Closed form for weighted sum involving product of binomial coefficients Is there a closed form for the following expression?$$\sum_iA^i \binom{K-j}{L-i}\binom{j}{i}$$
If $A=1$ this is equal to $\binom{K}{L}$ by Vandermonde's identity.  Further, I know that the sum in question is equal to the coefficient of $x^L$ in $(1+x)^{K-j}(1+Ax)^j$.  However, I do not see how to come up with a closed form.  
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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For the purpose of the example, I'll assume $\ds{\ell = 0,1,\ldots,j}$:

\begin{align}
&\sum_{\ell = 0}^{j}A^{\ell}{K - j \choose L - \ell}{j \choose \ell} =
\sum_{\ell = 0}^{j}A^{\ell}{j \choose \ell}\ \overbrace{%
\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{K - j} \over z^{L - \ell + 1}}
\,{\dd z \over 2\pi\ic}}^{\ds{K - j \choose L - \ell}}
\\[5mm] = &\
\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{K - j} \over z^{L + 1}}
\sum_{\ell = 0}^{j}{j \choose \ell}\pars{Az}^{\ell}\,{\dd z \over 2\pi\ic} =
\oint_{\verts{z}\ =\ 1}
{\pars{1 + z}^{K - j}\,\,\pars{1 + Az}^{\,j} \over z^{L + 1}}
\,{\dd z \over 2\pi\ic}
\\[5mm] = &\
\,\bbox[#ffe,15px,border:2px dotted navy]{\ds{\bracks{z^{L}}\bracks{\pars{1 + z}^{K - j}\,\,\pars{1 + Az}^{\, j}}}}
\end{align}

which agrees with the claimed OP solution.

CAS yields the cumbersome answer:
$$
A^{L}{j \choose L}\
{}_{2}\mrm{F}_{1}\pars{j - K,-L;j - L + 1;{1 \over A}}
$$
