For the positive integer n, the rational number r satisfies \( \binom{2n}{n} - \binom{2n}{n + 1} = r \cdot \binom{2n}{n}.\) Express r in terms of n (your problem should be as simplified as possible).
(2n)! (2n)! (2n)!
________ - ___________________ = r * _________ rearrange as
(2n -n)! n! (2n - (n + 1))! (n + 1)! (2n - n)! n!
(2n)! (1 -r) (2n)!
_________ = __________________ (divide out (2n)! and multiply through by -1 )
(2n - n)! n! (2n - (n + 1))! (n + 1)!
(1 - r) 1
_________ = _____________
(n!) (n!) (n - 1)! (n + 1)!
( r - 1) -1
______ = ________________
(n!)(n!) (n -1)! ( n + 1)!
(r -1) = - (n! (n!)
_____________
(n - 1)! (n + 1)!
r - 1 = - (n)! / ( n -1)! * n! / (n + 1)!
r -1 = - n * 1/(n + 1)
r - 1 = - n / (n + 1)
r = 1 - [ n / (n + 1) ]
r = ( n + 1 - n) / ( n + 1)
r = 1 / ( n + 1)