∵ 2 4n2?1 = 2 (2n+1)(2n?1) = 1 2n?1 ? 1 2n+1 ，∴数列{ 2 4n2?1 }的前n项和为：Sn=1- 1 3 + 1 3 ? 1 5 +…+ 1 2n?1 ? 1 2n+1 =1- 1 2n+1 = 2n 2n+1 ．故选：A．
