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Question
Prove by method of induction, for all n ∈ N:
12 + 22 + 32 + .... + n2 = `("n"("n" + 1)(2"n" + 1))/6`
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Solution
Let P(n) ≡ 12 + 22 + 32 + .... + n2 = `("n"("n" + 1)(2"n" + 1))/6`, for all n ∈ N
Step I:
Put n = 1
L.H.S. = 12 = 1
R.H.S. = `(1(1 + 1)[2(1) + 1])/6 = 6/6` = 1
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1
Step II:
Let us consider that P(n) is true for n = k
∴ 12 + 22 + 32 + .... + k2 = `("k"("k" + 1)(2"k" + 1))/6` ...(i)
Step III:
We have to prove that P(n) is true for n = k + 1
i.e., to prove that
12 + 22 + 32 + …. + (k + 1)2
= `(("k" + 1)[("k" + 1) + 1][2("k" + 1) + 1])/6`
= `(("k" + 1)("k" + 2)(2"k" + 3))/6`
L.H.S. = 12 + 22 + 32 + …. + (k + 1)2
= 12 + 22 + 32 + …. + k2 + (k + 1)2
= `("k"("k" + 1)(2"k" + 1))/6 + ("k" + 1)^2` ...[From (i)]
= `("k" + 1)[("k"(2"k" + 1))/6 + ("k" + 1)]`
= `("k" + 1)[(2"k"^2 + "k" + 6"k" + 6)/6]`
= `(("k" + 1)(2"k"^2 + 7"k" + 6))/6`
= `(("k" + 1)("k" + 2)(2"k" + 3))/6`
= 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.
∴ 12 + 22 + 32 + .... + n2 = `("n"("n" + 1)(2"n" + 1))/6` for all n ∈ N.
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