Question

设数列{an}的前n项和为Sn,a1=1,an=Sn/n+2(n-1）

2）设数列｛1/（anan+1）｝前n项和为Tn，证明：1/5≤Tn＜1/4
（3）是否存自然数n，使S1+（S2/2）+（S3/3)+...+（Sn/n）-（n-1)^2=2009
若存在，求n，若不存在，说明。

Answer 1

(1)
an=Sn/n+ 2(n-1)
Sn = nan-2n(n-1)
an = Sn-S(n-1)
= nan-2n(n-1) -(n-1)a(n-1) +2(n-1)(n-2)
an - a(n-1) =4
an - a1= 4(n-1)
an = 4n-3
(2)
bn = 1/[an.a(n+1)]
= (1/4)[ 1/(4n-3) - 1/(4n+1)]
Tn = b1+b2+...+bn
=(1/4)[ 1 - 1/(4n+1)]
= n/(4n+1)
= 1/4 - 1/[4(4n+1)]
Tn < 1/4
min Tn = T1 = 1/4 - 1/20 = 1/5
ie
1/5≤Tn<1/4

(3)
an=Sn/n+ 2(n-1)
Sn/n = an-2(n-1)
= 4n-3 -2(n-1)
= 2n-1

(S1/1)+(S2/2)+...+(Sn/n）-(n-1)^2
=n^2 -(n-1)^2
=2n-1 =2009
=> n=1005