A sum involving powers of binomial coefficients. Find the formula for the following sum of binomial coefficients:
$$ \sum_{m\ge 0} (-1)^m {\binom{n}{m} }^3 .$$ Could you find the formula for $\sum\limits_{m\ge 0} (-1)^m{\binom{n}{m}}^4$?
 A: Your sum is a special case of Dixon's well-poised sum
$$\begin{align*}
\sum_{k=0}^\infty (-1)^k \binom{n}{k}^3&={}_3 F_2\left({{-n,-n,-n}\atop{1,1}}\mid 1\right)\\
&=\frac{\Gamma\left(1-\frac{n}{2}\right)\Gamma \left(\frac{3 n}{2}+1\right)}{n!\Gamma(1-n)\Gamma\left(\frac{n}{2}+1\right)^2}=\frac{\cos\left(\frac{\pi n}{2}\right)\left(\frac{3n}{2}\right)!}{\left(\left(\frac{n}{2}\right)!\right)^3}
\end{align*}$$
Generally,
$$\sum_{k=0}^\infty (-1)^k \binom{n}{k}^p={}_p F_{p-1}\left({{-n,\cdots,-n}\atop{1,\cdots,1}}\mid (-1)^{p+1}\right)$$
since $\dbinom{n}{k}=\dfrac{(-1)^k (-n)_k}{k!}$, and $(1)_k=k!$ and thus
$$\sum_{k=0}^\infty (-1)^k \binom{n}{k}^p=\sum_{k=0}^\infty (-1)^k \left(\frac{(-1)^k (-n)_k}{k!}\right)^p=\sum_{k=0}^\infty ((-1)^{p+1})^k \frac{((-n)_k)^p}{((1)_k)^{p-1} k!}$$
and the hypergeometric form is now noticeable.
A: Binomial coefficient sums are mechanical via Gosper's algorithm and extensions.
Wolfram Alpha gives $$\sum_{m=0}^{\infty}
(-1)^m \binom{n}{m}^3 = \frac{(1-n)^\overline{n} (1+\frac{n}{2})^\overline{n}}{n! (1-\frac{n}{2})^\overline{n}}$$ (where I've used slightly different notation for the Pochhammer symbol).
Similarly $$\sum_{m=0}^{\infty}
(-1)^m \binom{n}{m}^4 = {}_4F_3(-n, -n, -n, -n;1, 1, 1;-1)$$ is hypergeometric-summable.
Checking for $k=5$ and $k=6$ leads to the hypothesis that $$\sum_{m=0}^{\infty}
(-1)^m \binom{n}{m}^k = {}_kF_{k-1}(-n, \ldots, -n;1, \ldots, 1;-1^{k-1})$$ but I don't have time now to try to prove this.
