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Worked solution

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

Answer

∑r=1nr2=16n(n+1)(2n+1)\sum_{r=1}^nr^2=\tfrac16n(n+1)(2n+1).

Separate and evaluate the known sums

Let Q=∑r=1nr2Q=\sum_{r=1}^nr^2. Then 3Q+3n(n+1)/2+n=(n+1)3−1=n3+3n2+3n3Q+3n(n+1)/2+n=(n+1)^3-1=n^3+3n^2+3n.

How did this part go?

Factor before dividing by three

3Q=−3n(n+1)/2+n3+3n2+2n=−3n(n+1)/2+n(n+1)(n+2)=12n(n+1)[−3+2(n+2)]=12n(n+1)(2n+1)3Q=-3n(n+1)/2+n^3+3n^2+2n=-3n(n+1)/2+n(n+1)(n+2)=\tfrac12n(n+1)[-3+2(n+2)]=\tfrac12n(n+1)(2n+1). Thus Q=16n(n+1)(2n+1)Q=\tfrac16n(n+1)(2n+1) for positive integer nn.

How did this part go?

Common mistakes

There are n copies of the constant 1; do not replace their sum by 1.

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