Question

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

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

Answer 1

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 t_n be the n^th term

∴ t_n = 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]³ 

= 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.