Question

数列｛an｝，a1=1，an=2Sn*2/2Sn-1(n≥2)，（1）求证：数列｛1/Sn｝等差（2）求｛an｝通项

Answer 1

a1=1，an=2Sn²/(2Sn—1）（n≥2)
利用 an = Sn - S<n-1>

Sn - S<n-1> = 2Sn²/(2Sn -1)
(Sn - S<n-1>)(2Sn -1) = 2Sn²
2Sn² - Sn - 2SnS<n-1> + S<n-1> = 2Sn²
Sn + 2SnS<n-1> - S<n-1> = 0
两端同时除以 SnS<n-1>
1/Sn - 1/S<n-1> = 2
所以 1/Sn 是公差d=2的等差数列
1/Sn = 1/S1 + (n-1)*d
1/Sn = 1/a1 + 2(n-1)
1/Sn = 1 + 2n - 2
Sn = 1/(2n -1)

n=1 时 a1 = 1
n≥2 时
an = Sn - S<n-1> = 1/(2n-1) - 1/(2n-3)