Para os inteiros positivos $k$ e $n$, com $k\leq n$, sabe-se que $\dfrac{n+1}{k+1} {n\choose k}={n+1\choose k+1}$. Então, o valor de ${n\choose 0}+\dfrac{1}{2}{n\choose 1}+\dfrac{1}{3}{n\choose 2}+\cdot \cdot\cdot+\dfrac{1}{n+1}{n\choose n}$ é igual a


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ITA IIIT 26/11/2021 15:17
$-$ Vejamos: \begin{matrix} {n \choose 0} + \frac{1}{2}{n \choose 1} +...+ \frac{1}{n+1}{n \choose n} &\Rightarrow& \sum_{k=0}^{n} {n \choose k} \frac{1}{k+1} &\Rightarrow& \sum \frac{n!}{k!.(n-k)!}.\frac{1}{k+1}.\color{royalblue}{\frac{n+1}{n+1}} &\Rightarrow& \sum \frac{(n+1)!}{(k+1)!.(n-k)!}.\frac{1}{n+1} &\therefore& \frac{1}{n+1}. \sum {n+1 \choose k+1} \end{matrix}Com o conhecimento do $Teorema \ das \ Linhas$ \begin{matrix} {n \choose 0} + {n \choose 1} +...+ {n \choose n} = 2^n &\Rightarrow& {n \choose 1} +...+ {n \choose n} = 2^n -{n \choose 0}&\therefore& {n \choose 1} +...+ {n \choose n} = 2^n -1 \end{matrix}Dessa forma: \begin{matrix} \frac{1}{n+1}. \sum {n+1 \choose k+1} = {\Large{\frac{2^{n+1} - 1}{n+1} }}\\ \\ Letra \ (D) \end{matrix}
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