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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 $\alpha \in \mathbb{N}_0$?

amWhy
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Jakob W.
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2 Answers2

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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.

Jam
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2

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

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

user153012
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