1) Simplifique a expressão
[(n+2)! - (n+1)!] .n! / [(n+1)!]², n ∈ N.
Soluções para a tarefa
Respondido por
3
Olá Maslip,
Organizando a equação:
![\mathsf{\dfrac{[(n+2)! - (n+1)!]\cdot n!} {[(n+1)!]^2}~~~~~~~~~~~~~~~ ~\forall ~n\in \mathbb{N}}\\\\=\\\\\mathsf{\dfrac{[(n+2)\cdot(n+1)!-(n+1)!]\cdot n!}{(n+1)!\cdot(n+1)!}}\\\\=\\\\\mathsf{\dfrac{[(n+1)!\cdot(n+2-1)]\cdot \diagup\!\!\!\!n!}{(n+1)!\cdot (n+1)\cdot \diagup\!\!\!\!n!}}\\\\=\\\\\mathsf{\dfrac{(n+1)!\cdot(n+1)}{(n+1)!\cdot (n+1)}}\\\\=\\\\\boxed{\mathsf{1}}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdfrac%7B%5B%28n%2B2%29%21+-+%28n%2B1%29%21%5D%5Ccdot+n%21%7D+%7B%5B%28n%2B1%29%21%5D%5E2%7D%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E+%7E%5Cforall+%7En%5Cin+%5Cmathbb%7BN%7D%7D%5C%5C%5C%5C%3D%5C%5C%5C%5C%5Cmathsf%7B%5Cdfrac%7B%5B%28n%2B2%29%5Ccdot%28n%2B1%29%21-%28n%2B1%29%21%5D%5Ccdot+n%21%7D%7B%28n%2B1%29%21%5Ccdot%28n%2B1%29%21%7D%7D%5C%5C%5C%5C%3D%5C%5C%5C%5C%5Cmathsf%7B%5Cdfrac%7B%5B%28n%2B1%29%21%5Ccdot%28n%2B2-1%29%5D%5Ccdot+%5Cdiagup%5C%21%5C%21%5C%21%5C%21n%21%7D%7B%28n%2B1%29%21%5Ccdot+%28n%2B1%29%5Ccdot+%5Cdiagup%5C%21%5C%21%5C%21%5C%21n%21%7D%7D%5C%5C%5C%5C%3D%5C%5C%5C%5C%5Cmathsf%7B%5Cdfrac%7B%28n%2B1%29%21%5Ccdot%28n%2B1%29%7D%7B%28n%2B1%29%21%5Ccdot+%28n%2B1%29%7D%7D%5C%5C%5C%5C%3D%5C%5C%5C%5C%5Cboxed%7B%5Cmathsf%7B1%7D%7D "\mathsf{\dfrac{[(n+2)! - (n+1)!]\cdot n!} {[(n+1)!]^2}~~~~~~~~~~~~~~~ ~\forall ~n\in \mathbb{N}}\\\\=\\\\\mathsf{\dfrac{[(n+2)\cdot(n+1)!-(n+1)!]\cdot n!}{(n+1)!\cdot(n+1)!}}\\\\=\\\\\mathsf{\dfrac{[(n+1)!\cdot(n+2-1)]\cdot \diagup\!\!\!\!n!}{(n+1)!\cdot (n+1)\cdot \diagup\!\!\!\!n!}}\\\\=\\\\\mathsf{\dfrac{(n+1)!\cdot(n+1)}{(n+1)!\cdot (n+1)}}\\\\=\\\\\boxed{\mathsf{1}}")
O resultado da simplificação é igual a 1!
Dúvidas? comente
Organizando a equação:
O resultado da simplificação é igual a 1!
Dúvidas? comente
Perguntas interessantes