Computing $\sum_{i=1}^k{i\cdot c^i}$

Question

I would like to compute the following sum : $$\sum_{i=1}^k{i\cdot c^i}$$ where $k$ and $c$ are any real numbers.

I know how to compute $i$ and $c^i$ separately but I don't know how to do it when they are multiplied together.

Thanks a lot for the help!

Answer 1

Hint:

Do some analysis: $$\sum_{i=1}^k ic^i=c\sum_{i=1}^kic^{i-1}=c\Bigl(\sum_{i=0}^k c^i\Bigr)'. $$

Answer 2

Let $Z=\sum_{i=1}^kc^i$, then (derivative with respect to $c$), $Z'=\sum_{i=1}^kic^{i-1}$. When multiplied with $c$, we get your question, i.e. we want $cZ'$. You can easily substitute $Z=\frac{1-c^{k+1}}{1-c}-1$.

Answer 3

Hint: Let $a,q$ be real numbers with $q\ne 1$: $$a + aq + aq^2 + \ldots +aq^{n-1} = a\frac{q^n-1}{q-1}.$$