Could you help me with a question I've tried to solve but stuck in the middle?
i need to calculate the coefficient $x^{2m}$ in every part of the algebraic identity $\frac{(1-x^2)^n}{(1-x)^n}=(1+x)^n$ in order to obtain the some sort of binomial identity: $\sum _{k=0} ^? ?? = \binom{n}{2m}$
what i did, i first verified that it holds true, and then got that the binom identity is $(1+x)^n=\sum_{i=o}^{n} \binom{n}{i}x^i$, but i don't think it's true. i also got that $x^{2m}=\binom {n}{m}$.
if relevant, i used the identity: $\frac{1}{(1-x)^n}=(1+x+x^2+x^3+...)^n=\sum_{k=0}^{\infty}D(n,k)x^k$ where $D(n,k) = \binom{n+k-1}{k}$
can you help me correct it please? how would you show it's true for m=2,n=5 for instance?
thank you very much for helping. i'm stuck on this one.