Question

n!/(n+2)!+(n+1)!=1/48 me ajudem pfv ​

Answer 1

n! / (n + 2)! + (n + 1)!  =  1/48

n! / (n + 2)*(n + 1)*n!  +  (n + 1)*n!   =   1/48

n! / n! [ (n + 2)*(n + 1) + (n + 1) ]  =  1/48

1 / (n + 2)*(n + 1) + (n + 1)  =  1/48

1 / n² + n + 2n + 2 + n + 1  =  1/48

1 / n² + 4n + 3  =  1/48

n² + 4n + 3 = 48

n² + 4n + 3 - 48 = 0

n² + 4n - 45 = 0

Δ = b² - 4ac

Δ = (4)² - 4(1)(-45)

Δ = 16 + 180

Δ = 196

n = - b ± √Δ / 2a

n = - 4 ± 14 / 2

n' = 10/2

n' = 5

n'' = - 18/2

n'' = - 9  

Para n = 5 temos:

5! / (5 + 2)! + (5 + 1)! = 1 /48

5! / 7! + 6! = 1 /48

5! / 7 * 6 * 5! + 6 * 5! = 1 /48

5! / 5! * (42 + 6) = 1 /48

1 / 48 = 1 /48

Para n = - 9 , temos:

(-9)! / ( -9 + 2)! + ( -9 + 1)! = 1 /48

Não existe fatorial de número negativo.

Então n = 5.