Question

Simplifique a expressão: (n+2)!+(n+1) vezes (n-1)! dividido por (n+1) vezes (n-1)!

Answer 1

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

Note que (n + 2)! = (n + 2)(n + 1)n(n - 1)!, então:

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

Cancelando (n + 1) e (n - 1)!, temos

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