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，（1）求{an}

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，（1）求{an}的通项公式（2）若a1，a2，a5恰为等比数列{bn}的前三项，记数列cn=1/（log3）4bn+1*（log3）4bn+2，数列{cn}的前n项和为Tn，求Tn

Answer 1

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

2,
a1 = 1/4
a2 = 3/4
a5 = 9/4
所以等比数列{bn}公比q =3

{bn}的通项公式是

bn = 1/4 * 3^(n-1)

cn = 1/ （log3）4bn+1*（log3）4bn+2
分母部分 = （log3）4 *1/4*3^n * （log3）4 *1/4*3^(n+1) = n * (n+1)
{cn}的通项公式是

1/(n*(n+1))
{cn}的前n项和 Tn
= 1/(1*2) + 1/(2*3) + .. + 1/(n*(n+1)) = (1-1/2) +(1/2-1/3) + ... + (1/n - 1/(n+1))

= 1- 1/(n+1)

Answer 2

设公差为d，首项a1
Sn=na1+n(n-1)d/2
√Sn=√(S1)+(n-1)d=√(a1)+(n-1)d 平方
Sn=a1+2√(a1)*(n-1)d+(n-1)^2d^2

na1+n(n-1)d/2=a1+2√(a1)*(n-1)d+(n-1)^2d^2
(n-1)a1+n(n-1)d/2=2√(a1)*(n-1)d+(n-1)^2d^2
a1+nd/2=2√(a1)d+(n-1)d^2
nd/2-nd^2=2√(a1)d-d^2-a1
n(d/2-d^2)=2√(a1)d-d^2-a1
因为对任意n∈N+都成立，所以
d/2-d^2=0
d=0或d=1/2
(1) d=0时 2√(a1)d-d^2-a1=0 a1=0 正项数列 舍
(2) d=1/2 2√(a1)d-d^2-a1=0 a1=1/4