How do I re-index $\sum_{i=0}^t \sum_{j=0}^t \sum_{k=0}^\infty \binom{t}{i} \binom{t}{j} \binom{k+t-1}{t-1} x^{i+j+k}$ How do I re-index 
$$\sum_{i=0}^t \sum_{j=0}^t \sum_{k=0}^\infty \binom{t}{i} \binom{t}{j} \binom{k+t-1}{t-1} x^{i+j+k}$$
to compare coefficients with $\displaystyle \sum_{n=0}^t \binom{t}{n} x^n$?
I am trying to extract the coefficient for this power series identity:
$$\frac {(1+x)^t} {(1-x^2)^t} = \frac 1 {(1-x)^t}$$
after replacing with respective binomial series, I get this triple summation but I don't understand how to re-index $i+j+k = n$ correctly to compare the coefficients of $x^n$.
Any help appreciated, thanks!
 A: The derivation of the triple sum is not that clear to me, but here is an alternate calculation which might be helpful.

From the given representation
  \begin{align*}
  (1+x)^t=\sum_{n=0}^t\binom{t}{n}x^n\tag{1}
  \end{align*}
we compare the coefficients of the generating functions of
  \begin{align*}
  (1+x)^t=\frac{(1-x^2)^t}{(1-x)^t}
  \end{align*}
We obtain
  \begin{align*}
\frac{(1-x^2)^t}{(1-x)^t}&=\left(\sum_{j=0}^\infty\binom{t}{j}(-x^2)^j\right)\left(\sum_{k=0}^\infty\binom{-t}{k}(-x)^k\right)\\
&=\sum_{n=0}^\infty\left(\sum_{{2j+k=n}\atop{j,k\geq 0}}\binom{t}{j}\binom{-t}{k}(-1)^{j+k}\right)x^n\\
&=\sum_{n=0}^\infty\left(\sum_{j=0}^n\binom{t}{j}\binom{-t}{n-2j}(-1)^{j+n}\right)x^n\tag{2}\\
&=\sum_{n=0}^\infty\left(\sum_{j=0}^n\binom{t}{j}\binom{t+n-2j-1}{t-1}(-1)^{j+n}\right)x^n\tag{3}\\
\end{align*}

Comment:


*

*In (2) we replace the index $k$ by $n-2j$ and use $n$ as upper limit of the inner sum by noting that $\binom{p}{q}=0$ if $0\leq p<q$.

*In (3) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{p-1}(-1)^q$.

We conclude by comparing the coeffcients of $x^n$ of (1) and (3)
  \begin{align*}
(-1)^n\binom{t}{n}=\sum_{j=0}^n(-1)^{j}\binom{t}{j}\binom{t+n-2j-1}{n-2j}\qquad\qquad 0\leq n\leq t
\end{align*}

A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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:
$$
{2t \choose n}\ \mbox{}_{2}\mathrm{F}_{1}\pars{-n,t;-n + 2t + 1;-1}
$$
