# How to compute this sum

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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Have you seen formulas for power sums? – Dylan Moreland Oct 14 '11 at 14:27
I changed the title "How to solve this summation". – Américo Tavares Oct 14 '11 at 20:39

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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Awesome, just what I needed :) – Timo Willemsen Oct 14 '11 at 20:25
@Timo Willemsen: I changed the title of your question. Glad to help. – Américo Tavares Oct 14 '11 at 20:37
The other way to "remove the summation sign from the formula" is to just calculate the 500 summands and do the 500 additions, preferably with electronic assistance. This has the advantage of being perfectly general, in that it applies to $a+\sum_{k=m}^nf(k)$ for any $a$, $m$, $n$, and $f$, provided only that you can compute $f(k)$ for all $k$ between $m$ and $n$. – Gerry Myerson Oct 14 '11 at 21:15
@Gerry Myerson: I changed "How to solve this summation" to "How to compute this sum" because I assumed OP wanted to evaluate the given sum without computing all its terms. But your remark is quite right. – Américo Tavares Oct 14 '11 at 21:37
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. – Gerry Myerson Oct 14 '11 at 21:48