{[2(n+1)]!/[(n+1)!]^2}/[(2n)!/(n!)^2]得多少
2个回答

[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