# Computing $\sum_{i=1}^k{i\cdot c^i}$ [duplicate]

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!

## marked as duplicate by Hans Lundmark, Community♦Jan 27 at 21:06

• Welcome to Maths SX! I suppose you mean you want to compute this sum? – Bernard Jan 27 at 18:23
• Oh yes thank you ! – Louis-Simon Cyr Jan 27 at 19:31

Do some analysis: $$\sum_{i=1}^k ic^i=c\sum_{i=1}^kic^{i-1}=c\Bigl(\sum_{i=0}^k c^i\Bigr)'.$$
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 $$Z=\frac{1-c^{k+1}}{1-c}-1$$.
Hint: Let $$a,q$$ be real numbers with $$q\ne 1$$: $$a + aq + aq^2 + \ldots +aq^{n-1} = a\frac{q^n-1}{q-1}.$$
• This is a true fact, but it is irrelevant to the question. OP is trying to compute the sum $$1c^1 + 2 c^2 + 3 c^3 + \cdots + k c^k$$ The coefficients on the powers of $c$ are not constant. – Blue Jan 27 at 20:15