I want to show that the geometric series $$1+r+r^2 + \cdots + r^n = \frac{r^{n+1} - 1}{r-1}, r \neq 1$$
I first started with the following using Gauss method:
$$S = 1+r+r^2 + \cdots + r^n$$
$$S=r^n + r^{n-1} + r^{n-2} + ... + 1$$
Adding both S's:
$$2S = (1+r^n) + (r+r^{n-1})+(r^2+r^{n-2})+...+(1+r^n)$$
Am I on the right track?
PS: I initially tried to follow this question, however, there was not enough to go on, so I attempted to do this on my own.
Update:
I originally assumed that I can use Gauss method which proved (when he was 10 years old!) $$\sum_{k=1}^nk= \frac{n(n+1)}{2}$$
but from user comments below, Gauss method cannot be applied here, as my questions is regarding a geometric sum not an arithmetic sum.
If $S = 1 + r + r^2 + \ldots + r^n$ then $rS$ = $r+r^2 + \ldots + r^n + r^{n+1}$, so you just have to subtract them.
We are trying to find a formula for $S_n=\sum_{k=0}^{n}r^k$, where $r \neq 1$. We have: $$S_n+r^{n+1}=1+\sum_{k=1}^{n+1}r^k=1+\sum_{k=0}^{n}r^{k+1}= 1+r\sum_{k=0}^{n}r^{k}=1+rS_n,$$ so $(q-1)S_n=q^{n+1}-1$ and $S_n=\frac{q^{n+1}-1}{q-1}$.
Here is another example of this method: we want to find out what $G_n=\sum_{k=0}^{n}k2^k$ really is. It turns out that
\begin{align} L & = G_n+(n+1)2^{n+1}=0\cdot 2^0+\sum_{k=1}^{n+1}k2^k = \sum_{k=0}^n(k+1)2^{k+1} \\ & =2 \sum_{k=0}^n k 2^k + 2 \sum_{k=0}^n 2^k = 2G_n + 2(2^{n+1}-1), \end{align} which we can transform into $$G_n=(n+1)2^{n+1}-2\cdot (2^{n+1}-1)=(n-1)2^{n+1}+2.$$