Binomio de newton

por thiago ro Dom 09 Dez 2012, 16:26

usando a relaçâo de Euler prove que

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$Binomio de newton Gif.latex?%5Cbinom%7B2n%7D%7Bn%7D=%5Cbinom%7Bn%7D%7B0%7D%5E%7B2%7D+%5Cbinom%7Bn%7D%7B1%7D%5E%7B2%7D+..$

por **Robson Jr.** Dom 09 Dez 2012, 18:27

Fórmula de Euler:

$\small \binom{m}{0}\binom{h}{p}+\binom{m}{1}\binom{h}{p-1}+\binom{m}{2}\binom{h}{p-2}+...+\binom{m}{p}\binom{h}{0}=\binom{m+h}{p}$

Fazendo m = p = h = n, vem:

$\small \binom{n}{0}\binom{n}{n}+\binom{n}{1}\binom{n}{n-1}+\binom{n}{2}\binom{n}{n-2}+...+\binom{n}{n}\binom{n}{0}=\binom{n+n}{n}=\binom{2n}{n}$

Aplicando a combinação complementar C(x,x-y) = C(x,y) em todos os segundos fatores de cada parcela:

$\small \binom{n}{0}\binom{n}{0}+\binom{n}{1}\binom{n}{1}+\binom{n}{2}\binom{n}{2}+...+\binom{n}{n}\binom{n}{n}=\binom{2n}{n} \\\\ \Leftrightarrow \ \binom{n}{0}^2+\binom{n}{1}^2+\binom{n}{2}^2+...+\binom{n}{n}^2=\binom{2n}{n}$

CqD