# Formula for $r+2r^2+3r^3+...+nr^n$ [duplicate]

Is there a formula to get $r+2r^2+3r^3+\dots+nr^n$ provided that $|r|<1$? This seems like the geometric "sum" $r+r^2+\dots+r^n$ so I guess that we have to use some kind of trick to get it, but I cannot think of a single one. Can you please help me with this problem?

• If you take it to infinity then yes. May 6, 2015 at 0:02
• This is surely a duplicate. May 6, 2015 at 0:06
• @Gamamal There is a well-known formula for the finite sum, too. May 6, 2015 at 0:06
• You know the sum of a finite geometric progression. Differentiate and multiply by $r$ May 6, 2015 at 0:07
• Another way: Knowing that $(n+2)-2(n+1)+(n)=0$, calculate $S-2S+S$ in a clever way. May 6, 2015 at 0:09

We have $$1+r+r^2 + \cdots + r^n = \dfrac{r^{n+1}-1}{r-1}$$ Differentiating this once, we obtain $$1+2r + 3r^2 + \cdots + nr^{n-1}= \dfrac{nr^{n+1}-(n+1)r^n + 1}{(r-1)^2}$$ Multiply the above by $r$ to obtain what you want.

• Note that this works even if $|r| \ge 1$ May 6, 2015 at 0:43

HINT: $r+2r^2+3r^3+... +nr^n=(r+r^2+\dots+r^n)+(r^2+r^3+\dots+r^n)+\dots+(r^n)$ and compute values in parentheses.

• Oh dayum, this is nice. May 6, 2015 at 0:09
• @Gamamal Thank you! Without differentiation it is possible to understand on a lower level of mathematical education. May 6, 2015 at 0:12

You can use the following approach: $$\sum_{k=1}^n k r^k = r\sum_{k=1}^n k r^{k-1} = r\left(\sum_{k=1}^n r^{k}\right)^\prime = r\left(r\frac{1-r^n}{1-r}\right)^\prime$$

What I am thinking of is this: $$r+2r^2+3r^3+...+nr^n = \sum_{k=1}^{n}r^k + \sum_{k=2}^nr^k + ... \sum_{k=n}^nr^k$$ Each is a geometric series, so the sum would be $$\frac{1-r^n}{1-r} + r\frac{1-r^{n-1}}{1-r} + ... + r^n$$.

Hope this helps.

\begin{align*}r+2r^2+3r^3+\ldots+nr^n&=r(1+2r+3r^2+\ldots+nr^{n-1})=r(1+r+r^2+\dots+r^n)'\\&=r\biggl(\frac{1-r^{n+1}}{1-r}\biggl)'=r\,\frac{nr^{n+1}-(n+1)r^n+1}{(1-r)^2}. \end{align*}