Porisma

Worked solution

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

Answer

13574772761357477276.

Find the last index and substitute

The first term is (2(1)+3)2=52(2(1)+3)^2=5^2. From 2n+3=20112n+3=2011, n=1004n=1004. Therefore the sum is 10043[4(1004)2+24(1004)+47]=1357477276\tfrac{1004}3[4(1004)^2+24(1004)+47]=1357477276.

How did this part go?

Common mistakes

The first term is 5 squared, so the sequence does not begin with r=0.

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