Exercise 2087Combinatorics › Binomial coefficientsM1Added 3 Nov 2026

Gauss's binomial theorem

#Gauss binomial#hypergeometric#identity

Statement

State Gauss's ${}_2F_1(1)$ evaluation:
${}_2F_1\!\left(\begin{matrix}a,b\\c\end{matrix};1\right)=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)},$
and deduce $\sum_{k=0}^{n}\binom{n}{k}^2=\binom{2n}{n}$ .

Hint

Specialize to appropriate parameters.

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Solution

Gauss. The evaluation above holds for $\operatorname{Re}(c-a-b)>0$ .

Deduce the binomial identity. The sum $\sum_k\binom{n}{k}^2=\sum_k\binom{n}{k}\binom{n}{n-k}$ is exactly Vandermonde's identity, giving $\binom{2n}{n}$ .

Conclusion. Gauss's theorem generalizes Vandermonde's identity and reduces to it for specific parameters.

Gauss's binomial theorem

Statement

Solution

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