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 1^{3} + 2^{3} + 3^{3} + … + n^{3} = \(\frac{n^{2}(n+1)^{2}}{4}\), ∀ n ∈ N

Solution:

Let p(n) be the statement:

1^{3} + 2^{3} + 3^{3} + … + n^{3} = \(\frac{n^{2}(n+1)^{2}}{4}\) and let S(n) be the sum on the L.H.S.

Since S(1) = 1^{3} = 1^{3} = \(\frac{1^{2}(1+1)^{2}}{4}\) = 1 = 1^{3}

∴ The formula is true for n = 1.

Assume that the statement p(n) is true for n = k

(i.e.,) S(k) = 1^{3} + 2^{3} + 3^{3} + … + k^{3} = \(\frac{k^{2}(k+1)^{2}}{4}\)

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) = 1^{3} + 2^{3} + 3^{3} + … + k^{3} + (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

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

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

∴ The formula is true for n = 1

Assume that the statement p(n) is true for n = k

(i.e.,) S(k) = 1^{2} + 3^{2} + 5^{2} + … + (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) = 1^{2} + 3^{2} + 5^{2} + ….. + (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 + 4.2^{2} + … upto n terms = n.2^{n}, ∀ n ∈ N. (May ’07)

Solution:

Let p(n) be the statement:

2 + 3.2 + 4.2^{2} + … + (n + 1) . 2^{n-1} = n.2^{n} and let S(n) be the sum on the L.H.S.

∵ S(1) = 2 = (1) . 2^{1}

∵ 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 + 4.2^{2} + … + (k + 1).2^{k-1} = k . 2^{k}

We show that the formula is true for n = k + 1 (i.e.,) we show that S(k + 1) = (k + 1). 2^{k + 1}

We observe that

S(k + 1)= 2 + 3.2 + 4.2^{2} + …(k + 1)2^{k + 1} + (k + 2) . 2^{k}

= S(k) + (k + 2) . 2^{k}

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

= (k + k + 2) 2^{k}

= 2(k + 1) . 2^{k} = 2^{k + 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 + 4.2^{2} + … + (n + 1)2^{n – 1}

= n.2^{n} ∀ 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 n^{th} term is

1 + (n – 1)3 = 3n – 2

4, 7, 10, … are in A.R, whose n^{th} term is

4 + (n – 1)3 = 3n + 1

∴ The n^{th} 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 ≤ 2^{2n – 2} for all n ≥ 5, n ∈ N.

Solution:

Let P(n) be the statement:

2n – 3 ≤ 2^{n – 2}, ∀ n ≥ 5, n ∈ N.

Here we note that the basis of induction is 5.

Since 2.5 – 3 ≤ 2^{5 – 2}, the statement is true for n = 5.

Assume the statement is true for n = k, k ≥ 5.

i.e., 2K – 3 ≤ 2^{k – 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)

≤ 2^{k – 2} + 2^{k – 2} for k ≥ 5

= 2.2^{k – 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 + x^{2} > 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 + kx^{2}

> 1 + (k + 1)x, (since kx^{2} > 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 x^{n} – y^{n} is divisible by x – y for all n ∈ N. (June ’04)

Solution:

Let p(n) be the statement:

x^{n} – y^{n} is divisible by (x – y)

Since x^{1} – y^{1} = 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) x^{k} – y^{k} is divisible by x – y

Then x^{k} – y^{k} = (x – y)p, for some integer p (1)

We show that the statement is true for n = k + 1. (i.e) we show that x^{k + 1} – y^{k + 1} is divisible by x – y from (1), we have

x^{k} – y^{k} = (x – y) p

⇒ x^{k} = (x – y)p + y^{k}

∴ x^{k + 1} = (x – y)p x + y^{k} . x

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

= (x – y)p x + y^{k}(x – y)

= (x – y)[px + y^{k}],

(where px + y^{k} is an integer)

∴ x^{k + 1} – y^{k + 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) x^{k} – y^{k} is divisible by (x – y) for all n **Σ** N.

Question 8.

Using mathematical induction, show that x^{m} + y^{m} 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:

x^{2n + 1} + y^{2n + 1} is divisible by (x + y)

Since x^{1} + y^{1} = x + y is divisible by x + y

∴ The statement is true for n = 0 and since x^{2.1 + 1} + y^{2.1 + 1} = x^{3} + y^{3} = (x + y)(x^{2} – xy + y^{2}) 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.) x^{2k + 1} + y^{2k + 1} is divisible by x + y. Then x^{2k + 1} + y^{2k + 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

x^{2k + 3} + y^{2k + 3} is divisible by (x + y)

From (1)

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

∴ x^{2k + 1} = (x + y) p – y^{2k + 1}

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

∴ x^{2k + 3} = (x + y) p x^{2} – y^{2k + 1} . x^{2}

∴ x^{2k + 3} + y^{2k + 3} = (x + y) p. (x^{2}) – y^{2k + 1}.x^{2} + y^{2k + 3}

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

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

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

Here p x^{2} – y^{2k + 1} (x – y) is an integer.

∴ x^{2k + 3} + y^{2k + 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.,) x^{2n + 1} + y^{2n + 1} is divisible by (x + y), for all non-negative integers.

(i.e.,) x^{m} + y^{m} is divisible by (x + y), if m is an odd natural number.

Question 9.

Show that 49^{n} +16n – 1 is divisible by 64 for all positive integers n. (May ’05)

Solution:

Let p(n) be the statement:

49^{n} + 16n – 1 is divisible by 64

Since 49^{1} + 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.,) 49^{k} + 16k – 1 is divisible by 64

Then (49^{k} + 16k – 1) = 64t, for some t ∈ N ——– (1)

we show that the statement is true for

n = k + 1.

(i.e.,) we show that

49^{k + 1} + 16 (k + 1) – 1 is divisible by 64

From (1), we have

49^{k }+ 16k – 1 = 64t

∴ 49^{k} = 64t – 16k + 1

∴ 49^{k} . 49 = (64t – 16k + 1). 49

(64t – 16k + 1)49 + 16(k + 1) – 1

∴ 49^{k + 1} + 16(k + 1) – 1

= (64t – 16 K + 1)49 + 16(k + 1) – 1

∴ 49^{k + 1} + 16(k + 1) – 1 = 64(49t – 12k + 1)

here (49t — 12k + 1) is an integer

∴ 49^{k + 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.,) 49^{n} + 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)} = 2.4^{3} + 3^{4}

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