combinatorial proof $\sum_{i=0}^m x^i=\frac{x^{m+1}-1}{{x-1}}$ I would like a proof by counting two ways that for positive integers $x,m $ we have $\sum_{i=0}^m x^i=\frac{x^{m+1}-1}{{x-1}}$
 A: Suppose you have a coin that has probability $p$ of coming up heads, and you toss it up to $m+1$ times, stopping when you see a tail.


*

*On the one hand, the probability that you saw a tail at all (before you ran out of tosses) is the sum over all the different ways in which this might happen. The probability of seeing a tail after $i$ heads is $p^i (1-p)$, so the total probability is $\sum_{i=0}^m p^i (1-p)$

*On the other hand, the probability that you saw a tail is the complement of the probability that all $m+1$ tosses were heads, and is thus $1 - p^{m+1}$.
This shows that 
$$\sum_{i=0}^m p^i (1-p) = 1 - p^{m+1}$$ 
and therefore 
$$\sum_{i=0}^m p^i = \frac{1 - p^{m+1}}{1-p}$$
which is your identity with $p$ in place of $x$. Of course, if your $x$ is not between $0$ and $1$, you need to argue further (e.g. argue that since the polynomial identity holds for infinitely many $x$ it should hold for all $x$). For something like negative $x$, it's not obvious to me what a combinatorial interpretation could possibly be, so I suspect this is as good as it gets.

Edit: Here's a combinatorial (counting-only, no probability) variant of the same. Suppose you have a an alphabet of $c$ symbols (say the symbols are 0, 1, ...), and you want to count the number of strings of length $m+1$ in which the last character that is not a 0 is a 1. Then we can count it two ways:


*

*If the number of characters before the last non-0 character 1 is $i$, then these $i$ can be any of the $c$ symbols, and therefore there are $c^i$ such strings. Thus the number of strings is $\sum_{i=0}^m c^i$.

*Alternatively, we can take just any string of length $m+1$ that contains at least one character that is not a 0 (any of the $c^{m+1}$ strings of length $m+1$, except the one that contains only 0s), and change the last non-0 character to a 1. Each string will be counted $c-1$ times in this method (as there are $c-1$ possible non-0 symbols), so the number of distinct strings is $\dfrac{c^{m+1}-1}{c-1}$.
Equating the two gives your identity, with $c$ in place of $x$. This works for $c$ being a positive integer.
