Prove an equation with summation and binomial coefficients Good day everybody,
I am looking to prove this:
$$\delta_{a,b}=\left(\frac{1}{2}\right)^{2\left(a-b\right)}\sum_{\gamma=0}^{a-b}\left(-1\right)^{a-b-\gamma}\frac{b+\gamma}{b}\left(\begin{array}{c}
2a\\
a-b-\gamma
\end{array}\right)\left(\begin{array}{c}
2b-1+\gamma\\
2b-1
\end{array}\right)$$
for $a \ge 0 $ and $b \ge 1$.
The case $b=0$ is an exception to the rule but is also of interest for me. Therefore I would actually most appreciate a proof of matrices $D$ and $C$ to be inverse:
$$D_{mn}=\begin{cases}
\left(-1\right)^{\frac{m-n}{2}}\left(\begin{array}{c}
m\\
\frac{m-n}{2}
\end{array}\right)\left(\frac{1}{2}\right)^{m} & \text{ if }n\leq m\text{ and }n-m\text{ is even}\\
0 & \text{else}
\end{cases}
$$
and 
$$C_{ij}=\begin{cases}
1 & \text{if }i=j=0\\
2^{j}\left[\left(\begin{array}{c}
\frac{i+j}{2}-1\\
j-1
\end{array}\right)+2\left(\begin{array}{c}
\frac{i+j}{2}-1\\
j
\end{array}\right)\right] & \text{else if }i\geq j\text{ and }i-j\text{ is even}\\
0 & \text{else}
\end{cases}$$
where indexing starts from zero (and thus contains also $b=0$ case).
Thank you.

We write $n$ instead  of    $\gamma$, focus on the essentials and skip the constant $(-1)^{a-b}2^{-2(a-b)}$.

We obtain for integers $a\geq  b>0$
  \begin{align*}
\color{blue}{\sum_{n=0}^{a-b}}&\color{blue}{(-1)^n\frac{b+n}{b}\binom{2a}{a-b-n}\binom{2b-1+n}{n}}\tag{1}\\
&=\sum_{n=0}^{a-b}\frac{b+n}{b}\binom{2a}{a-b-n}\binom{-2b}{n}\tag{2}\\
&=\sum_{n=0}^{a-b}\binom{2a}{a-b-n}\binom{-2b}{n}
-2\sum_{n=1}^{a-b}\binom{2a}{a-b-n}\binom{-2b-1}{n-1}\tag{3}\\
&=\binom{2a-2b}{a-b}-2\sum_{n=0}^{a-b-1}\binom{2a}{a-b-n-1}\binom{-2b-1}{n}\tag{4}\\
&=\binom{2a-2b}{a-b}-2\binom{2a-2b-1}{a-b-1}[[a>b]]\tag{5}\\
&\color{blue}{=[[a=b]]}
\end{align*}

Comment:


*

*In (1) we use the binomial  identity $\binom{p}{q}=\binom{p}{p-q}$.

*In (2) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

*In (3) we split the sum and apply $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$ to  the right-hand sum.

*In (4) we apply Chu-Vandermonde's identity
to the left sum and shift the index of the right sum by one to start with $n=0$. 

*In (5) we use  Iverson brackets in  the right expression, since the corresponding sum in the line before is zero if $a=b$. We also use the binomial identity $\binom{2p}{p}=2\binom{2p-1}{p-1}$.
