Let $(l_1,K)$ be integers such that $0 \le l_1$ and $0 \le K \le 2 l_1$ . Consider a following multiple sum: \begin{eqnarray} &&{\mathcal S}_{l_1}^{(K)}:= \frac{(-1)^K}{2^K} \sum\limits_{k=0}^{l_1} \sum\limits_{k_1=0}^{l_1} \sum\limits_{k_2=0}^{l_1-2 k-1} \sum\limits_{l=0}^{l_1}\\ && \left(\binom{l_1+2}{l+2} \binom{k}{K-k-k_1-k_2}+4 \binom{l_1+1}{l+2} \binom{k}{K-k-k_1-k_2-1} \right)2^{2 k+2 k_1+k_2} \cdot \\ &&(-1)^l \frac{l!}{(2k+1)!(k_2)!(l-k_2-2 k-1)!} \cdot \frac{l!}{k_1!(l-k_1)!} \end{eqnarray}
We have checked numerically that the following results hold: \begin{eqnarray} &&{\mathcal S}_{l_1}^{(K)} = \left\{ \begin{array}{rr} -l_1 & \mbox{if $K=0$}\\ 0 & \mbox{if $1 \le K \le \lfloor \frac{l_1}{2} \rfloor$}\\ \cdots & \mbox{if $\lfloor \frac{l_1}{2} \rfloor < K \le 2l_1 -1$}\\ 0 &\mbox{if $2 l_1 \le K$} \end{array} \right. \end{eqnarray}
Now, the question is how do we prove this result? In particular can we fill in the gaps in the formula above and find a closed form expression for the sum whenever it is non-zero?
Motivation: This sum appeared in the course of simplifying the following expression: \begin{eqnarray} && \sum\limits_{\xi=\pm} \frac{\xi e^{\imath \xi \phi}(1-e^{\imath \xi \phi})(1-a e^{-\imath \xi \phi})^{l_1+1}}{2 \imath \sin(\phi) \cdot a (1-a)^{1+\lfloor \frac{l_1}{2} \rfloor}}=\\ && \frac{1}{2^{l_1+1}}\sum\limits_{\xi=0}^1 \sum\limits_{p=l_1\%(2-\xi)}^{l_1-\lfloor \frac{l_1}{2}\rfloor} \binom{l_1+1}{2 p-l_1\%2+\xi} (-1)^{\xi+\lceil \frac{l_1}{2}\rceil-p} (1-a)^p a^{l_1+l_1\%2-2 p} (2+a)^{2 p+1-l_1\%2} (3+a)^{\lceil \frac{l_1}{2}\rceil-p} =\\ && \sum\limits_{K=\lfloor \frac{l_1}{2}\rfloor+1}^{2 l_1} {\mathcal S}_{l_1}^{(K)} \cdot (1-a)^{K-\lfloor \frac{l_1}{2}\rfloor-1} \end{eqnarray} where $\phi=\arccos((1+a)/2)$. By using Mathematica I checked that the left-hand side is a polynomial in the variable $1-a$ with the sum in question being a coefficient at a given power of $1-a$. Therefore I need to find the sum in question in closed form as a function of $l_1$ and $K$.
First of all note that the sum can only be non-zero iff the binomial factors involving $K$ in the parentheses satisfy certain constraints. To be exact we must have $k \ge K-k-k_1-k_2-1$ and $K - k - k_1-k_2 \ge 0$. Working out $k_2$ from those inequalities and taking into account the already existing constraints on that variable we have: \begin{equation} 0 \vee (K-2 k-k_1-1) \le k_2 \le (l_1-2k-1)\wedge (K-k-k_1) \end{equation}
2.Now let us take $K=2 l_1-1$ and $k_1=l_1-p$. Then the inequalities on top of this post read: \begin{eqnarray} &&0 \vee (l_1-2 k+p-2) \le k_2 \le (l_1-2 k-1) \wedge (l_1-k+p-1)\\ &&\Leftrightarrow\\ &&0 \vee ((l_1-2 k-1) + (p-1)) \le k_2 \le (l_1-2 k-1) \end{eqnarray} This implies that $p=0,1$. Therefore we have now three cases.
A. Firstly we have $(k_1=l_1-1,k_2=l_1-2k-1)$. Inserting this into the term in sum implies that $l=l_1$ and this in turn implies the term in sum being zero.
B. Secondly we have $(k_1=l_1-0,k_2=l_1-2k-1)$. Again, inserting this into the term in the sum implies that $l=l_1$ and this in turn implies that the term in the sum reads: \begin{equation} (-1)^{l_1+1} 2^{1-l_1} \frac{l_1!}{(2k+1)!(l_1-2k-1)!} \end{equation}
C. Thirdly we have $(k_1=l_1-0,k_2=l_1-2k-2)$. Again, inserting this into the term in sum implies that $l=l_1$ and this in turn implies the term in sum being zero.
Therefore we conclude:
\begin{equation} {\mathcal S}_{l_1}^{(2 l_1-1)} = \sum\limits_{k=0}^{l_1} (-1)^{l_1+1} 2^{1-l_1} \frac{l_1!}{(2k+1)!(l_1-2k-1)!} = (-1)^{l_1+1} \end{equation}
3.Now let us take $K=2 l_1-2$ and $k_1=l_1-p$. Then in the same manner as before we get:
\begin{eqnarray} &&0 \vee (l_1-2 k+p-3) \le k_2 \le (l_1-2 k-1) \wedge (l_1-k+p-2)\\ &&\Leftrightarrow\\ &&0 \vee ((l_1-2 k-1) + (p-2)) \le k_2 \le (l_1-2 k-1) \end{eqnarray} except the negligible case $(k,p)=(0,0)$. This implies that $p=0,1,2$. This gives us six cases.
A. $(k_1=l_1-2,k_2=l_1-2k-1)$.
B1. $(k_1=l_1-1,k_2=l_1-2k-1)$.
B2. $(k_1=l_1-1,k_2=l_1-2k-2)$.
C1. $(k_1=l_1-0,k_2=l_1-2k-1)$.
C2. $(k_1=l_1-0,k_2=l_1-2k-2)$.
C3. $(k_1=l_1-0,k_2=l_1-2k-3)$.
Again, inserting this into the term in the sum we get that in all cases except B2. we have to have $l=l_1$ for the term in sum not to vanish. In addition to this the case B2. allows $l=l_1-1$. Now we write down all the seven cases, firstly A,B1,B2,C1,C2,C3 all at $l=l_1$ and then B2 at $l=l_1-1$. In this order we have: \begin{eqnarray} &&{\mathcal S}_{l_1}^{(2 l_1-2)} = \sum\limits_{k=0}^{l_1} \left(\right.\\ &&\left. 0 + \right.\\ &&\left.\frac{(-1)^{\text{l1}} 2^{1-\text{l1}} (\text{l1}!)^2}{(2 k+1)! (\text{l1}-1)! (-2 k+\text{l1}-1)!} + \right.\\ &&\left.0 +\right.\\ &&\left. \frac{k (-1)^{\text{l1}} 2^{3-\text{l1}} \text{l1}!}{(2 k+1)! (-2 k+\text{l1}-1)!} +\right.\\ &&\left. \frac{(-1)^{\text{l1}} 2^{2-\text{l1}} \text{l1}!}{(2 k+1)! (-2 k+\text{l1}-2)!} +\right.\\ &&\left. 0 \right)+\\ && \sum\limits_{k=0}^{l_1} \left( -\frac{(-1)^{\text{l1}} 2^{2-\text{l1}} (\text{l1}-1)!}{(2 k+1)! (-2 k+\text{l1}-2)!} \right)=\\ &&(-1)^{l_1} (-2+3 l_1) +\\ &&(-1)^{l_1}(-1)=\\ &&(-1)^{l_1} (-3+3 l_1) \end{eqnarray} Therefore we conclude that
\begin{equation} {\mathcal S}_{l_1}^{(2 l_1-2)} = (-1)^{l_1} \cdot 3(l_1-1) \end{equation}