Question

设正项数列﹛An﹜的前n项和为Sn,若﹛An﹜和﹛√Sn﹜都是等差数列，且公差相等

设正项数列﹛An﹜的前n项和为Sn,若﹛An﹜和﹛√Sn﹜都是等差数列，且公差相等若a1，a2，a5恰为等比数列{bn}的前三项，记数列cn=24bn/(12bn-1)²,数列{cn}的前n项和为Tn，求证：对任意n为正整数，都有Tn<2

Answer 1

,
设﹛An﹜首项为 a 且公差为 d
Sn = na + n(n-1)/2 *d

S2 = 2a + d
S3 = 3a + 3d

因﹛√Sn﹜是等差数列

√S1 = √a

√S2 = √a +d

√S3 = √a +2d

所以
S2 = a + 2d√a +d^2

S3 = a + 4d√a +4d^2
得方程
2a + d = a + 2d√a +d^2 ...(1)

3a + 3d = a + 4d√a +4d^2 ...(2)

(2)-(1)得
a + 2d = 2d√a + 3 d^2 即 a + 2d√a +d^2 = a + 2d +a -2d^2
代回(1)
2a + d = a + 2d +a -2d^2

所以 d=1/2, a = 1/4
或d=0, a=0(舍去)

{an}的通项公式是

an = 1/4 + (n-1) /2