Given that $\sum_{r=1}^{n} r^2=\frac{n}{6}(n+1)(2n+1)$, find $\sum_{r=1}^{n}(4r+1)(6r+1)$ i

Question

Given that $\sum_{r=1}^{n} r^2=\frac{n}{6}(n+1)(2n+1)$, find $\sum_{r=1}^{n}(4r+1)(6r+1)$ in terms of $n$.[3] Hence find $\sum_{r=8}^{2n}(4r+1)(6r+1)$ in terms of $n$.[3]

Accepted Answer

$8n^3+17n^2+10n$; $64n^3+68n^2+20n-3647$

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