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](/uploads/image/z/3968657-17-7.jpg?t=%E8%AE%BE%E6%AD%A3%E9%A1%B9%E6%95%B0%E5%88%97%EF%B9%9BAn%EF%B9%9C%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E8%8B%A5%EF%B9%9BAn%EF%B9%9C%E5%92%8C%EF%B9%9B%E2%88%9ASn%EF%B9%9C%E9%83%BD%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E4%B8%94%E5%85%AC%E5%B7%AE%E7%9B%B8%E7%AD%89%E8%8B%A5a1%2Ca2%2Ca5%E6%81%B0%E4%B8%BA%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%7Bbn%7D%E7%9A%84%E5%89%8D%E4%B8%89%E9%A1%B9%2C%E8%AE%B0%E6%95%B0%E5%88%97cn%3D24bn%2F%2812bn-1%29%26%23178%3B%2C%E6%95%B0%E5%88%97%7Bcn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BATn%2C%E6%B1%82%E8%AF%81%EF%BC%9A%E5%AF%B9%E4%BB%BB%E6%84%8Fn%E4%B8%BA%E6%AD%A3%E6%95%B4%E6%95%B0%2C%E9%83%BD%E6%9C%89Tn)
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
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