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I'm trying to remove the summation sign from this formula, is this possible?

$1+\sum_{k=1}^{500}(4(2k+1)^2-12k)$

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    I cha$n$ged the title "How to solve this summation".2011-10-14

1 Answers 1

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Hints: expand $4(2k+1)^{2}-12k$ and use

$\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$

$\sum_{k=1}^{n}k^{2}=\frac{n(n+1)(2n+1)}{6}.$

You can find a proof of the second formula in this post of mine (in Portuguese).

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    Undoubtedly your assumption was correct. My comment was mostly made in jest, though perhaps it will alert OP to the importance of wording problems carefully when there are mathematicians in the audience.2011-10-14