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Question
Prove by method of induction, for all n ∈ N:
13 + 33 + 53 + .... to n terms = n2(2n2 − 1)
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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]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.
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