Is there a simple proof of the following binomial Identity (part 2)? This is a related question to the one I posted on MO earlier:
Is there a simple proof of the following Identity for $\sum_{k=m-1}^l(-1)^{k+m}\frac{k+2}{k+1}{\binom l k}\binom{k+1}m$?
It arose in the same context: the degeneracy of umbilic points on Weingarten surfaces.
For all $l,m\in{\mathbb N}$ with $l\geq m\geq0$ the following identities appear to hold:
\begin{eqnarray}
&(1-(2m+1)(m+1)){\textstyle{{l+1 \choose m}}}\nonumber\\
+\sum\limits_{k=m+1}^{l+1}&(-1)^{\scriptstyle{{k+m}}}{\textstyle{{l+1 \choose k}}}\left[(1-(2k+1)(m+2)){\textstyle{\frac{2m+2}{2k+1}{k \choose m+1}}}+(1-(2k+1)(m+1)){\textstyle{\frac{2m+1}{2k+1}{k \choose m}}}\right]\nonumber\\
&=
\left\{\begin{array}{ccl}
                  0&if& l>m\\
                  2(l+1)(l+2) &if& l=m
                \end{array}.
              \right.\nonumber
\end{eqnarray}
Obviously the $l=m$ case is trivial (I include it for completeness). So, any suggestions for a proof of this?
 A: Following the hint @darijgrinberg   stated  in the comment section with respect to the beauty inside the square brackets we  focus on the sum and

we obtain
  \begin{align*}
\color{blue}{\sum_{k=m+1}^{l+1}}&\color{blue}{(-1)^{k+m}\binom{l+1}{k}
\left[(1-(2k+1)(m+2))\frac{2m+2}{2k+1}\binom{k}{m+1}\right.}\\
&\qquad\qquad\qquad\qquad\quad
\color{blue}{\left.+(1-(2k+1)(m+1))\frac{2m+1}{2k+1}\binom{k}{m}\right]}\\
&=\sum_{k=m+1}^{l+1}(-1)^{k+m}\binom{l+1}{k}[m-2k(m+2)]\binom{k}{m}\tag{1}\\
&=\binom{l+1}{m}\sum_{k=m+1}^{l+1}(-1)^{k+m}\binom{l+1-m}{k-m}[m-2k(m+2)]\tag{2}\\
&=\binom{l+1}{m}\sum_{k=1}^{l+1-m}(-1)^{k}\binom{l+1-m}{k}[-2k(m+2)-m(2m+3)]\tag{3}\\
&=-2(m+2)\binom{l+1}{m}\sum_{k=1}^{l+1-m}(-1)^{k}\binom{l+1-m}{k}k\\
&\qquad-m(2m+3)\binom{l+1}{m}\left([[l+1=m]]-1\right)\tag{4}\\
&=-2(m+2)\binom{l+1}{l+1-m}(l+1-m)\sum_{k=1}^{l+1-m}(-1)^{k}\binom{l-m}{k-1}\\
&\qquad-m(2m+3)\binom{l+1}{m}\left([[l+1=m]]-1\right)\tag{5}\\
&=2(m+2)(l+1)\binom{l}{m}\sum_{k=0}^{l-m}(-1)^{k}\binom{l-m}{k}\\
&\qquad-m(2m+3)\binom{l+1}{m}\left([[l+1=m]]-1\right)\tag{6}\\
&\color{blue}{=2(l+1)(l+2)[[l=m]]}\\
&\qquad\color{blue}{-(1-(2m+1)(m+1))\binom{l+1}{m}\left([[l+1=m]]-1\right)}\tag{7}\\
\end{align*}
  in accordance with OPs claim.

Comment:


*

*In (1) we use @darijgrinbergs simplified bracketed beauty.

*In (2) we use the binomial identity
$$\binom{p}{q}\binom{q}{r}=\binom{p}{r}\binom{p-r}{q-r}$$

*In (3) we shift the index to start with $k=1$.

*In (4) we split the sum and do some simplifications regarding $(1-1)^{l+1-m}$ using Iverson brackets.

*In (5) and (6) we use the binomial identity
$$\binom{p+1}{q+1}=\frac{p+1}{q+1}\binom{p}{q}$$
and we shift the index to start with $k=0$.

*In (7) we do some final simplifications and adaptions to better see the relationship with OPs identity.
