Question

设数列{an}的前n项和Sn=n^2,数列{bn}满足bn=an/an+m(m属于N*)，（1）若b1,b2,b8成

等比数列，试求m的值；（2）是否存在m，使得数列｛bn｝中存在某项bt满足b1,b4,bt(t属于N*,t大于等于5)成等差数列？若存在，指出m的个数，若不存在，说明理由。

Answer 1

n>=2
Sn=n^2
Sn-1=(n-1)^2
Sn-Sn-1=n^2-(n-1)^2=2n-1=an
(n>=2)
n=1,a1=S1=1,
a1=2-1=1
an=2n-1(n,N*)
bn=an/(an+m)=（2n-1)/(2n-1+m)
b1=1/(1+m),b2=3/(3+m),b8=15/(15+m)

b2^2=b1b8
9/(3+m)^2=1/(1+m)*15/(15+m)
3/(3+m）^2=5/(1+m)(15+m)
3(1+m)(15+m)=5(3+m)^2
3(15+16m+m^2)=5(9+6m+m^2)
45+48m+3m^2=45+30m+5m^2
2m^2-18m=0
m^2-9m=0
m=0,m=9
m,N*
m=9
(2)若存在
b1=1/(1+m),b4=7/(7+m),bt=(2t-1）/(2t-1+m)
2b4=b1+bt
2*7/(7+m)=1/(1+m)+(2t-1)/(2t-1+m)
14/(7+m)=1/(1+m)+(2t-1)/(2t-1+m)
14(1+m)(2t-1+m)=(7+m)(2t-1+m)+(2t-1)(7+m)(1+m)