(1)
a1=1
(2n+1)an = (2n-1) a(n+1)
a(n+1)/an = (2n+1)/(2n-1)
an/a(n-1) = (2n-1)/(2n-3)
an/a1 =(2n-1)
an = 2n-1
=>{an } 是等差数列, d=2
(2)
bn = 1/[an.a(n+1)]
= 1/[(2n-1)(2n+1)]
=(1/2) [ 1/(2n-1) -1/(2n+1)]
Tn =b1+b2+...+bn
=(1/2) [ 1/1 - 1/(2n+1) ]
=n/(2n+1)