Propriedades dos somatórios.Calcule o somatório k · C(n, k)ondeC(n, k) = n!/[k! · (n − k)!]é o coefi.... Pergunta de ideia deLukyo

December 2019 1 170 Report

Propriedades dos somatórios.

Calcule o somatório

$\mathsf{\displaystyle\sum_{k=0}^n}$ k · C(n, k)

onde

C(n, k) = n!/[k! · (n − k)!]

é o coeficiente binomial.

Lista de comentários

superaks

Verified answer

Olá Lukyo.

Propriedades utilizadas

$\star~\boxed{\boxed{\mathsf{\displaystyle\sum_{k=p}^{n-p}f(k)=\displaystyle\sum_{k=p}^{n-p} f(n-k)}}}\\\\\\\\\star~\boxed{\boxed{\mathsf{(a+b)^n=\displaystyle\sum_{k=0}^n\binom{n}{k} a^{n-k}\cdot b^k}}}\\\\\\\\~\star\boxed{\boxed{\mathsf{\binom{n}{k}=\binom{n}{n-k}}}}$

_____________________

Iguale o somatório a S

$\mathsf{S=\displaystyle\sum_{k=0}^n k\cdot \binom{n}{k}}$

Pela troca de variáveis, temos que

$\mathsf{\displaystyle\sum_{k=0}^{n}k\cdot \binom{n}{k}=\displaystyle\sum_{k=0}^{n}(n-k)\cdot \binom{n}{n-k}}$

Portanto, se somarmos os somatórios, temos

$\mathsf{2S=\mathsf{\displaystyle\sum_{k=0}^{n}k\cdot \binom{n}{k}+\displaystyle\sum_{k=0}^{n}(n-k)\cdot \binom{n}{n-k}}}$

Sabendo que:

$\mathsf{\displaystyle\binom{n}{k}=\binom{n}{n-k}}$

$\mathsf{2S=\mathsf{\displaystyle\sum_{k=0}^{n}k\cdot \binom{n}{k}+\displaystyle\sum_{k=0}^{n}(n-k)\cdot \binom{n}{k}}}\\\\\\\\\mathsf{2S=\mathsf{\displaystyle\sum_{k=0}^{n}k\cdot \binom{n}{k}+\displaystyle\sum_{k=0}^{n}n\cdot\binom{n}{k}-\displaystyle\sum_{k=0}^{n}k\cdot \binom{n}{k}}}\\\\\\\\\mathsf{2S=\mathsf{\displaystyle\sum_{k=0}^{n}n\cdot \binom{n}{k}}}\\\\\\\\\mathsf{2S=\mathsf{n\cdot\displaystyle\sum_{k=0}^{n}\binom{n}{k}}}$

Achando o somatório $\mathsf{\displaystyle\sum_{k=0}^n\binom{n}{k}}$

Não é difícil mostrar que esse somatório é igual a $\mathsf{2^n}$

$\mathsf{(1+1)^n=\displaystyle\sum_{k=0}^n\binom{n}{k}\cdot 1^{n-k}\cdot 1^k}\\\\\\\mathsf{2^n=\displaystyle\binom{n}{0}+1\binom{n}{1}+1^2\binom{n}{2}+...+1^n\binom{n}{n}}$

Portanto, temos

$\mathsf{2S=n\cdot \displaystyle\sum_{k=0}^n\binom{n}{k}}\\\\\\\mathsf{2S=n\cdot 2^n}\\\\\\\mathsf{S=\dfrac{n2^n}{2}}\\\\\\\boxed{\mathsf{S=n2^{n-1}}}$

Concluindo o que queríamos achar

Dúvidas? Comente.

4 votes Thanks 5

Lukyo Excelente! Muito obrigado. :)

superaks :D