Question

Cum se rezolva acest exercitiu(n+1)!+5n!-40(n-1)=0?

Answer 1

n! =n(n-1)!
(n+1)! = (n+1)xnx(n-1)! = (n^2 +n)(n-1)!
Vom avea : 
(n^2 +n)(n-1)! + 5n(n-1)!  - 40(n-1)! =0
(n-1)!(n^2 +n +5n-40 ) =0
=> n^2 +6n -40 =0      
Delta = 36+160 =196 
n1 = (-6+14)/2 =8/2 =4 
n2 = (-6-14)/2 = -20/2 =-10  nu apartine lui N
deci n=4

Answer 2

C.E. n∈N*

(n-1)!(n(n+1)+5n-40)=0
n²+n+5n-40=0
n²+6n-40=0
n²+10n-4n-40=0
n(n+10)-4(n+10)=0
(n+10)(n-4)=0
n1=-10∉N*
n2=4 care convine
 verificare
 5!+5*4!=40*3!
120+120=40*6
240=240 adevarat , bine rezolvat