The value of ∑r=115 r2r(r+2)! is equal to

r×2r(r+2)!=(r+2−2)2r(r+2)!=2r(r+1)!−2r+1(r+2)!=−2r+1(r+2)!−2r(r+1)!=−(V(r)−V(r−1))⇒ ∑r=115 r×2r(r+2)!=−(V(15)−V(0)) =−21617!−22! =1−216(17)!

The value of ∑r=115 r2r(r+2)! is equal to

r×2r(r+2)!=(r+2−2)2r(r+2)!=2r(r+1)!−2r+1(r+2)!=−2r+1(r+2)!−2r(r+1)!=−(V(r)−V(r−1))⇒ ∑r=115 r×2r(r+2)!=−(V(15)−V(0)) =−21617!−22! =1−216(17)!

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

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

$\frac{(18)! - 2^{17}}{(18)!}$

$\frac{(16)! - 2^{15}}{(16)!}$

$\frac{(15)! - 2^{14}}{(15)!}$

answer is A.

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Detailed Solution

$\begin{matrix} \frac{r \times 2^{r}}{(r + 2)!} = \frac{(r + 2 - 2) 2^{r}}{(r + 2)!} \\ = \frac{2^{r}}{(r + 1)!} - \frac{2^{r + 1}}{(r + 2)!} \\ = - (\frac{2^{r + 1}}{(r + 2)!} - \frac{2^{r}}{(r + 1)!}) \\ = - (V (r) - V (r - 1)) \end{matrix}$

$\begin{array}{l} \Rightarrow \sum_{r = 1}^{15} \frac{r \times 2^{r}}{(r + 2)!} = - (V (15) - V (0)) \\ = - (\frac{2^{16}}{17!} - \frac{2}{2!}) \\ = 1 - \frac{2^{16}}{(17)!} \end{array}$