Question 8.
Using mathematical induction, show that xm + ym is divisible by x + y, if m is an odd natural number and x, y are natural numbers.
Solution:
Since m is an odd natural number,
Let m = 2n + 1 where n is a non-negative integer.
∴ Let p(n) be the statement:
x2n + 1 + y2n + 1 is divisible by (x + y)
Since x1 + y1 = x + y is divisible by x + y
∴ The statement is true for n = 0 and since x2.1 + 1 + y2.1 + 1 = x3 + y3 = (x + y)(x2 – xy + y2) is divisible by x + y, the statement is true for n = 1.
Assume that the statement p(n) is true for n = k (i.e.) x2k + 1 + y2k + 1 is divisible by x + y. Then x2k + 1 + y2k + 1 = (x + y) p,
Where p is an integer ———— (1)
We show that the statement is true for
n = k + 1 (ie.) We show that
x2k + 3 + y2k + 3 is divisible by (x + y)
From (1)
x2k + 1 + y2k + 1 = (x + y)p
∴ x2k + 1 = (x + y) p – y2k + 1
∴ x2k + 1 . x2 = (x + y) p x2 – y2k + 1 . x2
∴ x2k + 3 = (x + y) p x2 – y2k + 1 . x2
∴ x2k + 3 + y2k + 3 = (x + y) p. (x2) – y2k + 1.x2 + y2k + 3
= (x + y) p.x2 – y2k + 1 (x2 – y2)
= (x + y)p x2 – y2k + 1 . (x + y)(x – y)
= (x + y)[p . x2 – y2k + 1(x – y)]
Here p x2 – y2k + 1 (x – y) is an integer.
∴ x2k + 3 + y2k + 3 is divisible by (x + y)
∴ The statement is true for n = k + 1
∴ By the principle of mathematical induction, p(n) is true for all n
(i.e.,) x2n + 1 + y2n + 1 is divisible by (x + y), for all non-negative integers.
(i.e.,) xm + ym is divisible by (x + y), if m is an odd natural number.

Question 9.
Show that 49n +16n – 1 is divisible by 64 for all positive integers n. (May ’05)
Solution:
Let p(n) be the statement:
49n + 16n – 1 is divisible by 64
Since 491 + 16(1) – 1 = 64 is divisible by 64
The statement is true for n = 1
Assume that the statement p(n) is true for n = k
(i.e.,) 49k + 16k – 1 is divisible by 64
Then (49k + 16k – 1) = 64t, for some t ∈ N ——– (1)
we show that the statement is true for
n = k + 1.
(i.e.,) we show that
49k + 1 + 16 (k + 1) – 1 is divisible by 64
From (1), we have
49k + 16k – 1 = 64t
∴ 49k = 64t – 16k + 1
∴ 49k . 49 = (64t – 16k + 1). 49
(64t – 16k + 1)49 + 16(k + 1) – 1
∴ 49k + 1 + 16(k + 1) – 1
= (64t – 16 K + 1)49 + 16(k + 1) – 1
∴ 49k + 1 + 16(k + 1) – 1 = 64(49t – 12k + 1)
here (49t — 12k + 1) is an integer
∴ 49k + 1 + 16(k + 1) – 1 is divisible by 64
∴ The statement is true for n = k + 1
∴ By the principle of mathematical induction,
∴ p(n) is true for all n ∈ N.
(i.e.,) 49n + 16n – 1 is divisible by 64, ∀ n ∈ N.

Question 10.
Use mathematical induction to prove that 2.4(2n + 1) + 3(3n + 1) is divisible by 11, ∀ n ∈ N.
Solution:
Let p(n) be the statement.
2.4(2n + 1) + 3(3n + 1) is divisible by 11
since 2.4(2.1 + 1) + 3(3.1 + 1) = 2.43 + 34
= 2(64) + 81
= 209 = 11 × 19 is divisible by 11
∴ The statement p(1) is true
Assume that the statement p(n) is true for n = k
(i.e.,) 24(2k + 1) + 3(3k + 1) is divisible by 11
Then 2.4(2k + 1) + 3(3k + 1) = (11)t, for some integer t ——– (1)
we show that the statement P(n) is true for n = K + 1
(i.e.,) we show that 2.4(2k + 3) + 3(3k + 4) is divisible by 11.
From (1), we have
2.4(2k + 1) + 3(3k + 1) = 11t
∴ 2.4(2k + 1) = 11t – 3(3k + 1)
∴ 2.4(2k + 1) . 42 = (11t – 3(3k + 1))
2.4(2k + 3) + 3(3k + 4) = (11t – 3(3k + 1)) 16 + 3(3k + 4)
= (11t)(16) + 3(3k + 1)[27 – 16]
= 11[16t + 33k + 1]
Here 16t + 3(3k + 1) is an integer
∴ 2.4(2k + 3) + 3(3k + 4) is divisible by 11
∴ The statement p(n) is true for n = k + 1.
∴ By the principle of mathematical induction, p(n) is true for all n ∈ N.
(i.e.,) 2.4(2n + 1) + 3(3n + 1) is divisible by 11, ∀ n ∈ N.