\( \begin{pmatrix} n \\ k \end{pmatrix}\) = \(\frac{n!}{k!·(n-k)!}\)
Induktionsschluss:
\(\sum\limits_{k=1}^{n+1} \) \( \begin{pmatrix} k \\ 2 \end{pmatrix}\)
= \(\sum\limits_{k=1}^{n} \) \( \begin{pmatrix} k \\ 2 \end{pmatrix}\) + \( \begin{pmatrix} n+1 \\ 2\end{pmatrix}\) =IV \( \begin{pmatrix} n+1 \\ 3 \end{pmatrix}\) + \( \begin{pmatrix} n+1 \\ 2 \end{pmatrix}\)
= \(\frac{(n+1)!}{3!·(n-2)!}\) + \(\frac{n+1)!}{2!·(n-1)!}\) = \(\frac{(n+1)!·(n-1)+3·(n+1)!}{3!·(n-1)!}\) = \(\frac{(n+1)!·(n-1+3)}{3!·(n-1)!}\)
= \(\frac{(n+1)!·(n+2)}{3!·(n-1)!}\) = \(\frac{(n+2)!}{3!·(n-1)!}\)
= \( \begin{pmatrix} n+2 \\ 3 \end{pmatrix}\)
Gruß Wolfgang