Using generating function for Legendre polynomials, prove that ∑0l(2l+1)Pl(x)tl=1−t2(1−2xt+t2)32

Introduction
The legendre polynomial denoted by ${\displaystyle P_n\left(x\right)}$.
The solution of physical problems is the class of function called legendre polynomials.
Explanation
To prove,
${\displaystyle \sum _0^l{P}_l\left(x\right)t^l(l+l+1)=\frac{xt-t^2+xt-t^2}{{(1-2xt+t^2)}^{\frac{3}{2}}}+\frac{1}{{(1-2xt+t^2)}^{\frac{1}{2}}}}$
Use generating function for legendre polynomial,
${\displaystyle \sum _{x=0}^{\mathrm{\infty }}{P}_n\left(x\right)t^n={(1-2xt+t^2)}^{-\frac{1}{2}}}$
Consider the limit 0 to l
${\displaystyle \sum _{n=0}^l{P}_l\left(x\right)t^l={(1-2xt+t^2)}^{-\frac{1}{2}}}$
Differentiating with respect to t,
${\displaystyle \sum _0^l{P}_l\left(x\right){<}^{l-1}=(-\frac{1}{2}){(1-2xt+t^2)}^{-\frac{3}{2}}(-2x+2t)}$
${\displaystyle \sum _0^l{P}_l\left(x\right){<}^{l-1}=\frac{(x-t)}{{(1-2xt+t^2)}^{\frac{3}{2}}}}$
Multiplying t on both sides we get,
${\displaystyle \sum _0^l{P}_l\left(x\right){<}^{l-1}t=\frac{(x-t)t}{{(1-2xt+t^2)}^{\frac{3}{2}}}}$
${\displaystyle \sum _0^l{P}_l\left(x\right){<}^l=\frac{xt-t^2}{{(1-2xt+t^2)}^{\frac{3}{2}}}}$
Let us consider,