# How to solve this summation

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!

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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.

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I thought that maybe you could form a telescopic sum but I couldn't figure it out! Thanks! – Csmity Nov 23 '13 at 19:39
@Csmity The reason why I thought about this solution is that you had $k+2$ in the denominator and $k$ at the top. Their difference is exactly the $2$ in $2^k$, thus it was natural to write $k+2-k=2$ or $k=(k+2)-2$ and see what we get.... – N. S. Nov 23 '13 at 19:41

## 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:

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

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