Question

(30 PONTOS) Sabemos que é valida a seguinte propriedade para os somátórios:
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$\mathbf{(P1)}~~~\displaystyle\sum_{k=0}^{n}{f(k)}=\sum_{k=0}^{n}{f(n-k)}$


Utilize $\mathbf{(P1)}$ para mostrar que

$S=\displaystyle\sum_{k=0}^{n}{k\dbinom{n}{k}}=n\cdot 2^{n-1}.$
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Answer 1

Considerando $f(k)=k\displaystyle\binom{n}{k},~k=0,1,...,n,$ temos:

$\displaystyle\sum\limits_{k=0}^{n}f(k)=\sum\limits_{k=0}^{n}k\binom{n}{k}=\sum\limits_{k=0}^{n}(n-k)\binom{n}{n-k}=\sum\limits_{k=0}^{n}f(n-k)$
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Porém, note que $\binom{n}{k}=\binom{n}{n-k}$:

$\displaystyle\binom{n}{n-k}=\dfrac{n!}{(n-k)!(n-[n-k])!}\\\\\\\binom{n}{n-k}=\dfrac{n!}{(n-k)!(n-n+k)!}\\\\\\\binom{n}{n-k}=\dfrac{n!}{k!(n-k)!}\\\\\\\boxed{\boxed{\binom{n}{n-k}=\binom{n}{k}}}$

Então:

$\displaystyle\sum\limits_{k=0}^{n}k\binom{n}{k}=\sum\limits_{k=0}^{n}(n-k)\binom{n}{n-k}=\sum\limits_{k=0}^{n}(n-k)\binom{n}{k}\\\\\\\displaystyle\sum\limits_{k=0}^{n}k\binom{n}{k}=\sum\limits_{k=0}^{n}\left[n\binom{n}{k}-k\binom{n}{k}\right]\\\\\\\sum\limits_{k=0}^{n}k\binom{n}{k}=\sum\limits_{k=0}^{n}n\binom{n}{k}-\sum\limits_{k=0}^{n}k\binom{n}{k}\\\\\\\sum\limits_{k=0}^{n}k\binom{n}{k}+\sum\limits_{k=0}^{n}k\binom{n}{k}=n\sum\limits_{k=0}^{n}\binom{n}{k}$

$2\displaystyle\sum\limits_{k=0}^{n}k\binom{n}{k}=n\sum\limits_{k=0}^{n}\binom{n}{k}1^{k}1^{n-k}$

Na direita, temos a expansão de um Binômio de Newton:

$2^{n}=(1+1)^{n}=\displaystyle\sum\limits_{k=0}^{n}\binom{n}{k}1^{k}1^{n-k}$

Logo:

$2\displaystyle\sum\limits_{k=0}^{n}k\binom{n}{k}=n\cdot2^{n}\\\\\\\sum\limits_{k=0}^{n}k\binom{n}{k}=n\cdot\dfrac{2^{n}}{2}\\\\\\\boxed{\boxed{\sum\limits_{k=0}^{n}k\binom{n}{k}=n\cdot2^{n-1}}}$