Find the sum _ r=1 ^n(r+1)(2 r-1). - Vidyadip

Question

Find the sum $\sum_{r=1}^n(r+1)(2 r-1)$.

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Answer

$\begin{aligned} & \sum_{r=1}^n(r+1)(2 r-1) \\ & =\sum_{i=1}^n\left(2 r^2+r-1\right) \\ & =2 \sum_{r=1}^n r^2+\sum_{r=1}^n r-\sum_{r=1}^n 1 \\ & =2 \cdot \frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}-n \\ & =\frac{n}{6}\left[2\left(2 n^2+3 n+1\right)+3(n+1)-6\right] \\ & =\frac{n}{6}\left(4 n^2+6 n+2+3 n+3-6\right) \\ & =\frac{n}{6}\left(4 n^2+9 n-1\right)\end{aligned}$

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