Determine o conjunto verdade da equação:
(n+1)!-n! / (n-1)! = 8n
A resposta é 8, mas queria saber como chegar a esse resultado.
Respostas
respondido por:
4
Basta abrir os fatoriais, assim:
![\frac{(n+1)! - n!}{(n-1)!} = 8n \\ \\ \frac{[(n+1)((n+1)-1)((n+1)-2)\cdot...\cdot1] - n!}{(n-1)((n-1)-1)\cdot...\cdot1} = 8n \\ \\ \frac{[(n+1)(n)(n-1)(n-2)\cdot...\cdot1]-n\cdot(n-1)(n-2)...}{(n-1)(n-2)\cdot...\cdot1} = 8n \\ \\ (n+1)n - n = 8n \\ \\ (n+1) - 1 = 8 \\ \\ \boxed{\boxed{n = 8}}](https://tex.z-dn.net/?f=+%5Cfrac%7B%28n%2B1%29%21+-+n%21%7D%7B%28n-1%29%21%7D+%3D+8n+%5C%5C++%5C%5C++%5Cfrac%7B%5B%28n%2B1%29%28%28n%2B1%29-1%29%28%28n%2B1%29-2%29%5Ccdot...%5Ccdot1%5D+-+n%21%7D%7B%28n-1%29%28%28n-1%29-1%29%5Ccdot...%5Ccdot1%7D+%3D+8n+%5C%5C++%5C%5C++%5Cfrac%7B%5B%28n%2B1%29%28n%29%28n-1%29%28n-2%29%5Ccdot...%5Ccdot1%5D-n%5Ccdot%28n-1%29%28n-2%29...%7D%7B%28n-1%29%28n-2%29%5Ccdot...%5Ccdot1%7D+%3D+8n+%5C%5C++%5C%5C+%28n%2B1%29n+-+n+%3D+8n+%5C%5C++%5C%5C+%28n%2B1%29+-+1+%3D+8+%5C%5C++%5C%5C+%5Cboxed%7B%5Cboxed%7Bn+%3D+8%7D%7D "\frac{(n+1)! - n!}{(n-1)!} = 8n \\ \\ \frac{[(n+1)((n+1)-1)((n+1)-2)\cdot...\cdot1] - n!}{(n-1)((n-1)-1)\cdot...\cdot1} = 8n \\ \\ \frac{[(n+1)(n)(n-1)(n-2)\cdot...\cdot1]-n\cdot(n-1)(n-2)...}{(n-1)(n-2)\cdot...\cdot1} = 8n \\ \\ (n+1)n - n = 8n \\ \\ (n+1) - 1 = 8 \\ \\ \boxed{\boxed{n = 8}}")
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