Question

alguém desenvolve pfv! Mas bem desenvolvido, passo a passo

Answer 1

$n\cdot2^{n-1}=n\cdot(1+1)^{n-1}\\\\n\cdot2^{n-1}=n\cdot\displaystyle\sum\limits_{k=0}^{n-1}\binom{n-1}{k}1^{k}1^{n-1-k}\\\\\\n\cdot2^{n-1}=\sum\limits_{k=0}^{n-1}n\binom{n-1}{k}$

Vamos manipular a seguinte expressão:

$n\displaystyle\binom{n-1}{k}=n\cdot\dfrac{(n-1)!}{k!(n-1-k)!}\\\\\\n\binom{n-1}{k}=\dfrac{n\cdot(n-1)!}{k!(n-[k+1])!}\\\\\\n\binom{n-1}{k}=\dfrac{n!}{k!(n-[k+1])!}\\\\\\n\binom{n-1}{k}=(k+1)\cdot\dfrac{n!}{(k+1)\cdot k!\cdot(n-[k+1])!}\\\\\\n\binom{n-1}{k}=(k+1)\cdot\dfrac{n!}{(k+1)!\cdot(n-[k+1])!}\\\\\\\boxed{\boxed{n\binom{n-1}{k}=(k+1)\binom{n}{k+1}}}$

Então:

$n\cdot2^{n-1}=\displaystyle\sum\limits_{k=0}^{n-1}n\binom{n-1}{k}\\\\\\n\cdot2^{n-1}=\sum\limits_{k=0}^{n-1}(k+1)\binom{n}{k+1}\\\\\\n\cdot2^{n-1}=1\binom{n}{1}+2\binom{n}{2}+...+n\binom{n}{n}\\\\\\n\cdot2^{n-1}=0\binom{n}{0}+1\binom{n}{1}+...+n\binom{n}{n}\\\\\\\boxed{\boxed{n\cdot2^{n-1}=\sum\limits_{k=0}^{n}k\binom{n}{k}}}$
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$s=\displaystyle\sum\limits_{k=0}^{n}k\binom{n}{k}+2^{n}\\\\\\s=\sum\limits_{k=0}^{n}k\binom{n}{k}+\sum\limits_{k=0}^{n}2^{n}\\\\\\s=n\cdot2^{n-1}+(n+1)\cdot2^{n}\\\\s=2^{n}\cdot2^{-1}\cdot n+2^{n}\cdot(n+1)\\\\s=2^{n}\cdot\left(\frac{n}{2}+n+1\right)\\\\\\s=2^{n}\cdot\dfrac{3n+2}{2}\\\\\\\boxed{\boxed{s=2^{n-1}\cdot(3n+2)}}$