Binômio de newton

Prove que:

[latex]2*1\binom{n}{2}+3*2\binom{n}{3}+4*3\binom{n}{4}+...+n(n-1)\binom{n}{n}=n(n-1)*2^{n-2}[/latex]

Temos a equação:


[latex]S = \sum_{k=2}^{n}\left [ k\cdot (k-1)\binom{n}{k} \right ][/latex]


Observe que:


[latex]\binom{n}{j} = \frac{n}{j}\cdot \binom{n-1}{j-1}[/latex]


[latex]\binom{n-1}{j-1} = \frac{n-1}{j-1}\cdot \binom{n-2}{j-2}[/latex]


Aplicando o Substituição:


[latex]\binom{n}{j} = \frac{n}{j}\cdot \binom{n-1}{j-1} = \frac{n}{j}\cdot \frac{n-1}{j-1}\cdot \binom{n-2}{j-2}[/latex]


[latex]j\cdot (j-1)\cdot \binom{n}{j} = n\cdot (n-1)\cdot \binom{n-2}{j-2}[/latex]


Com isso,


[latex]S = \sum_{k=2}^{n}\left [ k\cdot (k-1)\binom{n}{k} \right ][/latex]


[latex]S = \sum_{k=2}^{n}\left [n\cdot (n-1)\cdot \binom{n-2}{k-2} \right ][/latex]


[latex]S = n\cdot (n-1)\cdot \sum_{k=2}^{n}\left [\binom{n-2}{k-2} \right ][/latex]


[latex]S = n\cdot (n-1)\cdot \sum_{k=0}^{n-2}\left [\binom{n-2}{k} \right ][/latex]


[latex]S = n\cdot (n-1)\cdot 2^{n-2}[/latex]