Finding $\sum_{k=1}^n k \text{ } 2^k$

Question

I need to calculate this sum: $$\sum_{k=1}^n k \text{ } 2^k$$ I tried to meddle with double sums using $$\left ( \sum_{k=1}^n k\right )\left( \sum_{j=1}^n 2^j \right)$$ but It doesn't seem to be the most fitting approach.

Answer 1

Hint:

Compute $\;\displaystyle \sum_{k=1}^n k x^k=x\sum_{k=1}^n k x^{k-1}=x\biggl(\sum_{k=1}^n x^k\biggr)'$ instead, then substitute $2$ for $x$.