On a combinatorial identity of Chaundy and Bullard

Abstract

We give two new proofs of the Chaundy-Bullard formula $$(1-x{)}^{n+1}\sum _{k=0}^m(\genfrac{}{}{0ex}{}{n+k}{k})x^k+x^{m+1}\sum _{k=0}^n(\genfrac{}{}{0ex}{}{m+k}{k})(1-x{)}^k=1$$ 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.

8 pages