# Is there a closed form for the sum of the cubes of the binomial coefficients?

We know that $$\sum_{k=0}^n \binom{n}{k} = 2^n\;\; \text{ and }\;\; \sum_{k=0}^n \binom{n}{k}^2 = \binom{2n}{n}$$ hold for all $$n\in \mathbb{N}_0$$. Now I tried to find a similar expression for $$\sum_{k=0}^n \binom{n}{k}^3$$ but didn't get anywhere at all. What I found were only asymptotic estimates (see Sum of cubes of binomial coefficients or Asymptotics of $$\sum_{k=0}^{n} {\binom n k}^a$$).

Now is there a closed form for this sum or, what would be even better, for $$\sum_{k=0}^n \binom{n}{k}^\alpha$$ with any $$\alpha \in \mathbb{N}_0$$?

These numbers are called the Franel Numbers. It's proven in (Petkovšek, M., Wilf, H. and Zeilberger, D. (1996). A=B. Wellesley, MA: A K Peters. p. 160) that there is no closed form for these numbers, in terms of the sum of a fixed number of hypergeometric terms.

However, as @Robert_Israel points out, the expression could possibly be represented by different types of closed form.

• ... if "closed form" is defined as "the sum of a fixed number of hypergeometric terms". There could be other types of "closed form". – Robert Israel Nov 28 '18 at 18:24
• @Robert Was just thinking that. Thanks for the suggestion. – Jam Nov 28 '18 at 18:26

The binomial coefficient for a given pair of $$n \geq k \geq 0$$ integers can be expressed in terms of a Pochhammer symbol as the following.

$$\binom n k = \frac{(-1)^k(-n)_k} {k!}.$$

The expression is valid even if $$n$$ is an arbitrary real number.

Here we note two things.

1. The Pochhammer symbol $$(-n)_k$$ is zero, if $$n \geq 0$$ and $$k > -n$$.
2. The factorial $$k!$$ can be written as $$(1)_k$$.

Using these observations, we can express your sums in terms of a generalized hypergeometric function $$_pF_q$$ as the following. For the sum of the binomial coefficients, we have

$$\sum_{k=0}^n \binom n k = \sum_{k=0}^n \frac{(-1)^k(-n)_k}{k!} = \sum_{k=0}^\infty (-n)_k{\frac{(-1)^k}{k!}} = {_1F_0}\left({{-n}\atop{-}}\middle|\,-1\right).$$ For the sum the square of the binomial coefficients, we have $$\sum_{k=0}^n {\binom n k}^2 = \sum_{k=0}^n \left(\frac{(-1)^k(-n)_k}{k!}\right)^2 = \sum_{k=0}^\infty \frac{\left((-n)_k\right)^2}{k!} \cdot \frac{1}{k!} = {_2F_1}\left({{-n, -n}\atop{1}}\middle|\,1\right).$$ And for the sum of the cube of the binomial coefficients $$-$$ also known as Franel numbers $$-$$, we have $$\sum_{k=0}^n {\binom n k}^3 = \sum_{k=0}^n \left(\frac{(-1)^k(-n)_k}{k!}\right)^3 = \sum_{k=0}^\infty \frac{\left((-n)_k\right)^3}{(k!)^2} \cdot \frac{(-1)^k}{k!} = {_3F_2}\left({{-n, -n, -n}\atop{1, 1}}\middle|\,-1\right).$$ In general, for a positive integer $$r$$, we have the binomial sum

\begin{align*} \sum_{k=0}^n {\binom n k}^r &= \sum_{k=0}^n \left(\frac{(-1)^k(-n)_k}{k!}\right)^r = \sum_{k=0}^\infty \frac{\left((-n)_k\right)^r}{(k!)^{r-1}} \cdot \frac{(-1)^{rk}}{k!} \\ &= {_rF_{r-1}}\left({{-n, -n, \dots, -n}\atop{1, \dots, 1}}\middle|\,(-1)^r\right). \end{align*}