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Question
Prove by method of induction, for all n ∈ N:
12 + 32 + 52 + .... + (2n − 1)2 = `"n"/3 (2"n" − 1)(2"n" + 1)`
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Solution
Let P(n) ≡ 12 + 32 + 52 + .... + (2n − 1)2 = `"n"/3 (2"n" − 1)(2"n" + 1)`, for all n ∈ N
Step I:
Put n = 1
L.H.S. = 12 = 1
R.H.S. = `1/3[2(1) - 1][(2(1) + 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
∴ 12 + 32 + 52 + .... + (2k − 1)2
= `"k"/3(2"k" - 1) (2"k" + 1)` ...(i)
Step III:
We have to prove that P(n) is true for n = k + 1
i.e., to prove that
12 + 32 + 52 + …. + [2(k + 1) − 1]2
= `(("k" + 1))/3[2("k" + 1) - 1][2("k" + 1) + 1]`
= `(("k" + 1)(2"k" + 1)(2"k" + 3))/3`
L.H.S. = 12 + 32 + 52 + …. + [2(k + 1) − 1]2
= 12 + 32 + 52 + …. + (2k − 1)2 + (2k + 1)2
= `"k"/3(2"k" - 1)(2"k" + 1) + (2"k" + 1)^2` ...[From (i)]
= `(2"k" + 1)[("k"(2"k" - 1))/3 + (2"k" + 1)]`
= `(2"k" + 1)[(2"k"^2 - "k" + 6"k" + 3)/3]`
= `((2"k" + 1))/3(2"k"^2 + 2"k" + 3"k" + 3)`
= `((2"k" + 1))/3[2"k"("k" + 1) + 3("k" + 1)]`
= `((2"k" + 1)("k" + 1)(2"k" + 3))/3`
= 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 + 32 + 52 + .... + (2n − 1)2 = `"n"/3 (2"n" − 1)(2"n" + 1)`, for all n ∈ N.
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