russ9 wrote:
Bunuel wrote:
Hraisha wrote:
Hi Bunuel! In your explanation for statement 1, I understand that (x+2)! equals to (x+2)*(x+1)*x in the nominator but I don't understand how you got rid of (x-1)! in the denominator. Could you please explain?
Thanks for your help!!
Thanks for your help!!
Not so.
(x+2)!=1*2*3*...*(x-1)*x*(x+1)*(x+2)=(x-1)!*x*(x+1)*(x+2)
\frac{(x+2)!}{(x-1)!}=\frac{(x-1)!*x*(x+1)*(x+2)}{(x-1)!} =x*(x+1)*(x+2)
...
