Matemática
manucarneiro1
3

1a) 3!5!/4!6! d) h!/(n - 2! e) (n+1)!/(n+2) f) (n+3)!/(n-2)! . (n-1)!/n+1)!

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(1) Respostas
joanesmel

a) [latex]\frac{3!5!}{4!6!}=\frac{3!5!}{4.3!.6.5!}=\frac{1}{24}[/latex] d) [latex]\frac{n!}{(n-2)!}=\frac{n.(n-1).(n-2)!}{(n-2)!}=n(n-1)[/latex] e) [latex]\frac{(n+1)!}{(n+2)!}=\frac{(n+1)!}{(n+2)(n+2)!}=\frac{1}{(n+2)}[/latex] f) [latex]{\frac{(n+3)!}{(n-2)!}}.{\frac{(n-1)!}{(n+1)!}}=\\{\frac{(n+3)(n+2)(n+1)n(n-1)(n-2)!}{(n-2)!}}.{\frac{(n-1)!}{(n+1)n(n-1)!}}=\\{(n+3)(n+2)(n+1)n(n-1)}.{\frac{1}{(n+1)n}=\\(n+3)(n+2)(n-1)[/latex] O Mozean espera ter ajudado!

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