# Prove by method of induction, for all n ∈ N: 13 + 33 + 53 + .... to n terms = n2(2n2 − 1) - Mathematics and Statistics

Sum

Prove by method of induction, for all n ∈ N:

13 + 33 + 53 + .... to n terms = n2(2n2 − 1)

#### Solution

Let P(n) ≡ 13 + 33 + 53 + .... to n terms = n2(2n2 − 1), for all n ∈ N

But 1, 3, 5, …… are in A.P.

∴ a = 1, d = 2

Let tn be the nth term

∴ tn = a + (n − 1)d = 1 + (n − 1)2 = 2n − 1

∴ P(n) ≡ 13 + 33 + 53 + …. + (2n − 1)3 = n2(2n2 − 1)

Step I:

Put n = 1

L.H.S. = 13 = 1

R.H.S. = 12 [2(1)2 − 1] = 1 = L.H.S.

∴ P(n) is true for n = 1.

Step II:

Let us consider that P(n) is true for n = k

∴ 13 + 33 + 53 + … + (2k − 1)3 = k2(2k2 − 1)  …(i)

Step III:

We have to prove that P(n) is true for n = k + 1

i.e., to prove that

13 + 33 + 53 + .… + [2(k + 1) − 1]3

= (k + 1)2 [2(k + 1)2 – 1]

= (k2 + 2k + 1) (2k2 + 4k + 1)

L.H.S. = 13 + 33 + 53 + .… + [2(k + 1) − 1]

= 13 + 33 + 53 + … + (2k − 1)3 + (2k + 1)3

= k2 (2k2 − 1) + (2k + 1)3   …[From (i)]

= 2k4 − k2 + 8k3 + 12k2 + 6k + 1

= 2k4 + 8k3 + 11k2 + 6k + 1

= 2k2 (k2 + 2k + 1) + 4k3 + 9k2 + 6k + 1

= 2k2 (k2 + 2k + 1)+ 4k (k2 + 2k + 1) + (k2 + 2k + 1)

= (k2 + 2k + 1) (2k2 + 4k + 1)

= R.H.S.

∴ P(n) is true for n = k + 1

Step IV:

From all steps above by the principle of mathematical induction, P(n) is true for all n ∈ N.

∴ 13 + 33 + 53 + .... to n terms = n2(2n2 − 1) for all n ∈ N.

Concept: Principle of Mathematical Induction
Is there an error in this question or solution?

#### APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 4 Methods of Induction and Binomial Theorem
Exercise 4.1 | Q 5 | Page 73