# The binomial theorem and other summation formulas

## The binomial theorem and other summation formulas

$\displaystyle {\displaystyle \index{Binomial theorem}\index{Summation formula!Binomial theorem} \HyperpFq{1}{0}@@{a}{-}{z}=\sum_{n=0}^{\infty}\frac{\pochhammer{a}{n}}{n!}z^n=(1-z)^{-a} }$

Constraint(s): $\displaystyle {\displaystyle |z|<1}$

$\displaystyle {\displaystyle \HyperpFq{1}{0}@@{-n}{-}{z}=\sum_{k=0}^n\frac{\pochhammer{-n}{k}}{k!}z^k =\sum_{k=0}^n\binomial{n}{k}(-z)^k=(1-z)^n }$

Constraint(s): $\displaystyle {\displaystyle n=0,1,2,\ldots}$

$\displaystyle {\displaystyle \index{Gauss summation formula}\index{Summation formula!Gauss} \HyperpFq{2}{1}@@{a,b}{c}{1}= \frac{\EulerGamma@{c}\EulerGamma@{c-a-b}}{\EulerGamma@{c-a}\EulerGamma@{c-b}} }$

Constraint(s): $\displaystyle {\displaystyle \realpart{c-a-b}>0}$

$\displaystyle {\displaystyle \index{Vandermonde summation formula}\index{Summation formula!Vandermonde}\index{Chu-Vandermonde summation formula}\index{Summation formula!Chu-Vandermonde} \HyperpFq{2}{1}@@{-n,b}{c}{1}=\frac{\pochhammer{c-b}{n}}{\pochhammer{c}{n}} }$

Constraint(s): $\displaystyle {\displaystyle n=0,1,2,\ldots}$

$\displaystyle {\displaystyle \index{Saalsch\"{u}tz summation formula}\index{Summation formula!Saalsch\"{u}tz}\index{Pfaff-Saalsch\"{u}tz summation formula}\index{Summation formula!Pfaff-Saalsch\"{u}tz} \HyperpFq{3}{2}@@{-n,a,b}{c,1+a+b-c-n}{1}=\frac{\pochhammer{c-a}{n}\pochhammer{c-b}{n}}{\pochhammer{c}{n}\pochhammer{c-a-b}{n}} }$

Constraint(s): $\displaystyle {\displaystyle n=0,1,2,\ldots}$

$\displaystyle {\displaystyle \index{Summation formula!for a very-well-poised \HyperpFq{5}{4}} \HyperpFq{5}{4}@@{1+a/2,a,b,c,d}{a/2,1+a-b,1+a-c,1+a-d}{1} {}=\frac{\EulerGamma@{1+a-b}\EulerGamma@{1+a-c}\EulerGamma@{1+a-d}\EulerGamma@{1+a-b-c-d}} {\EulerGamma@{1+a}\EulerGamma@{1+a-b-c}\EulerGamma@{1+a-b-d}\EulerGamma@{1+a-c-d}} }$
$\displaystyle {\displaystyle \HyperpFq{4}{3}@@{1+a/2,a,b,c}{a/2,1+a-b,1+a-c}{-1} =\frac{\EulerGamma@{1+a-b}\EulerGamma@{1+a-c}}{\EulerGamma@{1+a}\EulerGamma@{1+a-b-c}} }$
$\displaystyle {\displaystyle \index{Dougall's bilateral sum}\index{Summation formula!Dougall} \sum_{n=-\infty}^{\infty}\frac{\EulerGamma@{n+a}\EulerGamma@{n+b}}{\EulerGamma@{n+c}\EulerGamma@{n+d}} =\frac{\EulerGamma@{a}\EulerGamma@{1-a}\EulerGamma@{b}\EulerGamma@{1-b}\EulerGamma@{c+d-a-b-1}} {\EulerGamma@{c-a}\EulerGamma@{d-a}\EulerGamma@{c-b}\EulerGamma@{d-b}} }$
$\displaystyle {\displaystyle \EulerGamma@{n+a}\EulerGamma@{n+b}=\frac{\EulerGamma@{a}\EulerGamma@{1-a}\EulerGamma@{b}\EulerGamma@{1-b}} {\EulerGamma@{1-a-n}\EulerGamma@{1-b-n}} }$

Constraint(s): $\displaystyle {\displaystyle n\in\mathbb{Z}}$

$\displaystyle {\displaystyle \sum_{n=-\infty}^{\infty}\frac{1}{\EulerGamma@{n+c}\EulerGamma@{n+d}\EulerGamma@{1-a-n}\EulerGamma@{1-b-n}} {}=\frac{\EulerGamma@{c+d-a-b-1}}{\EulerGamma@{c-a}\EulerGamma@{d-a}\EulerGamma@{c-b}\EulerGamma@{d-b}} }$