Ensino Médio ⇒ Arranjo Tópico resolvido

[owen123]

Resolva as seguintes equações:

1) 𝐴2𝑛,2 − 𝐶2𝑛+1,2𝑛−1 = 20

2) 𝐶𝑛,3 = 3 ∙ 𝐴𝑛,2

[petras]

owen123,

[tex3]1) \frac{2n!}{(2n-2)!}-\frac{(2n+1)!}{(2n-1)!(2n+1-2n+1)!}=20\\
\frac{(2n(2n-1)\cancel{(2n-2)!}}{\cancel{(2n-2)!}}-\frac{(2n+1)(2n)\cancel{(2n-1)!}}{\cancel{(2n-1)!}2!}=20\\
4n(2n-1)-(2n+1)(2n)=40\\
8n^2-4n-4n^2-2n=40\rightarrow 4n^2-6n- 40=0\rightarrow2n^2-3n-20=0 \\
\therefore n = \cancel{-\frac{5}{2}}~ ou~ \boxed{n =4}\\
2)\frac{\cancel{n!}}{3!(n-3)!}=3.\frac{\cancel{n!}}{(n-2)!}\\
(n-2)! = 3.3!.(n-3)!\rightarrow (n-2)\cancel{(n-3)!}=18\cancel{(n-3)!}\\
\therefore \boxed{n=20} [/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]1) \frac{2n!}{(2n-2)!}-\frac{(2n+1)!}{(2n-1)!(2n+1-2n+1)!}=20\\
\frac{(2n(2n-1)\cancel{(2n-2)!}}{\cancel{(2n-2)!}}-\frac{(2n+1)(2n)\cancel{(2n-1)!}}{\cancel{(2n-1)!}2!}=20\\
4n(2n-1)-(2n+1)(2n)=40\\
8n^2-4n-4n^2-2n=40\rightarrow 4n^2-6n- 40=0\rightarrow2n^2-3n-20=0 \\
\therefore n = \cancel{-\frac{5}{2}}~ ou~ \boxed{n =4}\\
2)\frac{\cancel{n!}}{3!(n-3)!}=3.\frac{\cancel{n!}}{(n-2)!}\\
(n-2)! = 3.3!.(n-3)!\rightarrow (n-2)\cancel{(n-3)!}=18\cancel{(n-3)!}\\
\therefore \boxed{n=20} [/tex3]