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}
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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
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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
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
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