∵等差数列{an}的通项公式an=2n-1， 1 an?an+1 = 1 (2n?1)(2n+1) = 1 2 ( 1 2n?1 ? 1 2n+1 )，∴Sn= 1 2 （1- 1 3 + 1 3 ? 1 5 +…+ 1 2n?1 ? 1 2n+1 ）= 1 2 (1? 1 2n+1 )= n 2n+1 ．故选：B．
