# An addendum to a Z talk

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Doron Zeilberger recently gave a talk at RUMA, the Rutgers Undergraduate Math Association, about mathematical gambling. In the talk he mentioned a lovely half-page article he wrote confirming a centuries-old conjecture by Newton and Pepys. The proof is very short but relies on some hidden machinery, which I will explain here.

The basic idea is that we want to prove the identity

$\begin{equation} \label{goal} \sum_{k = 0}^{n - 1} {6n \choose k} \left( \frac{1}{6} \right)^k \left( \frac{5}{6} \right)^{6n - k} = \sum_{k = 0}^{n - 1} R(k) {6k \choose k} \left( \frac{1}{6} \right)^k \left( \frac{5}{6} \right)^{5k} \end{equation}$

where $R(k)$ is some explicit rational function. The reason is that the sum on the right has no $n$ in it, and after seeing $R(k)$ written out, is obviously monotonic because the terms are positive. The sum on the left contains a bunch of $n$’s, so it isn’t obvious if it gets bigger or smaller as $n$ grows.

First, a quick lesson on WZ theory. A sequence $a(n, k)$ is called hypergeometric provided that $a(n + 1, k) / a(n, k)$ and $a(n, k + 1) / a(n, k)$ are both explicit rational functions in $n$ and $k$. Certain nice hypergeometric sequences have a “partner” $b(n, k)$ which satisfies the identity

$a(n + 1, k) - a(n, k) = b(n, k + 1) - b(n, k).$

In this case $(a, b)$ is called a WZ pair. It is simple to algorithmically determine whether any explicit sequence is hypergeometric or part of a WZ pair. Our summand happens to be both.

# The theory

Say that you have a WZ pair $(a, b)$. That is,

$\begin{equation} \label{wz} a(n + 1, k) - a(n, k) = b(n, k + 1) - b(n, k). \end{equation}$

Let

$S(n) = \sum_{k = 0}^{n - 1} a(n, k)$

be the sum that we are interested in. Summing over \eqref{wz} from $k = 0$ to $k = n$ gives

$S(n + 1) - S(n) - a(n, n) = b(n, n + 1) - b(n, 0).$

If we move $a(n, n)$ to the right-hand side, then the left-hand side is telescoping, so summing again recovers $S(n)$ itself. The end-result will be

$S(n) = \sum_k (a(k, k) + b(k, k + 1) - b(k, 0)),$

where the bounds are too tedious to determine. In our case it turns out that $b(k, 0) = 0$, and by some standard theory about hypergeometric sequences $b(k, k + 1)$ is a rational multiples of $a(k, k)$, so we can write

$S(n) = \sum_k R(k) a(k, k),$

the kind of sum we desired.

# The proof

The proof now is merely a matter of executing the steps outlined in the previous section. All of these have been implemented in Maple or Mathematica for decades. If we download EKHAD and read it into Maple, then executing zeillim(a(n, k), k, n, N, 0, 1) produces output that implies \eqref{goal}. The rational function is

$R(k) = \frac{\frac{4375}{36} k^{4}+\frac{59675}{216} k^{3}+\frac{285955}{1296} k^{2}+\frac{281215}{3888} k +\frac{15625}{1944}}{\left(5 k +1\right) \left(k +1\right) \left(5 k +4\right) \left(5 k +3\right) \left(5 k +2\right)}.$