设数列an的前n项和为Sn,a1=1 ,an = 2Sn²/2Sn-1 （n≥2）

其实很简单 还是用公式 an = Sn-S(n-1)
2Sn²/(2Sn-1) = an = Sn- Sn-1
→ S(n-1) = Sn - 2Sn²/(2Sn-1) = -Sn/(2Sn-1)
分子分母颠倒 1/S(n-1) = 1/Sn - 2
于是 1/Sn = 1/S(n-1) +2
1/Sn 是以1 为首项,2为公差的等差数列 1/Sn = 2n-1
Tn = [ 1+(2n-1)]*n/2 = n^2
an = Sn-S(n-1) = 1/(2n-1) - 1/(2n-3) n≥2
a1 = 1

分母是2Sn-1还是(2sn)-1