Question

设数列{求an}的前n项和为Sn，已知a1=a，Sn+1=2Sn+n+1 数列{an}的同乡公式

设数列{an}的前n项和为Sn，已知a1=a，Sn+1=2Sn+n+1 1求数列{an}的同乡公式2当a=1 时 若bn=n/an+1-an，数列{bn}的前项和为Tn 证明 Tn<2

Answer 1

解：1）(Sn+1) + n + 3 = 2[(Sn) + n + 2]，而且S1 + 3 = a + 3 当a ≠ -3时，数列{(Sn) + n + 2}构成一个以a + 3为首项，2为公比的等比数列，因此(Sn) + n + 2 = (a + 3)*2n – 1 => Sn = (a + 3)*2n – 1 – n – 2 => 当n ≥ 2，n∈N时，an = Sn – Sn-1 = (a + 3) *2n – 2 – 1检查a1是否符合上式，可得，当a = 1时，an = (a + 3) *2n – 2 – 1 = 2n – 1 ，(n∈Z+) ；当a ≠ 1而且a ≠ -3时，an ={ a ，(n = 1) { (a + 3) *2n – 2 – 1，(n ≥ 2，n∈N) ；当a = - 3时，S2 + 4 = 0 => S2 = -4 => a2 = -1 ；S3 = 2S2 + 3 = -5 => a3 = -1，Sn+1 = (2Sn) + n + 1 => 当n ≥ 2，n∈N时，Sn = (2Sn-1) + n => an+1 = 2an + 1 => (an+1) + 1 = 2[(an) + 1]，而a2 = -1，所以以后的每一项都是 -1 ，可得当a = - 3时，an = { -3 ，(n = 1) { -1 ，(n ≥ 2，n∈N) ；2）当a = 1时，an = 2n – 1 => bn = n/an+1-an = n/[2n+1 – 2n] = n/2n所以{bn}的前n项和为1/21 + 2/22 + 3/23 + …… + (n - 1)/2n -1 + n/2n = (1/21 + 1/22 + 1/23 + …… + 1/2n -1 + 1/2n ) + [1/22 + 2/23 + …… + (n - 2)/2n -1 + (n - 1)/2n] = (1/2)*[1 - (1/2)n] / [1 – (1/2)] + [1/22 + 2/23 + …… + (n - 2)/2n -1 + (n - 1)/2n] = 1 – (1/2)n + [1/22 + 2/23 + …… + (n - 2)/2n -1 + (n - 1)/2n] = 1 – (1/2)n + [1/22 + 2/23 + …… + (n - 2)/2n -1 + (n - 1)/2n] = …… = 1 – (1/2)n + (1/2)[1 – (1/2)n - 1] + (1/4) [1 – (1/2)n - 2] + …… + (1/2)n-2[1 – (1/2)2] + (1/2)n-1[1 - (1/2)1] = 1 + (1/2) + (1/2)2 + …… + (1/2) n-1 – n(1/2)n = 1[1 – (1/2)n]/ [1 – (1/2)] – n(1/2)n = 2[1 – (1/2)n] – n(1/2)n = 2 – (n +2) (1/2)n < 2 ，得证。

Answer 2

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