Find the value of this series what is the value of this series $$\sum_{n=1}^\infty \frac{n^2}{2^n} = \frac{1}{2}+\frac{4}{4}+\frac{9}{8}+\frac{16}{16}+\frac{25}{32}+\cdots$$
I really tried, but I couldn't, help guys?

Assuming as @Soke does, your series is $$\sum_{n=1}^\infty \frac{n^2}{2^n}$$ we can use the geometric series in order to find a closed form solution. The trick is to take derivatives and multiply by $x$.
Note that $$f(x) = \sum_{n=1}^\infty x^n = \frac{x}{1-x}$$ yields a derivative of $$\sum_{n=1}^\infty n x^{n-1} = \frac{d}{dx} \frac{x}{1-x} = \frac{1}{(1-x)^2}.$$ We multiply it by $x$ to get: $$\sum_{n=1}^\infty nx^n = \frac{x}{(1-x)^2}.$$
If we take another derivative we find $$\sum_{n=1}^\infty n^2 x^{n-1} = \frac{d}{dx} \frac{x}{(1-x)^2} = \frac{(1-x)^2+2(1-x)x}{(1-x)^4}= \frac{1+x}{(1-x)^3}.$$
Finally we multiply by $x$ again: $$\sum_{n=1}^\infty n^2 x^n = \frac{x(1+x)}{(1-x)^3}$$ and the answer you are looking for corresponds to $x=1/2$.
