• Matéria: Matemática
  • Autor: mariasaopaulo
  • Perguntado 9 anos atrás

Alguém saberia resolver esse somatório?

Anexos:

Respostas

respondido por: carlosmath
1
\displaystyle
S=\sum\limits_{k=0}^n(k+1)\binom{n}{k}\\ \\ \\
S=\sum\limits_{k=0}^nk\binom{n}{k}+\sum\limits_{k=0}^n\binom{n}{k}\\ \\ \\
S=\sum\limits_{k=0}^nk\binom{n}{k}+2^n\\ \\ \\
S=\sum\limits_{k=0}^n\frac{n!}{(n-k)!}+2^n

\displaystyle
S=n!\sum\limits_{k=0}^n\frac{1}{k!}+2^n\\ \\ \\
\text{Debes saber que:}\\ \\
\Gamma(n,x)=(n-1)!\;e^{-x}\sum\limits_{k=0}^{n-1}\frac{x^k}{k!}\Longrightarrow
\Gamma(n+1,1)=n!\;e^{-1}\sum\limits_{k=0}^{n}\frac{1}{k!}\\ \\ \\
\Large\boxed{S=e\cdot\Gamma(n+1,1)+2^n}

mariasaopaulo: Obrigadaaa :)
Perguntas similares