partial sum of $\sum_{i=1}^k \frac{i}{a^i}$

Question

From "How to find the partial sum of n/2n?" we know that $S_n=2\left(1-\frac1{2^n}-\frac n{2^{n+1}}\right)$ so how can we prove that $\sum_{i=1}^k \frac{i}{a^i} = \frac{(a^{(k+1)}-a(k+1)+k)}{a^k(a-1)^2}$

Answer 1

$$S_n=~~r+2r^2+3r^3+4r^4+...+nr^n~~~~(1)$$ $$rS_n=~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~r^2+2r^3+3r^4+...+(n-1)r^n+nr^{n+1}~~~~(2)$$Suntrcting (2) from (1), we get $$(1-r)S_n=r+r^2+r^3+r^4+...r^n-nr^{n+1}$$ $$(1-r)S=r[1+r+r^2+r^3+...+r^{n-1}]-nr^{n+1}$$ $$(1-r)S_n=r\frac{1-r^n}{1-r}-nr^{n+1}$$ $$S_n=\frac{r-(n+1)r^{n+1}+nr^{n+2}}{(1-r)^2}$$ Let $r=1/a$, then $$S_n=\frac{a^{n+1}-(n+1)a+n}{(1-a)^2a^n}=\sum_{k=1}^{n} \frac{k}{a^k}$$