Find _ r=1 ^n (3 r^2-2 r+1 )

Question

Find $\sum_{r=1}^n\left(3 r^2-2 r+1\right)$

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Answer

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

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