On a combinatorial identity of Chaundy and Bullard

Abstract

We give two new proofs of the Chaundy-Bullard formula $$ (1-x)^{n+1} \sum_{k=0}^m {n+k\choose k} x^k +x^{m+1}\sum_{k=0}^n {m+k\choose k} (1-x)^k=1 $$ and we prove the "twin formula" $$ \frac{ (1-x)^{(n+1)}}{(n+1)!} \sum_{k=0}^m \frac{n+1}{n+k+1} \frac{ x^{(k)}}{k!} + \frac{ x^{(m+1)}}{(m+1)!} \sum_{k=0}^n \frac{m+1}{m+k+1} \frac{ (1-x)^{(k)}}{k!}=1, $$ where $z^{(n)}$ denotes the rising factorial. Moreover, we present identities involving the incomplete beta function and a certain combinatorial sum.