Formula for $\sum_{k=1}^n k2^k$

Question

How can I find a formula for $$\sum_{k=1}^n k2^k$$

I'd also appreciate if someone could indicate some materials about this subject.

Answer 1

Hint:

$$\sum_{k=1}^nx^k=\frac{x(x^n-1)}{x-1}$$

$$\sum_{k=1}^nkx^{k-1}=\frac{d}{dx}\left(\frac{x(x^n-1)}{x-1}\right)$$

$$\sum_{k=1}^nkx^{k}=x\frac{d}{dx}\left(\frac{x(x^n-1)}{x-1}\right)$$

Put $x=2$.