Simple way to compute the finite sum $\sum\limits_{k=1}^{n-1}k\cdot x^k$ I'm looking for an elementary method for computing a finite geometric-like sum,
$$\sum_{k=1}^{n-1} k\cdot3^k$$
I have a calculus-based solution: As a more general result, I replace $3$ with $x$ and denote the sum by $f(x)$. Then
$$f(x) = \sum_{k=1}^{n-1} k\cdot x^k = x\sum_{k=1}^{n-1}k\cdot x^{k-1} = x\frac{\mathrm d}{\mathrm dx}\left[\sum_{k=1}^{n-1}x^k + C\right]$$
for some constant $C$. I know that
$$\sum_{k=1}^{n-1}x^k = \frac{x-x^n}{1-x}$$
so it follows that
$$\begin{align}
f(x) &= x\frac{\mathrm d}{\mathrm dx}\left[\frac{x-x^n}{1-x}+C\right] \\[1ex]
&= x\cdot\frac{(1-x)\left(1-nx^{n-1}\right) + x-x^n}{(1-x)^2} \\[1ex]
&= \frac{x-nx^n(1-x)-x^{n+1}}{(1-x)^2}
\end{align}$$
which means the sum I started with is
$$\sum_{k=1}^{n-1}k\cdot3^k = \frac{3+2n\cdot3^n-3^{n+1}}4$$
I am aware of but not particularly adept with summation by parts, and I was wondering if there was another perhaps simpler method I can employ to get the same result?

Hint:
It suffices to notice that
$$S:=\sum_{k=1}^{n-1} k\cdot3^k=\sum_{k=0}^{n-2} (k+1)\cdot3^{k+1}=3\sum_{k=1}^{n-1} (k+1)\cdot3^k+3-n\cdot3^n=3S+\text{terms}.$$
Now subtract the RHS from three times the LHS (which gives you twice the LHS), simplify and fiddle with the terms.
