Chu-Vandermonde Identity/Examples/2 from e + pi/Proof 2

Example of Use of Chu-Vandermonde Identity

Let $r = e$, $s = \pi$ and $n = 2$

$\ds \binom {e + \pi} 2 = \sum_{k \mathop = 0}^2 \binom e k \binom \pi {2 - k}$

$\ds \binom {e + \pi} 2$

$=$

$\ds \dfrac {\paren {e + \pi} \times \paren {e + \pi - 1} } {2 \times 1}$

$\ds $

$=$

$\ds \dfrac 1 2 \paren {\paren {e + \pi}^2 - \paren {e + \pi} }$

multiplying out

$\ds $

$=$

$\ds \dfrac 1 2 \paren {e^2 + 2e\pi + \pi^2 - e - \pi}$

simplifying

$\ds \sum_{k \mathop = 0}^2 \binom e k \binom \pi {2 - k}$

$=$

$\ds \binom e 0 \binom {\pi} 2 + \binom e 1 \binom {\pi} 1 + \binom e 2 \binom {\pi} 0$

Definition of Summation

$\ds $

$=$

$\ds 1 \times \dfrac {\paren {\pi} \paren {\pi - 1} } 2 + e \times \pi + \dfrac {\paren {e} \paren {e - 1} } 2 \times 1$

$\ds $

$=$

$\ds \dfrac 1 2 \paren {\pi^2 - \pi + 2e\pi + e^2 - e}$

simplifying

$\ds $

$=$

$\ds \dfrac 1 2 \paren {e^2 + 2e\pi + \pi^2 - e - \pi}$

rearranging

$\ds $

$=$

$\ds \binom {e + \pi} 2$

from above

$\blacksquare$