Proving Bonnets' Recursion with Rodrigues' Formula I would like to show that $(n+1)P_{n+1}(x)+nP_{n-1}(x)=(2n+1)xP_{n}(x)$ using Rodrigues' formula, not the generating function.
I got to this point, but have not been able to progress further.
$$(n+1)P_{n+1}(x)=\frac{x(n+1)(D^{n}(x^{2}-1)^{n})+n(n+1)(D^{n-1}(x^{2}-1)^{n})}{2^{n}n!}$$
$D^{n}$ denotes the $n^{th}$ derivative with respect to $x$.
I see that part of the $(2n+1)xP_{n}(x)$ term is formed in the first term of the fraction, but do not see how to recover the rest of Bonnets' recursion.
Any help would be appreciated.
Thanks.

Using Rodrigues' formula $\displaystyle P_n(x)=\frac{1}{2^n n!}D^n\left(x^2-1\right)^n$, we can write
\begin{align*}
&\color{blue}{\left(n+1\right)P_{n+1}(x)}-\color{green}{\left(2n+1\right)xP_n(x)}+\color{red}{nP_{n-1}(x)}=\\
=\;&\frac{1}{2^{n+1}n!}D^{n-1}\biggl\{\color{blue}{D^2\left(x^2-1\right)^{n+1}}+\color{red}{4n^2\left(x^2-1\right)^{n-1}}\biggr\}-\color{green}{\left(2n+1\right)xP_n(x)}=\\
=\;&\frac{1}{2^{n+1}n!}D^{n-1}\biggl\{\left(x^2-1\right)^{n-1}\biggl[
\color{blue}{4n\left(n+1\right)x^2+2\left(n+1\right)\left(x^2-1\right)}+\color{red}{4n^2}\biggr]\biggr\}-\color{green}{\left(2n+1\right)xP_n(x)}=\\
=\;&\frac{2n+1}{2^{n}n!}D^{n-1}\biggl\{\left(x^2-1\right)^{n-1}\biggl[
\left(n+1\right)x^2+n-1\biggr]\biggr\}-\color{green}{\frac{2n+1}{2^{n}n!}xD^{n}\left(x^2-1\right)^n}.
\end{align*}
Now observe that $[D,x]=1$ implies $[D^n,x]=nD^{n-1}$. Using this to replace $xD^n$ in the above expression by $D^n\circ x-nD^{n-1}$, we can rewrite it as
\begin{align}
&\frac{2n+1}{2^{n}n!}D^{n-1}\biggl\{\left(x^2-1\right)^{n-1}\biggl[
\left(n+1\right)x^2+n-1\biggr]\color{green}{-D\left[x\left(x^2-1\right)^n\right]+n\left(x^2-1\right)^{n}}\biggr\}=\\
=\;&\frac{2n+1}{2^{n}n!}D^{n-1}\biggl\{\left(x^2-1\right)^{n-1}\cdot 0\biggr\}=0.
\end{align}
