How to calculate $\sum_{k=1}^n k*2^{n-k}$

Question

Is it possible to calculate this sum in such a way that it only depends on n? $\sum_{k=1}^n k*2^{n-k}$

Answer 1

Hint The derivative of $x^{k}$ of $kx^{k-1}$

Can you continue from here?

Answer 2

Take your sum and multiply by $2^{-n}$. Then you have $$ \sum_{k=1}^n \dfrac{k}{2^k}= \sum_{k=0}^n \dfrac{k}{2^k} $$ There are many ways to compute this sum. For example, you could use Taylor Series. For a few ways, see this page or this page. Then simply multiply your final formula by $2^n$ to 'undo' the $2^{-n}$ you multiplied by in the beginning.