Question

(50 PONTOS) Obter uma fórmula fechada para a seguinte soma:
$\displaystyle\sum\limits_{n=0}^{k}{n\cdot n!}$
-----------------------------------------------------------------------------------------------------
Resposta: $(k+1)!-1$

Answer 1

Definindo a sequência $a_{n}=n!$, temos que

$a_{n+1}-a_{n}=(n+1)!-n!\\\\a_{n+1}-a_{n}=(n+1)\cdot n!-n!$

Colocando n! em evidência, temos

$a_{n+1}-a_{n}=n!\cdot[(n+1)-1]\\\\a_{n+1}-a_{n}=n!\cdot[n+1-1]\\\\\boxed{\boxed{a_{n+1}-a_{n}=n\cdot n!}}$

Portanto:

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

Pelo resultado $\displaystyle\sum\limits_{n=N_{0}}^{N}(a_{n+1}-a_{n})=a_{N+1}-a_{N_{0}}$, tomando N₀ = 0 e N = k:

$\displaystyle\sum\limits_{n=0}^{k}n\cdot n!=a_{k+1}-a_{0}\\\\\\\displaystyle\sum\limits_{n=0}^{k}n\cdot n!=(k+1)!-0!\\\\\\\boxed{\boxed{\displaystyle\sum\limits_{n=0}^{k}n\cdot n!=(k+1)!-1}}$