Question

simplifique: a)(n+1)!-n! b) (n+2)+(n+1)! / (n+1)!=​

Answer 1

Resposta:

$\boxed{\boxed{\mathbf{a)\,\,n\cdot n!}}}$

$\boxed{\boxed{\mathbf{b)\,\,n+3}}}$

Solução:

Sabendo que $n!=n(n-1)(n-2)(n-3)...$, a cada termo você subtrai 1 do fatorial anterior, de posse disso, segue que

a)

$(n+1)!-n!\\\\\\=(n+1)(n+1-1)!-n!\\\\\\=(n+1)n!-n!\\\\\\=n\cdot n!+\not\!\!{n!}-\not\!\!{n!}\\\\\\=n\cdot n!$

b)

$\dfrac{(n+2)!+(n+1)!}{(n+1)!}\\\\\\=\dfrac{(n+2)!}{(n+1)!}+\dfrac{(n+1)!}{(n+1)!}\\\\\\=\dfrac{(n+2)(n+2-1)!}{(n+1)!}+\dfrac{(n+1)!}{(n+1)!}\\\\\\=\dfrac{(n+2)(n+1)!}{(n+1)!}+\dfrac{(n+1)!}{(n+1)!}\quad\mbox{simplificando}\,\,\, (n+1)!\\\\\\=(n+2)+1\\\\\\=n+3$

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