Prove that $ \sum_{i=0}^{2k}( {n+R-1\choose R+i} + (-1)^{i+1}{ n+R+i\choose R+i } )\sum_{j=0}^i {i\choose j}(-1)^j(i+1-j)^{2k}=0 $ For all $k,R \in \mathbb{N}$ fixed, prove that $ \sum_{i=0}^{2k}( {n+R-1\choose R+i} + (-1)^{i+1}{ n+R+i\choose R+i } )\sum_{j=0}^i {i\choose j}(-1)^j(i+1-j)^{2k}=0 $. I'm quite sure this is true but I'm not very sharp in seeing how to go about proving it. I was studying some stuff in some Pascal triangles is how I stumbled on this. Any ideas?
 A: First off, there should be $(-1)^{i+1}$ not $(-1)^{R+i}$ (now it's corrected in the question). UPDATE. Argument below is simplified and streamlined.
The identity generalizes the previous question, which essentially represents the case of $R=1$. In fact, this generalized identity has a somewhat easier proof (given below) than my earlier proof for $R=1$.
Again, we start with noticing that the inner sum equals $i!\cdot S(2k+1,i+1)$, where $S(\cdot,\cdot)$ are Stirling numbers of second kind.
Let $k>0$ is fixed, and $LHS(2k,n,R)$ denote the identity left-hand side, i.e.
$$LHS(2k,n,R) := \sum_{i=0}^{2k} \big( \binom{n+R-1}{R+i} + (-1)^{i+1} \binom{n+R+i}{R+i}\big) i! S(2k+1,i+1).$$
We start to prove the identity in two cases: $R=0$ and $n=0$.
In the case $R=0$, we have
\begin{split}
LHS(2k,n,0) &= \sum_{i=0}^{2k} \big( \binom{n-1}i + (-1)^{i+1} \binom{n+i}i\big) i! S(2k+1,i+1) \\
&= \sum_{i=0}^{2k} \big( (n-1)_i - (-n-1)_i \big) S(2k+1,i+1) \\
&= n^{2k} - (-n)^{2k} = 0.
\end{split}
In the case $n=0$ and $R>0$, we have
\begin{split}
LHS(2k,0,R) &= \sum_{i=0}^{2k} (-1)^{i+1} i! S(2k+1,i+1) \\
&= -\sum_{i=0}^{2k} (-1)_i S(2k+1,i+1) \\
&= -0^{2k} = 0.\end{split}
Now, we are ready to prove the identity by induction on $n+R$. From above, it follows that $LHS(2k,n,R)=0$ when $(n,R)=(0,1)$ or $(n,R)=(1,0)$, i.e. when $n+R=1$. Now, when $n+R>1$, the cases $n=0$ or $R=0$ are addressed above, while in the case $n>0$ and $R>0$, we use  Pascal's rule to conclude that
$$LHS(2k,n,R) = LHS(2k,n,R-1) + LHS(2k,n-1,R) = 0$$
by the induction assumption.
