Question

(50 PONTOS) Mostre que
$\displaystyle\sum\limits_{n=1}^{k}{(1+n^{2})\cdot n!}=k\cdot (k+1)!$

Answer 1

Seja $a_{n}=(n-1)\cdot n!$ uma sequência de números reais

Avaliando a diferença entre dois termos consecutivos dessa sequência:

$a_{n+1}-a_{n}=(n+1-1)(n+1)!-(n-1)n!\\\\a_{n+1}-a_{n}=n\cdot(n+1)!-(n-1)n!\\\\a_{n+1}-a_{n}=n\cdot(n+1)\cdot n!-(n-1)n!\\\\a_{n+1}-a_{n}=(n^{2}+n)n!-(n-1)n!$

Colocando n! em evidência:

$a_{n+1}-a_{n}=n!\cdot[n^{2}+n-(n-1)]\\\\a_{n+1}-a_{n}=n!\cdot[n^{2}+n-n+1]\\\\\boxed{\boxed{a_{n+1}-a_{n}=(1+n^{2})\cdot n!}}$

Então:

$\displaystyle\sum\limits_{n=1}^{k}(1+n^{2})\cdot n!=\sum\limits_{n=1}^{k}(a_{n+1}-a_{n})$

E, como vimos,

$\displaystyle\sum\limits_{n=1}^{k}(1+n^{2})\cdot n!=a_{k+1}-a_{1}\\\\\\\displaystyle\sum\limits_{n=1}^{k}(1+n^{2})\cdot n!=(k+1-1)(k+1)!-(1-1)1!\\\\\\\displaystyle\sum\limits_{n=1}^{k}(1+n^{2})\cdot n!=k(k+1)!-0\cdot1\\\\\\\boxed{\boxed{\displaystyle\sum\limits_{n=1}^{k}(1+n^{2})\cdot n!=k\cdot(k+1)!}}$