1. ## Obtain an expression

Given that $\sum^{n}_{r=1}r^2=\frac{1}{6}n(n+1)(2n+1)$ , obtain an expression , in its simplest term , for

$
\sum^{2n}_{r=n+1}(2r-1)^2
$

Thanks .

2. $
\begin{gathered}
\sum\limits_{r = n + 1}^{2n} {\left( {2r - 1} \right)^2 } = \sum\limits_{r = 1}^n {\left( {2(r + n) - 1} \right)^2 } = \sum\limits_{r = 1}^n {\left( {2r + 2n - 1} \right)^2 } = \hfill \\
\sum\limits_{r = 1}^n {4r^2 + 4n^2 + 1 + 8rn - 4r - 4n} = 4\sum\limits_{r = 1}^n {r^2 } + \left( {4n^2 - 4n + 1} \right)\sum\limits_{r = 1}^n 1 + \left( {8n - 4} \right)\sum\limits_{r = 1}^n r \hfill \\
\end{gathered}
$

Can you continue?

3. You may also rewrite it as $\sum_{k=n+1}^{2n}(2r-1)^2=\sum_{k=1}^{2n}(2r-1)^2-\sum_{k=1}^{n}(2r-1)^2$, expand the cuadratic binomy and apply your formula.