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Recover the square-sum formula from a cubic identity

Given ∑r=1n(3r2+3r+1)=(n+1)3−1\sum_{r=1}^n(3r^2+3r+1)=(n+1)^3-1, show ∑r=1nr2=16n(n+1)(2n+1)\sum_{r=1}^nr^2=\tfrac16n(n+1)(2n+1). marks

[2013/SAJC/II/3(a)(modified)]

CJC · Review problems · Question 2

Use the arithmetic sum and the number of constant terms to isolate the square sum.

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