C(2n,n) - C(2n,n-1) = (2n)!/[n!(n+1)!] = ___________.?

　(2n)!/[n!n!] - (2n)!/[(n-1)!(n+1)!]
= (n+1)(2n)!/[n!(n+1)n!] - n(2n)!/[n(n-1)!(n+1)!]
= (n+1)(2n)!/[n!(n+1)!] - n(2n)!/[n!(n+1)!]
= (2n)!/[n!(n+1)!]
= {catalan number}.

Please read :

i) By definition of C(n, r) = n!/{r!*(n - r)!}
C(2n, n) = (2n)!/{(n!)*(n!)} and C(2n, n - 1) = (2n)!/{(n - 1)!*(n + 1)!}

ii) Thus left side = (2n)!/{(n!)*(n!)} - (2n)!/{(n - 1)!*(n + 1)!}

= (2n)!*[(n + 1)/{(n!)*(n + 1)!} - n/{(n!)*(n + 1)!}

= (2n!)/{(n!)*(n + 1)!}*[(n + 1) - n} = (2n!)/{(n!)*(n + 1)!} = Right side [Proved]

[NOTE: Please find below the detailed solution to your question
Ref. https://in.answers.yahoo.com/question/index?qid=20160227011737AAqdMUC]

i) sin(3) = sin(18 - 15) = sin(18)*cos(15) - cos(18)sin(15)

ii) sin(18) = (√5 - 1)/4
So cos(18) = √{1 - sin²18) = √{10 + 2√5}/4
sin(15) = (√6 - √2)/4 and cos(15) = (√6 + √2)/4

iii) So substituting all these in sin(3),
sin(3) = {(√5 - 1)/4}*{ (√6 + √2)/4} - {√{10 + 2√5}/4}*{(√6 - √2)/4}

= (1/16)*[(√5 - 1)(√6 + √2) - √(10 + 2√5)*(√6 - √2)]

= (1/16)*[(√30 + √10 - √6 - √2) - 2(√3 - 1)*√(5 + √5)]

C(2n,n) - C(2n,n-1) = (2n)!/[n!(n+1)!] = 1/(2n+1) * (2n+1)! / [n! (n+1)!] = 1/(2n+1) * C(2n+1,n)

(or equivalently, 1/(2n+1) * C(2n+1,n+1) )