Question

O número e, uma das constantes mais importantes da Matemática, pode ser
definido por:

$\mathsf{e=\displaystyle\sum_{k = 0}^{\infty} \dfrac{1}{k!}=\dfrac{1}{0!}+\dfrac{1}{1!}+\dfrac{1}{2!}+...}$

em que 0! = 1 e n! = 1 · 2 · 3 · … · n para n > 0.

Então o número

$\mathsf{\displaystyle\sum_{k=0}^{\infty}\dfrac{(k+1)^2}{k!}=\dfrac{1^2}{0!}+\dfrac{2^2}{1!}+\dfrac{3^2}{2!}+...}$

É igual a:

A) 2e

B) 4e

C) 5e

D) e²

E) (e + 1)²

__________________

Por favor responder de forma detalhada.

Answer 1

É dada a série:

$\displaystyle S=\sum_{k=0}^{\infty}\dfrac{(k+1)^2}{k!}$

Abrindo o quadrado presente no termo geral:

$\displaystyle S=\sum_{k=0}^{\infty}\dfrac{k^2+2k+1}{k!}\\\\ S=\sum_{k=0}^{\infty}\left(\dfrac{k^2}{k!}+\dfrac{2k}{k!}+\dfrac{1}{k!}\right)\\\\ S=\underbrace{\sum_{k=0}^{\infty}\dfrac{k^2}{k!}}_{S_1}+\underbrace{\sum_{k=0}^{\infty}\dfrac{2k}{k!}}_{S_2}+\underbrace{\sum_{k=0}^{\infty}\dfrac{1}{k!}}_{S_3}$

Do enunciado, temos diretamente que $S_3=e$. Vamos calcular a série $S_2$:

$\displaystyle S_2=\sum_{k=0}^{\infty}\dfrac{2k}{k!}=2\sum_{k=0}^{\infty}\dfrac{k}{k!}\\\\ S_2=2\left(\dfrac{0}{0!}+\sum_{k=1}^{\infty}\dfrac{k}{k!}\right)\\\\ S_2=2\left(0+\sum_{k=1}^{\infty}\dfrac{1}{(k-1)!}\right)\\\\ S_2=2\sum_{k=1}^{\infty}\dfrac{1}{(k-1)!}$

Fazendo a troca $k-1\to n$:

$\displaystyle S_2=2\sum_{(n-1)=1}^{\infty}\dfrac{1}{n!}\\\\ S_2=2\sum_{n=0}^{\infty}\dfrac{1}{n!}\\\\ S_2=2e$

Resta apenas analisar $S_1$:

$\displaystyle S_1=\sum_{k=0}^{\infty}\dfrac{k^2}{k!}\\\\ S_1=\dfrac{0^2}{0!}+\sum_{k=1}^{\infty}\dfrac{k^2}{k!}\\\\ S_1=\sum_{k=1}^{\infty}\dfrac{k}{(k-1)!}\\\\ S_1=\sum_{k=1}^{\infty}\left(\dfrac{k-1}{(k-1)!}+\dfrac{1}{(k-1)!}\right)\\\\ S_1=\sum_{k=1}^{\infty}\dfrac{k-1}{(k-1)!}+\sum_{k=1}^{\infty}\dfrac{1}{(k-1)!}\\\\ S_1=\dfrac{1-1}{(1-1)!}+\sum_{k=2}^{\infty}\dfrac{k-1}{(k-1)!}+\sum_{k=1}^{\infty}\dfrac{1}{(k-1)!}\\\\ S_1=\sum_{k=2}^{\infty}\dfrac{1}{(k-2)!}+\sum_{k=1}^{\infty}\dfrac{1}{(k-1)!}$

Fazendo as trocas $k-2\to n$ e $k-1\to n'$ nas primeira e segunda séries acima, respectivamente:

$\displaystyle S_1=\sum_{(n-2)=2}^{\infty}\dfrac{1}{n!}+\sum_{(n'+1)=1}^{\infty}\dfrac{1}{n'!}\\\\ S_1=\sum_{n=0}^{\infty}\dfrac{1}{n!}+\sum_{n'=0}^{\infty}\dfrac{1}{n'!}\\\\ S_1=e+e\\\\ S_1=2e$

Portanto, o resultado final é:

$S=S_1+S_2+S_3\\\\ S=2e+2e+e\\\\ S=5e\\\\ \boxed{\displaystyle S=\sum_{k=0}^{\infty}\dfrac{(k+1)^2}{k!}=5e}\Longrightarrow\text{Letra }\bold{C}$