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.

See e.g. the answers here – Robert Israel Oct 07 '18 at 22:35 — Robert Israel, Oct 07 '18 at 22:35

score 3 · Accepted Answer · answered Oct 07 '18 at 22:32

3

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 $x$.

answered Oct 07 '18 at 22:32

Bernard

175,478

Thanks for the hint, I was able to continue from there. – fuzzypixelz Oct 07 '18 at 22:40
Glad to have been helpful! – Bernard Oct 07 '18 at 22:42

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

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