The value of _r=16^30(r+2)(r-3) is equal to:

Question

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

Accepted Answer

$$\sum_{r=16}^{30}(r^2-r-6)=\sum_{r=16}^{30}r^2-\sum_{r=16}^{30}r-\sum_{r=16}^{30}6.$$

$\sum_{r=16}^{30}r^2=\sum_{1}^{30}r^2-\sum_{1}^{15}r^2=9455-1240=8215$; $\sum_{r=16}^{30}r=\dfrac{(16+30)(15)}{2}=345$; $\sum 6$ over 15 terms $=90$.

$8215-345-90=7780.$

Solutions · 1