Question

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

• A. 7785
• B. 7780
• C. 7775
• D. 7770

Answer 1

Answer: (B)

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