Question

A respeito das combinações: an = (2n n) e bn = (2n n-1) temos que, para cada n =1,2,3..., a diferença an-bn é: a) n!/n+1an b) 2n/n+1an c) n/n+1an d) 2/n+1an e) 1/n+1an

Answer 1

2n! / n! (2n - n)! - 2n! / (n-1)! [2n - (n-1)]

2n! / n! . n! - 2n! / (n-1)! . (n+1)!

2 / n! - 2 / (n-1)! . (n+1)

2 / n (n-1)! - 2 / (n-1)!(n+1)


2(n+1) - 2n
------------------
n (n+1)(n-1)!


2 / (n+1)!