Porisma

A hint to get started

Match the final odd square to its summation index

Given ∑r=1n(2r+3)2=n3(4n2+24n+47)\sum_{r=1}^n(2r+3)^2=\tfrac n3(4n^2+24n+47), find 52+72+92+⋯+201125^2+7^2+9^2+\cdots+2011^2. marks

[2010/RVHS/Promo/4(modified)]

CJC · Review problems · Question 3

Solve 2n+3=2011 rather than counting all odd integers from 1.

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