Find the closed form of $$\sum_{k=1}^{n-1} k a^k $$
I was curious of this value and thought of two ways to accomplish it. One is essentially Euler's method for finding $\sum_{k=1}^n k^2$, the other is purely algebraic and which I came up with on my own after a lot of messing around. I chose to sum to $n-1$ rather than $n $ because I knew the geometric sum's closed form shifted the power up by one, and I figured this would be related. Moreover, I can always add on the $n^{th}$ term later.
Method 1: $$\sum_{k=0}^{n-1} k a^k = n\sum_{k=0}^{n-1} a^k - \sum_{k=0}^{n-1} (n-k)a^k$$ The left sum (on the RHS of the equality) is just the geometric sum with value: $$n\sum_{k=0}^{n-1} a^k = n\frac{a^{n}-1}{a-1}$$ The other sum can be expressed using a double sum: $$\sum_{k=0}^{n-1} (n-k) a^k = n + (n-1)a + (n-2)a^2 + ... + a^{n-1} $$
$$= 1 + (1 + a) + (1+a+a^2)+...+(1+a+a^2+...+a^{n-1})$$
$$ = \sum_{k=0}^{n-1} \sum_{m = 0}^k a^m = \sum_{k=0}^{n-1} (\frac{a^{k+1}-1}{a-1}) = \frac{a}{a-1} (\sum_{k=0}^{n-1}a^k) - \frac{n}{a-1}$$ So $$\sum_{k=0}^{n-1} (n-k)a^k = \frac{a}{a-1}(\frac{a^n-1}{a-1}) -\frac{n}{a-1}$$ Therefore, $$\sum_{k=0}^{n-1} ka^k =n\frac{a^{n}-1}{a-1} -\frac{a}{a-1}(\frac{a^n-1}{a-1}) +\frac{n}{a-1} = n\frac{a^n}{a-1} - \frac{a}{a-1}(\frac{a^n-1}{a-1})$$
Method 2: $$\sum_{k=0}^{n-1} a^k = \frac{a^n-1}{a-1}$$ Differentiating both sides with respect to $a$ and then multiplying by $a$ we reach: $$\sum_{k=1}^{n-1} ka^k = a \frac{d}{da}(\frac{a^n-1}{a-1}) = a (\frac{na^{n-1}(a-1)-(a^n-1)}{(a-1)^2}) =n \frac{a^n}{a-1}-\frac{a}{a-1}( \frac{a^{n}-1}{a-1})$$ Which is an identical result to Method 1.
My main motivation for not using the arguably more simple approach (method 2) is that I wanted something which relied only on math learned through pre-calculus (though if alternate proofs using any form of higher math exist, I'm quite open to it).
Disclaimer: I can't guarantee any sort of originality in method 1. The fact that I came up with it on my own does not imply it hasn't been done that way before (in fact, I feel it likely has, I just couldn't find it anywhere).
If anyone has any alternative methods to reaching this result, I'd love to see them!
This is an "arithmetic-geometric progression". Methods for summing this, and the more general form $\sum(rk+s)a^k$, used to be taught at school.
Let $S=\sum_{k=1}^{n-1}k a^k$ (why $n-1$ rather than $n$?!). Then $$(a-1)S=\sum_{k=1}^{n-1} k a^{k+1}-\sum_{k=1}^{n-1} k a^k =\sum_{k=2}^{n} (k-1) a^{k}-\sum_{k=1}^{n-1} k a^k =(n-1)a^n-\sum_{k=1}^{n-1}a^k$$ which reduces it to a GP.