Students must practice these TS Inter 1st Year Maths 1A Important Questions Chapter 2 Mathematical Induction to help strengthen their preparations for exams.

## TS Inter 1st Year Maths 1A Mathematical Induction Important Questions

Question 1.

By using mathematical induction show that ∀ n ∈ N

\(\frac{1}{1 \cdot 4}+\frac{1}{4 \cdot 7}+\frac{1}{7 \cdot 10}+\) ………….. upto n terms \(=\frac{n}{3 n+1}\)

Solution:

1, 4, 7. are in A.P whose n^{th} term is 1 (n – 1) 3 = 3n – 2

and 4, 7, lo are in A.P. whose n^{th} term is 4 (n – 1) 3 = 3n + 1

∴ The n^{th} term of the given series

Hence S(k + 1) is true ∀ n ∈ N = k + 1

∴ By the principle of MathematicaL

Induction, S(n) is true ∀ n ∈ N

Question 2.

Use mathematical induction to prove the statement 2 + 3 . 2 + 4 . 2^{2} + upto n terms = n 2^{n}∀n∈N

Solution :

∴ S(n) is true for n = k + 1 also

∴ By the principle of Mathematical Induction.

S(n) is true for all n ∈ N.

∴ 2+3 . 2 + 4 . 2^{2}+ ………………….. + (n+1) 2^{n-1}

= n . 2^{n} ∀ n ∈ N

Question 3.

If x and y are natural numbers and x ≠ y using mathematical induction show that x^{n}-y^{n} is divisible by x – y, ∀n∈N

Solution:

Let S(n) be the statement

”x^{n}-y^{n }is divisible by (x – y)”

For n = 1, x^{1} – y^{1} is divisible by x’ – y’

We have S(1) is true for n = 1

Assume S(n) is true for n = k

Then x^{k} – y^{k} is divisible by (x-y)

Then (x^{k} – y^{k}) = (x – y) p where p ∈ Z …………………. (1)

We have prove that the statement S(n) is true for n = k + 1 also

Question 4.

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

Solution:

Let S(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 S(n) is true for n = 1.

Suppose the statement is true for S = k then

49^{k} + 16k – 1 = 64t for t ∈ N

We have to prove that S(n) is true for

n = k + 1 also

We have to prove that S(n) is true for

n = k + 1 also

Consider 49^{k+1}. + 16 (k + 1) – 1

49^{k} . 49 + 16k+ 16 – 1

=(64t – 16k+ 1)49 + 16k + 15

= (49) 64t + 16k (1 – 49) + 64

= 64 [49t – 12k + 1] and 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 S(n) is true ∀ n ∈ N

∴ 49^{n} + 16n – 1 is divisible by 64, ∀ n∈ N

Question 5.

Use mathematical induction to prove the statement

\(1^3+2^3+3^3+\ldots \ldots+n^3=\frac{n^2(n+1)^2}{4}, \forall n \in N\)

Solution:

Let S(n) be the statement

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

∴ Statement is true for n = k + 1 also

∴ By the principle of Mathematical Induction, S(n) is true ∀ n ∈ N

Question 6.

Use mathematical induction to prove the statement.

\(\sum_{k=1}^n(2 k-1)^2=\frac{n(2 n-1)(2 n+1)}{3}, \forall n \in N\)

Solution:

Let S(n) be the statement

Question 7.

Use mathematical induction to prove that

2n – 3 ≤ 2^{n-2} ∀ n ≥ 5, n ∈ N

Solution:

Let S(n) be the statement

2n – 3 ≤ 2^{n-2} ∀ n ≥ 5, n ∈ N

∴ Since 2(5) – 3 ≤ 2^{5-2}

∴ The statement S(n) is true for

n k + 1, k ≥ 5

∴ By the principle of mathematical induction S(n) is true for all n ≥ 5, n ∈ N.

Question 8.

Use mathematical induction to prove that

(1 + x)^{n }> 1 + nx for n ≥ 2, x > – 1, x ≠ 0.

Solution:

Let S(n) be the statement

(1+ x)^{n} > 1 + nx for n ≥ 2, x > – 1, x ≠ 0.

The basis of induction is 2

and x ≠ 0, x > – 1 ⇒ x+1>0

For n = 2,

(1 + x)^{2} = 1 + 2x + x^{2} > 1 + 2x; the statement

S(n) is true for n = 2

Assume that the statement is true for n = k, k ≥ 2

∴ By the principle of mathematical induction the statement S(n) is true for all n≥2.

∴ (1 + x)^{n} > 1 + nx, ∀ n ≥ 2, X > 1, X ≠ 0

Question 9.

Using mathematical induction show that x^{m} + y^{m} is divisible by x + y, if in 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 S(n) be the statement

Now px^{2} – (x y) y^{2k+1} + 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 S(n) is true ∀n.

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

∴ x^{m} + y^{m} is divisible by (x + y) if n is an odd natural number.

Question 10.

Use mathematical induction to prove that 2.4^{(2n+1) }+ 3^{(3n+1)} + is divisible by 11, ∀ n ∈ N.

Solution:

Let S(n) be the statement

2.4^{(2n+1) }+ 3^{(3n+1)} is divisible by 11

For n = 1

S(I) is 2.4^{3} + 3^{4} = 128 + 81 = 209 is divisible by 11.

∴ S(n) is true is for n = I

Suppose S(n) is true for n = k then

2.4^{(2n+1) }+ 3^{(3n+1)} is divisible by 11