[2(n+1)]!/[(n+1)!]^2=[(n+2)(n+3)...(2n)(2n+1)(2n+2)]/[(n+1)!]
(2n)!/(n!)^2=[(n+1)(n+2)...(2n)]/(n!)
所以{[2(n+1)]!/[(n+1)!]^2}/[(2n)!/(n!)^2]
={[(n+2)(n+3)...(2n)(2n+1)(2n+2)]/[(n+1)(n+2)...(2n)]}*[(n!)/(n+1)!]
={[(2n+1)(2n+2)]/(n+1)}/(n+1)
=[(2n+1)(2n+2)]/(n+1)^2