Chu-Vandermonde Identity/Examples/2 from e + pi
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Example of Use of Chu-Vandermonde Identity
Let $r = e$, $s = \pi$ and $n = 2$
- $\ds \binom {e + \pi} 2 = \sum_{k \mathop = 0}^2 \binom e k \binom \pi {2 - k}$
Proof 1
From the Chu-Vandermonde Identity:
- $\ds \sum_{k \mathop = 0}^n \binom r k \binom s {n - k} = \binom {r + s} n$
The result follows on setting $r = e$, $s = \pi$ and $n = 2$.
$\blacksquare$
Proof 2
\(\ds \binom {e + \pi} 2\) | \(=\) | \(\ds \dfrac {\paren {e + \pi} \times \paren {e + \pi - 1} } {2 \times 1}\) | Definition of Binomial Coefficient | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 2 \paren {\paren {e + \pi}^2 - \paren {e + \pi} }\) | 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 | |||||||||||
\(\ds \) | \(=\) | \(\ds 1 \times \dfrac {\paren {\pi} \paren {\pi - 1} } 2 + e \times \pi + \dfrac {\paren {e} \paren {e - 1} } 2 \times 1\) | Definition of Binomial Coefficient | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 2 \paren {\pi^2 - \pi + 2e\pi + e^2 - e}\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 2 \paren {e^2 + 2e\pi + \pi^2 - e - \pi}\) | rearranging | |||||||||||
\(\ds \) | \(=\) | \(\ds \binom {e + \pi} 2\) | from above |
$\blacksquare$