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Question
Answer the following:
Prove, by method of induction, for all n ∈ N
12 + 42 + 72 + ... + (3n − 2)2 = `"n"/2 (6"n"^2 - 3"n" - 1)`
Solution
Let P(n) ≡ 12 + 42 + 72 + .... + (3n − 2)2 = `"n"/2 (6"n" − 3"n" − 1)`, for all n ∈ N
Step I:
Put n = 1
L.H.S. = 12 = 1
R.H.S. = `1/2[6(1)^2 - 3(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 + 42 + 72 + .... + (3k − 2)2
= `"k"/2(6"k"^2 - 3"k" - 1)` ...(i)
Step III:
We have to prove that P(n) is true for n = k + 1
i.e., to prove that
12 + 42 + 72 + …. + [3(k + 1) − 2]2
= `(("k" + 1))/3[6("k" + 1)^2 - 3("k" + 1) - 1]`
= `(("k" + 1))/2(6"k"^2 + 12"k" + 6 - 3"k" - 3 - 1)`
= `(("k" + 1))/2 (6"k"^2 + 9"k" + 2)`
L.H.S. = 12 + 42 + 72 + …. + [3(k + 1) − 2]2
= 12 + 42 + 72 + …. + (3k − 2)2 + (3(k + 1) − 2]2
= `"k"/2(6"k"^2 - 3"k" - 1) + (3"k" + 1)^2` ...[From (i)]
= `((6"k"^3 - 3"k"^2 - "k") + 2(9"k"^2 + 6"k" + 1))/2`
= `(6"k"^3 + 15"k"^2 + 11"k" + 2)/2`
= `(("k" + 1)(6"k"^2 + 9"k" + 2))/2`
= 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 + 42 + 72 + ... + (3n − 2)2 = `"n"/2 (6"n"^2 - 3"n" - 1)` for all n ∈ N
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