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.