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