[(8+n)!-(7+n)!]/(2n-1)=15!
tkhunny Moderator Staff member Joined Apr 12, 2005 Messages 11,339 Nov 20, 2019 #2 Have you expended any particular efforts? Write out the first two factors of (8+n)!
F fiuf New member Joined Nov 20, 2019 Messages 2 Nov 20, 2019 #3 So I'm at [(7+n)!(7+n)]/(2n-1)=15! right now, and don't know how to continue from there
tkhunny Moderator Staff member Joined Apr 12, 2005 Messages 11,339 Nov 20, 2019 #4 It may take a conceptual leap. No matter what n is, (7+n)/(2n-1) doesn't contribute much. What does that say about how (7+n)! and 15! need to compare?
It may take a conceptual leap. No matter what n is, (7+n)/(2n-1) doesn't contribute much. What does that say about how (7+n)! and 15! need to compare?
Steven G Elite Member Joined Dec 30, 2014 Messages 14,598 Nov 20, 2019 #5 I would get to \(\displaystyle \dfrac{(7+n)!}{15!} = 2-\dfrac{15}{n+7}\) Can you finish from here?
