Question

me ajudeem
a) (n+2)! / 2! (n+1)
b)(n+2)! + (n+1)! / (n+3)!
c) n! = 6
d) (n-2)! = 1
e) (n+1)! = 24​

Answer 1

Bem,vamos lá:

A)

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

(n+2)!×n+(n+2)!/2!=

(n+2)!×n+(n+2)!/2

B)

(n+2)! + (n+1)! / (n+3)!=

(n+2)×(n+1)!+(n+1)!/(n+3)!=

(n+2)×(n+1)!+ (n+1)!/(n+3)×(n+2)!=

(n+3)×1/(n+3)×(n+2)=

1/n+2

C)

n! = 6

6×n!

D)

(n-2)! = 1

N-2=0

N-2=1

N=2

N=3

(2-2)!=1

(3-2)!=1

N=2

N=3

E)

(n+1)! = 24​

24×(n+1)!

Espero ter ajudado!