Question

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

• A. $\frac{\left(17\right)!-{12}^{16}}{\left(17\right)!}$
• 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

Answer: (A)

$\frac{\left(17\right)!-{12}^{16}}{\left(17\right)!}$

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

$=\sum \frac{(r+2).2^r}{(r+2)!}-\sum \frac{2.2^r}{(r+2)!}$

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

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

$=[\frac{2}{2!}-\frac{2^2}{3!}+\frac{2^2}{3!}-\frac{2^3}{4!}+......+\frac{2^{15}}{16!}-\frac{2^{16}}{17!}]$

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

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

$=\frac{\left(17\right)!-2^{16}}{\left(17\right)!}$

Hence, the answer is $=\frac{\left(17\right)!-2^{16}}{\left(17\right)!}.$