How to solve this summation: $\sum_{k=1}^{n}\frac{k2^{k}}{(k+2)!}$

Question

How do I solve this summation?

I have tried simplifying k on the numerator with the factorial in the denominator but it just gets me nowhere.

$$\sum_{k=1}^{n}\frac{k2^{k}}{(k+2)!}$$

Thanks!

Answer 1

Hint

$$\frac{k2^k}{(k+2)!}=\frac{(k+2-2)2^k}{(k+2)!}=\frac{(k+2)2^k}{(k+2)!}-\frac{2\cdot2^k}{(k+2)!}=\frac{2^k}{(k+1)!}-\frac{2^{k+1}}{(k+2)!}$$

You get a telescopic sum.

Answer 2

Using the Hint given by N.S.:

$$\sum_{k=1}^{n}\frac{k2^{k}}{(k+1)!}=\sum_{k=1}^{n}\frac{(k+2-2)2^{k}}{(k+2)!}=\sum_{k=1}^{n}\frac{2^k}{(k+1)!}-\frac{2^{k+1}}{(k+2)!}$$

Using telescopic sum we get:

$$\frac{2}{2!}-\frac{2^{n+1}}{(n+2)!}=1-\frac{2^{n+1}}{(n+2)!}$$