# How to produce the closed form of finite summations of the form $\sum_{i=0}^n ia^i$.

I am having a hard time reasoning about how to produce the closed form of summations with the form: $$\sum_{i=0}^n ia^i$$

How do I produce the closed form for summations like this and why? This question seems relevant and suggests that I might start by transforming the summation into a double summation, but I'm a math baby and I don't really understand the reasoning involved as much as I'd like.

Thanks a bunch!

It is possible to evaluate it by converting it into a double summation and reversing the order of summation:

\begin{align*} \sum_{k=0}^nka^k&=\sum_{k=1}^nka^k\\ &=\sum_{k=1}^n\sum_{j=1}^ka^k\tag{1}\\ &=\sum_{1\le j\le k\le n}a^k\tag{2}\\ &=\sum_{j=1}^n\sum_{k=j}^na^k\tag{3}\\ &=\sum_{j=1}^n\frac{a^j-a^{n+1}}{1-a}\\ &=\sum_{j=1}^n\frac{a^j}{1-a}-\sum_{j=1}^n\frac{a^{n+1}}{1-a}\\ &=\frac1{1-a}\sum_{j=1}^na^j-\frac{na^{n+1}}{1-a}\\ &=\frac1{1-a}\cdot\frac{a-a^{n+1}}{1-a}-\frac{na^{n+1}}{1-a}\\ &=\frac{a-a^{n+1}}{(1-a)^2}-\frac{na^{n+1}}{1-a} \end{align*}

The summations $(1),(2)$, and $(3)$ are just three different ways of writing a sum over all pairs of indices $j$ and $k$ satisfying the inequalities $1\le j\le k\le n$.

If you examine this closely, you’ll see that it’s really just what bof did in his answer: he split $ka^k$ into $k$ copies of $a^k$ arranged in a column, and then instead of summing each column to get the sums $a^1,2a^2,\ldots,na^n$ and then adding those sums, he summed the rows, which were geometric series and hence easy to sum, and then added these row sums. In other words, he reversed the order of summation, just as I did. His is the presentation that I’d use to introduce the idea; mine shows how it can be expressed more compactly and how you might see it written if some familiarity with summation notation is assumed (though I did include a few extra steps than usual for greater clarity).

• Thanks a bunch! All three were great answers but this one had enough English explanation for me to get a grip on the reasoning for the others. – Nick Oct 1 '16 at 21:24
• @Nick: You’re welcome! – Brian M. Scott Oct 1 '16 at 21:26

Your sum is $$a \sum_{i=0}^n i a^{i-1} = a \dfrac{d}{da} \sum_{i=0}^n a^i$$ Now use the formula for a geometric series.

$$a+a^2+a^3+\cdots+a^n=\frac{a-a^{n+1}}{1-a}$$ $$\ \ \ \ \ \ \ a^2+a^3+\cdots+a^n=\frac{a^2-a^{n+1}}{1-a}$$ $$\ \ \ \ \ \ \ \ \ \ \ \ \ \ a^3+\cdots+a^n=\frac{a^3-a^{n+1}}{1-a}$$ $$\dots$$ $$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a^n=\frac{a^n-a^{n+1}}{1-a}$$ Add them up: $$a+2a^2+3a^3+\cdots+na^n=\frac{a+a^2+a^3+\cdots+a^n}{1-a}-\frac{na^{n+1}}{1-a}=\boxed{\frac{a-a^{n+1}}{(1-a)^2}-\frac{na^{n+1}}{1-a}}$$