find $\sum_{k=0}^{t}(-1)^k\binom{t}{k}^2$ for odd t then for even t find $$\sum_{k=0}^{t}(-1)^k\binom{t}{k}^2$$ for $t=2n$ then for $t=2n+1$
I tried by expand $(1-x)^n(1-x)^n$, with no result.
Any Help ?
 A: It is better to exploit symmetry. Since $\binom{t}{k}=\binom{t}{t-k}$, we have:
$$ \sum_{k=0}^{t}\binom{t}{k}^2(-1)^k = [x^t]\left[\left(\sum_{k=0}^{t}\binom{t}{k}(-1)^k x^k\right)\cdot\left(\sum_{k=0}^{t}\binom{t}{t-k}x^{t-k}\right)\right]$$
hence:
$$ \sum_{k=0}^{t}\binom{t}{k}^2(-1)^k = [x^t]\left[(1-x)^t(1+x)^t\right]=[x^t](1-x^2)^t$$
so:
$$ \sum_{k=0}^{t}\binom{t}{k}^2(-1)^k=\left\{\begin{array}{rcl}0&\text{if}&t\equiv 1\pmod{2},\\ \binom{t}{t/2}(-1)^{t/2}&\text{if}&t\equiv 0\pmod{2}.\end{array}\right.$$
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$\ds{\sum_{k = 0}^{t}\pars{-1}^{k}{t \choose k}^{2}:\ {\large ?}}$.

\begin{align}&\color{#66f}{\large\sum_{k = 0}^{t}\pars{-1}^{k}{t \choose k}^{2}}
=\sum_{k = 0}^{t}\pars{-1}^{k}{t \choose k}
\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{t} \over z^{k + 1}}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{t} \over z}
\sum_{k = 0}^{t}{t \choose k}\pars{-\,{1 \over z}}^{k}\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{t} \over z}
\bracks{1 + \pars{-\,{1 \over z}}}^{t}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\pars{-1}^{t}\oint_{\verts{z}\ =\ 1}{\pars{1 - z^{2}}^{t} \over z^{t + 1}}
\,{\dd z \over 2\pi\ic}
=\pars{-1}^{t}\sum_{k=0}^{t}{t \choose k}\pars{-1}^{k}
\oint_{\verts{z}\ =\ 1}{1 \over z^{t - 2k + 1}}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\pars{-1}^{t}\sum_{k=0}^{t}{t \choose k}\pars{-1}^{k}\delta_{k,t/2}
\end{align}

$$
\color{#66f}{\large\sum_{k = 0}^{t}\pars{-1}^{k}{t \choose k}^{2}}
=\color{#c00000}{\large%
\left\{\begin{array}{lcl}
\pars{-1}^{t/2}{t \choose t/2} & \mbox{if} & t\ \mbox{is even}
\\[2mm]
0                              &           & \mbox{otherwise}
\end{array}\right.}
$$
