Question

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

• A. $7782$
• B. $7787$
• C. $7790$
• D. $7780$

Answer 1

Answer: (D)

$7780$

Let $S=\sum _{r=16}^{30}(r+2)(r-3)=\sum _{r=1}^{30}(r+2)(r-3)-\sum _{r=1}^{15}(r+2)(r-3)$

$=\sum _{r=1}^{30}(r^2-r-6)-\sum _{r=1}^{15}(r^2-r-6)$

[By using ${\sum }_{i=1}^n{k}^2=1^2+2^2+3^2+\dots +n^2=\frac{n(n+1)(2n+1)}{6}$]

$={(\frac{r(r+1)(2r+1)}{6}-\frac{r(r+1)}{2}-6r)}_{r=30}-{(\frac{r(r+1)(2r+1)}{6}-\frac{r(r+1)}{2}-6r)}_{r=15}$

$=(\frac{30(30+1)(60+1)}{6}-\frac{30(30+1)}{2}-180)-(\frac{15(15+1)(30+1)}{6}-\frac{15(15+1)}{2}-90)$

$=7780$