Question

The value of ${\sum }_{r=1}^{15}\frac{r{2}^r}{(r+2)!}$ is equal to

• A. $\frac{\left(17\right)!-2^{16}}{\left(17\right)!}$
• B. $\frac{\left(18\right)!-2^{17}}{\left(18\right)!}$
• C. $\frac{\left(16\right)!-2^{15}}{\left(16\right)!}$
• D. $\frac{\left(15\right)!-2^{14}}{\left(15\right)!}$

Answer 1

The correct option is A $\frac{\left(17\right)!-2^{16}}{\left(17\right)!}$

${\sum }_{r=1}^{15}\frac{r{2}^r}{(r+2)!}$

${\sum }_{r=1}^{15}\frac{(r+2-2)2^r}{(r+2)!}$

${\sum }_{r=1}^{15}\frac{\left(\right(r+2)-2)2^r}{(r+2)!}$

${\sum }_{r=1}^{15}\frac{(r+2)2^r-2^{r+1}}{(r+2)!}$

${\sum }_{r=1}^{15}\frac{(r+2)2^r}{(r+2)(r+1)!}-\frac{2^{r+1}}{(r+2)!}$

${\sum }_{r=1}^{15}\frac{2^r}{(r+1)!}-\frac{2^{r+1}}{(r+2)!}$

$\text{Pluggin the values of r=1,2,3....}$

$=(\frac{2}{2!}+\frac{2^2}{3!}+\frac{2^3}{4!}+\frac{2^4}{5!}+......)-(\frac{2^2}{3!}+\frac{2^3}{4!}+\frac{2^4}{5!}+.........+\frac{2^{16}}{17!})$

$=\frac{2}{2!}+(\frac{2^2}{3!}-\frac{2^2}{3!})+(\frac{2^3}{4!}-\frac{2^3}{4!})+(\frac{2^4}{5!}-\frac{2^4}{5!})-\frac{2^{16}}{17!}$

$=1-\frac{2^{16}}{17!}$

$=\frac{17!-2^{16}}{17!}$