Proof by induction for a summation? Looking for some help with a proof by iduction. Im looking to proove the following summation holds true:
$$\frac{\langle W \vert (C)^{N} \vert V \rangle}{\langle W \vert \vert V \rangle}=\frac{\langle W \vert (D+E)^{N} \vert V \rangle}{\langle W \vert \vert V \rangle}= \sum_{p=0}^N\frac{p(2N-1-p)!}{N!(N-p)!}\frac{\beta^{-p-1}-\alpha^{-p-1}}{\beta^{-1}-\alpha^{-1}}$$
The problem is related to the following rules:
$$DE= D+E=C $$
$$D\vert V \rangle= \frac{1}{\beta} \vert V \rangle $$       
$$\langle W\vert E= \frac{1}{\alpha}\langle W \vert $$
Now using the base case of N=1 yields to, $\frac{1}{\alpha}+\frac{1}{\beta}$ which I have shown but struggling with the hypothesis step.
Any help would be most appreciated, many thanks.
 A: Here's a lead:
So $C^N$ is a sum over all $N$-length words in $\{D,E\}$. If I've interpreted your rules correctly, the only way to simplify words ending in $E$ is to use the rule $DE = D + E$. This allows one (in theory) to simplify any word coming from $C^N$ into a sum of words ending in $D$ and perhaps a word consisting of some power of $E$.
The practical induction will be a mess, but here's a thought: (to do the induction it suffices to look at $C^{N} E$)
\begin{align*}
C^N E = C^{N-1}(D + E)E = C^{N-1} DE + C^{N-1} E^2 = C^N + C^{N-1} E^2 \\
= C^N + C^{N-1}E + C^{N-2}E^3 = \cdots = \sum_{i = 0}^{N-1} C^{N-i} E^i + E^{N+1}
\end{align*}
if I'm not mistaken.
Now concentrate on how to reduce the terms in the sum: for instance, $C E^{N-1}$ equals $(D + E) E^{N-1} = C E^{N-2} + E^N = C E^{N-3} + E^{N-1} + E^N = \cdots = C + \sum_{j = 2}^N E^j$. Try to use this form to come up with a closed form for the summands $C^{N-i} E^i$ (I would just hit both sides with a factor of $C$ and see what comes out).
