Induction on nth polynomial proof. Question: Prove by induction that $ 1+r+r^2+\cdots+r^n = \dfrac {1-r^{n+1}} {1-r} $ where $ r \in \mathbb{R} $ 
When $n$ is odd, this is really easy as the right side breaks down to $\dfrac {(1-r^{n+1})(r+1)} {1-r}$ . Then you can fiddle with the first n variables on the left: put them into the same fraction and add; a buncha stuff cancels out and eventually the lhs breaks down into the rhs. 
The problems arise when $n$ is even. I can't seem to find a way to split the right side. Any ideas where to start? Or a way to do everything in one step?
EDIT: from the hint $$r^{n+1} + \frac{1-r^{n+1}}{1-r} = \frac{(1-r)r^{n+1} + (1-r^{n+1})}{1-r}$$
We get $  \dfrac{(1-r^{n+2} -r^{n+1}) + r^{n=1}}{1-r} =\dfrac {1-r^{n+2}} {1-r} $ hence the LHS turns into the right thanks!
 A: Hint:
$$r^{n+1} + \dfrac{1-r^{n+1}}{1-r} = \dfrac{(1-r)r^{n+1} + (1-r^{n+1})}{1-r}$$
Edit: What N.F. Taussig says is absolutely correct. If $r=1$, what is true is,
$$n+1 = \lim\limits_{r\to 1} (1+r+\cdots + r^{n}) = \lim\limits_{r\to 1} \dfrac{1-r^{n+1}}{1-r}$$
A: You can easily prove it for all $n>0$, even or not. But if you need the proof wnen $n$ is odd (and $r\ne1$, of course), here it is.
The base step ($n=1$) is obvious, as $1-r^2=(1-r)(1+r)$. Suppose it holds for $n$ odd; then the next odd exponent is $n+2$ and
\begin{align}
1+r+\dots+r^n+r^{n+1}+r^{n+2}
&=\frac{1-r^{n+1}}{1-r}+r^{n+1}+r^{n+2}\\
&=\frac{1-r^{n+1}+r^{n+1}+r^{n+2}-r^{n+2}-r^{n+3}}{1-r}
\end{align}
Can you conclude?
A: The equation holds for $n=1: 1 = (1-r)/(1-r) = 1$ (with $r \neq 1$).
Assume
$$\sum_{k=0}^n r^k = \frac{1-r^{n+1}}{1-r}.$$
Then for $n+1$:
$$\sum_{k=0}^{n+1} r^k = r^{n+1} + \sum_{k=0}^n r^k = r^{n+1} + \frac{1-r^{n+1}}{1-r} = \frac{1-r^{n+1} + (1-r)r^{n+1}}{1-r} = \frac{1-r^{(n+1)+1}}{1-r},$$
and we're done for all positive integers $n$, which includes all of the even ones.
