,设﹛An﹜首项为 a 且公差为 dSn = na + n(n-1)/2 *dS2 = 2a + dS3 = 3a + 3d因﹛√Sn﹜是等差数列√S1 = √a√S2 = √a +d√S3 = √a +2d所以S2 = a + 2d√a +d^2S3 = a + 4d√a +4d^2得方程2a + d = a + 2d√a +d^...
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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