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Q. The value of $ \displaystyle \sum^{30}_{r = 16} (r + 2)(r - 3)$ is equal to :

JEE MainJEE Main 2015Sequences and Series

Solution:

$\displaystyle \sum_{r=16}^{30}(r+2)(r-3)$
$= \displaystyle \sum_{r=16}^{30}\left(r^{2}-r-6\right)=\left( \displaystyle \sum_{r=1}^{30} r^{2}- \displaystyle \sum_{r=1}^{15} r^{2}\right)\left(\sum_{i=1}^{30} r-\sum_{r=1}^{15} r\right)-6 \displaystyle \sum_{r=16}^{30} 1$
$=\left[\frac{(30)(31)(61)}{6}-\frac{(15)(16)(31)}{6}\right]-\left[\frac{(30)(31)}{2}-\frac{(15)(16)}{2}\right]-6(15)$
$=[(5)(31)(61)-1240]-[465-120]-90$
$=[7780]$