Question

已知An=4n-2,若Bn=4/[An*A(n+1)],求Bn前n项和Tn

Answer 1

Bn=4/[An*A(n+1)] = 4/[(4n-2)(4n+2)] = [1/(4n - 2)] - [1/(4n + 2)] ,

Tn = 1/2 - 1/6 + 1/6 - 1/10······ - [1/(4n + 2)]

= 1/2 - [1/(4n + 2)] = n/(2n + 1)