Suppose we want to find the sum:
\[\begin{align} S &= \sum_{i=1}^n ir^i \\ S &= 1r^1 + 2r^2 +3r^3 + \ldots + nr^n \end{align}\]
If we multiply both sides by \(r\):
\[rS = 1r^2 + 2r^3 +3r^4 + \ldots + r^{n+1} \]
Let's subtract the second equation from the first:
\[S - rS = (1r^1 + 2r^2 +3r^3 + \ldots + r^n) - (1r^2 + 2r^3 +3r^4 + \ldots + nr^{n+1}) \]
We can put all the like terms together:
\[\begin{align} S - rS &= r + (2r^2 - 1r^2) + (3r^3 - 2r^3) + \ldots + (nr^n - (n-1)r^n) - nr^{n+1} \\ (1 - r) S &= (r + r^2 + r^3 + \ldots + r^n) - nr^{n+1} \end{align}\]
The terms in the brackets form a geometric series. We can use the formula for the sum of a finite geometric here:
\[\begin{align} (1 - r) S &= \frac{r(r^n - 1)}{r-1} - nr^{n+1} \\ (1 - r) S &= \frac{r(1-r^n)}{1-r} - nr^{n+1} \end{align}\]
Rearranging:
\[S = \frac{r(1-r^n)}{(1-r)^2} - \frac{nr^{n+1}}{1-r}\]
If the series ends at \(n-1\) instead of \(n\), then the sum would be:
\[S = \frac{r(1-r^{n-1})}{(1-r)^2} - \frac{(n-1)r^n}{1-r}\]