udowodnic

[cwelinho]

1 . \(\displaystyle{ {n\choose k}}\)=\(\displaystyle{ {n\choose n-k}}\) gdzie n,k \(\displaystyle{ \in}\)N i n \(\displaystyle{ \geq}\) k

2. k×\(\displaystyle{ {n\choose k}}\)=n×\(\displaystyle{ {n-1\choose n-k}}\) gdzie n,k \(\displaystyle{ \in}\)N i n \(\displaystyle{ \geq}\) k

3. \(\displaystyle{ {n\choose k}}\) + \(\displaystyle{ {n\choose k+1}}\)= \(\displaystyle{ {n+1\choose k+1}}\)
gdzie n,k \(\displaystyle{ \in}\)N i n >k

z góry dzieki za pomoc

[Calasilyar]

1)
\(\displaystyle{ {n\choose k}=\frac{n!}{k!(n-k)!}=\frac{n!}{(n-k)!k!}={n\choose n-k}}\)

2)
\(\displaystyle{ k\cdot {n\choose k}=\frac{kn!}{k!(n-k)!}=n\cdot \frac{k(n-1)!}{k!(n-k)!}=n\cdot \frac{(n-1)!}{(k-1)!(n-k)!}=n\cdot \frac{(n-1)!}{(n-k)!(k-1)!}=n\cdot {n-1 \choose n-k}}\)

3)
\(\displaystyle{ {n\choose k}+{n\choose k+1}=\frac{n!}{k!(n-k)!}+\frac{n!}{(k+1)!(n-k-1)!}=\frac{n!}{(k!(n-k-1)!)}(\frac{1}{n-k}+\frac{1}{k+1})=\\
=\frac{n!}{k!(n-k-1)!}(\frac{k+1+n-k}{(n-k)(k+1)})=\frac{n!(n+1)}{k!(n-k-1)!(k+1)(n-k)}=\frac{(n+1)!}{(k+1)!(n-k)!}=\frac{(n+1)!}{(k+1)!((n+1)-(k+1))!}={n+1\choose k+1}}\)

[cwelinho]

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