Prove that
2^{k}\begin{pmatrix} n\\0 \end{pmatrix}\begin{pmatrix} n\\k \end{pmatrix}-2^{k-1}\begin{pmatrix} n\\1 \end{pmatrix}\begin{pmatrix} n-1\\k-1 \end{pmatrix}+2^{k-2}\begin{pmatrix} n\\2 \end{pmatrix}\begin{pmatrix} n-2\\k-2 \end{pmatrix}-.......+(-1)^{k}\begin{pmatrix} n\\k \end{pmatrix}\begin{pmatrix} n-k\\0 \end{pmatrix}=\begin{pmatrix} n\\k \end{pmatrix}
-
UP 0 DOWN 0 0 3
3 Answers
21=\sum_{r=0}^{k}{(-1)^{r}.2^{k-r}.^{n}C_{r}.^{n-r}C_{k-r}}
=\sum_{r=0}^{k}{(-1)^{r}.2^{k-r}}.\frac{n!}{r!(n-r)!}.\frac{(n-r)!}{(k-r)!(n-k)!}
=\sum_{r=0}^{k}{(-1)^{r}.2^{k-r}}.\frac{n!}{k!(n-k)!}.\frac{k!}{r!(k-r)!}
=\sum_{r=0}^{k}{(-1)^{r}.2^{k-r}}.^{n}C_{k}.^{k}C_{r}
=2^{k}.^{n}C_{k}\left\{ \sum_{r=0}^{k}(-1)^{r}.\frac{1}{2^{r}}.^{k}C_{r} \right\}
=2^{k}.^{n}C_{k}\left(1-\frac{1}{2} \right) ^{k}
=^{n}C_{k}