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$n^2+(n+k)^2+(n+2k)^2+\ldots+(n+mk)^2$ have a general formula?
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down vote accepted  The sum can we written as
$\sum_{j=0}^m(n+jk)^2=\sum_{j=0}^mn^2+2jkn+j^2k^2$ and using the
formulas $\sum_{j=0}^mj=\frac{m(m+1)}2$,
$\sum_{j=0}^mj^2=\frac{m(m+1)(2m+1)}6$, we get \begin{align*}\su
m_{j=0}^m(n+jk)^2&=(m+1)n^2+knm(m+1)+k^2\frac{m(m+1)(2m+1)}6\\
&=(m+1)\left(n^2+knm+\frac{k^2m(2m+1)}6\right). \end{align*}
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