Is it possible to find the sum $\Sigma_{x=1}^n ((2x)(4x+1))$? If yes then can somebody please explain for me the formula?
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There are well-known formulas for $\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}6$, $\sum_{k=1}^n k=\frac{n(n+1)}{2}$, $\sum_{k=1}^n 1=n$, from which you can combine your result.
Hagen von Eitzen
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