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So I have a problem that looks like this:

Evaluate $\displaystyle\sum\limits_{r=1}^{25} 4r^2-2r+2$ using the Sum of Squares: $\displaystyle\sum\limits_{k=1}^n \frac{n(n+1)(2n+1)}{6}$.

I really have no idea what I'm supposed to do with this.

I saw this page on About.com regarding the some of squares, as well as a few others, but I still don't get it.

The first one uses variable $r$, but the second one uses $k$ and $n$

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Decompose it:

$$\sum\limits_{r=1}^{25} 4r^2-2r+2=4\sum\limits_{r=1}^{25}r^2-2\sum\limits_{r=1}^{25}r+\sum\limits_{r=1}^{25}2$$

Then apply the formulas for $\displaystyle\sum\limits_{r=1}^nr^2$ and $\displaystyle\sum\limits_{r=1}^nr$. It may help to know that you've misstated the sum-of-squares formula. The correct form is

$$\sum\limits_{r=1}^nr^2=\frac{n(n+1)(2n+1)}{6}$$

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  • $\begingroup$ See edit -- I'm a little confused because it actually said it on my sheet as k. I misstated in the last sentence, saying it was r (which I fixed). Was the one listed on my sheet wrong as well? $\endgroup$ – evamvid May 8 '14 at 11:39
  • $\begingroup$ Also, I don't understand where the some of squares is actually used here. Everything I see is coming from the first side here... $\endgroup$ – evamvid May 8 '14 at 11:42
  • $\begingroup$ @evamvid: It doesn't matter what you choose to name your variables. $k$, $r$, $potato$, whatever, it makes no difference. $\endgroup$ – user2357112 May 8 '14 at 17:38
  • $\begingroup$ Okay...I think I got it. Thanks! $\endgroup$ – evamvid May 9 '14 at 1:57

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