Matemática, perguntado por Lukyo, 1 ano atrás

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.

Soluções para a tarefa

Respondido por superaks
6
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.


Lukyo: Excelente! Muito obrigado. :)
superaks: :D
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