# Find ine indicated sum. sum_{i=1}^9i^{2}+2

Find ine indicated sum.
$\sum _{i=1}^{9}{i}^{2}+2$
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Note that
$\sum _{i=1}^{9}{i}^{2}+2=\sum _{i=1}^{9}{i}^{2}+\sum 2\left[\because \sum an+bn=\because \sum an+\because \sum bn\right]$
Recall the formula: $\sum _{k=1}^{9}{k}^{2}=1/6n\left(n+1\right)\left(2n+1\right)$
$=\left[1/6\cdot 9\left(9+1\right)\left(2\cdot 9+1\right)\right]+2\sum _{i=1}^{91}$
$=\left(285\right)+2\cdot 9=285+18=303$
$\sum _{i=1}^{9}\left(\left({i}^{2}\right)+2\right)=303$