Question

A simplificação da expressão  resulta em:

Answer 1

$\dfrac{(n+1)!-n!+(n-1)!}{n!}$

$(n+1)!=(n+1)n(n-1)!$

$n!=n(n-1)!$

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

$\dfrac{(n+1)!-n!+(n-1)!}{n!}=\dfrac{(n-1)![(n+1)n-n+1]}{n(n-1)!}$

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

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