Question

não consigo desenvolver esta expressão, a resposta é D. alguém ajuda?

Answer 1

Temos que,

$(n+1)!+n!=n!\cdot[(n+1)+1]=n!\cdot(n+2)$
 
$(n+2)!=(n+2)(n+1)n!$.

Com isso, $\dfrac{(n+1)!+n!}{(n+2)!}=\dfrac{n!(n+2)}{(n+2)(n+1)n!}=\dfrac{1}{n+1}$.

Answer 2

$\frac{(n+1)!+n!}{(n+2)!}=\\\\\\\frac{(n+1)\cdot\,n!+n!}{(n+2)\cdot(n+1)\cdot\,n!}=\\\\\\\frac{n![(n+1)+1]}{(n+2)\cdot(n+1)\cdot\,n!}=\\\\\\\frac{\cancel{n!}\cancel{(n+2)}}{\cancel{(n+2)}\cdot(n+1)\cdot\,\cancel{n!}}=\\\\\\\boxed{\frac{1}{n+1}}$