Proving the result for Legendre's polynomial $\frac{1-z^2}{(1-2xz+z^2)^3/2}=\sum_{n=0}^{\infty}(2n+1)P_n(x)z^n$.

Question

I am stuck in proving the following equality for Legendre's polynomial $P_n(x)$, which is:

Prove that $\frac{1-z^2}{(1-2xz+z^2)^3/2}=\sum_{n=0}^{\infty}(2n+1)P_n(x)z^n$.

One way to prove it by expanding the LHS i.e. $(1-z^2)[1-z(2x-z)]^{-3/2}$ and finding the coefficient of $z^n$ will give $P_n(x)$, but I think there must be a short method for this using the already known result $\frac{1}{(1-2xz+z^2)^1/2}=\sum_{n=0}^{\infty}P_n(x)z^n$. Any hint is appreciated. Thanks!!

Answer 1

Motivated by the comment of user1952009, I am posting the answer in order to not leave it unanswered.

$\frac{1}{(1-2xz+z^2)^1/2}=\sum_{n=0}^{\infty}P_n(x)z^n$ ....................................$(1)$

On differentiating it partially w.r.t. $z$, $\frac{x-z}{(1-2xz+z^2)^3/2}=\sum_{n=0}^{\infty}P_n(x)nz^{n-1}$

Multiplying by $2z$, $\frac{2xz-2z^2}{(1-2xz+z^2)^3/2}=\sum_{n=0}^{\infty}P_n(x)2nz^n$ ..........................$(2)$

Adding $(1)$ and $(2)$, we get the desired result.