Calculate sum of a series using telescopy $$A=\sum_{n=1}^\infty \frac{2^n n} {(n+2)!} $$

By using induction, one can show that $A = 1$ but I don't know how to solve this problem by using other methods.
Hint:

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

Now, use telescopy.
As written by Prasun Biswas you have:

$$A=\sum_{n=1}^\infty \frac{2^n n} {(n+2)!}$$

Observe that:

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

So your sum becomes:

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

Which is:

$$A=1-\frac 23+\frac 23-\frac 13+\frac 13-...=1$$