[Math] A sum involving powers of binomial coefficients.

Question

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 $\sum\limits_{m\ge 0} (-1)^m{\binom{n}{m}}^4$?

Answer 1

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.