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.
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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: $$ A^{L}{j \choose L}\ {}_{2}\mrm{F}_{1}\pars{j - K,-L;j - L + 1;{1 \over A}} $$
