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\begin{align}
&\color{#f00}{\sum_{i = 0}^{t}\sum_{j = 0}^{t}\sum_{k = 0}^{\infty}{t \choose i}{t \choose j} {k + t - 1 \choose t - 1}x^{i + j + k}} =
\sum_{i = 0}^{t}{t \choose i}x^{i}\sum_{j = 0}^{t}{t \choose j}x^{j}
\sum_{k = 0}^{\infty}{k + t - 1 \choose k}x^{k}
\\[5mm] = &\
\sum_{i = 0}^{t}{t \choose i}x^{i}\sum_{j = 0}^{t}{t \choose j}x^{j}
\sum_{k = 0}^{\infty}
{-\bracks{k + t - 1} + \bracks{k} - 1 \choose k}\pars{-1}^{k}x^{k}
\\[5mm] = &
\sum_{i = 0}^{t}{t \choose i}x^{i}\sum_{j = 0}^{t}{t \choose j}x^{j}\
\overbrace{\sum_{k = 0}^{\infty}{-t \choose k}\pars{-x}^{k}}
^{\ds{\pars{1 - x}^{-t}}}
=
\pars{1 - x}^{-t}\
\overbrace{\sum_{i = 0}^{t}{t \choose i}x^{i}}^{\ds{\pars{1 + x}^{t}}}\
\overbrace{\sum_{j = 0}^{t}{t \choose j}x^{j}}^{\ds{\pars{1 + x}^{t}}} =
\color{#f00}{\bracks{\pars{1 + x}^{2} \over 1 - x}^{t}}
\end{align}
Moreover,
\begin{align}
\bracks{\pars{1 + x}^{2} \over 1 - x}^{t} & =
\sum_{n = 0}^{2t}{2t \choose n}x^{n}\sum_{j = 0}^{\infty}{-t \choose j}
\pars{-x}^{j} =
\sum_{n = 0}^{\infty}\sum_{j = 0}^{\infty}
\pars{-1}^{j}{2t \choose n}{-t \choose j}x^{n + j}
\\[5mm] = &\
\sum_{j = 0}^{\infty}\sum_{n = j}^{\infty}
\pars{-1}^{j}{2t \choose n - j}{-t \choose j}x^{n}
=
\sum_{n = 0}^{\infty}
\bracks{\sum_{j = 0}^{n}\pars{-1}^{j}{2t \choose n - j}{-t \choose j}}x^{n}
\end{align}
such that
$$
\bracks{x^{n}}
\color{#f00}{\sum_{i = 0}^{t}\sum_{j = 0}^{t}\sum_{k = 0}^{\infty}{t \choose i}{t \choose j} {k + t - 1 \choose t - 1}x^{i + j + k}} =
\color{#f00}{\sum_{j = 0}^{n}\pars{-1}^{j}{2t \choose n - j}{-t \choose j}}
$$
CAS yields, for the last sum, an expression in terms of a Hypergeometric Function:
$$
{2t \choose n}\ \mbox{}_{2}\mathrm{F}_{1}\pars{-n,t;-n + 2t + 1;-1}
$$