设函数{an}的前n项和为Sn,且Sn=n²;,数列{bn}满足bn=an/an+m,m∈N

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设函数{an}的前n项和为Sn,且Sn=n²;,数列{bn}满足bn=an/an+m,m∈N
是否存在m,使得数列{bn}中存在某项bt满足b1、b4、bt（t∈N﹢,t≥5）成等差数列?若存在,请指出符合题意的m的个数；若不存在,请说明理由.

Sn=n^2 ,an=2n-1,bn=(2n-1)/(2n-1+m),当m=0时,bn=1,bt满足b1、b4、bt（t∈N﹢,t≥5）成等差数列;
当m>0时,2b4=b1 +bt ,整理,得 m =[36/(t-7)] +5,令t-7=1,2,3,4,6,9,12,18,36（36的正约数）,得m=6,7,8,9,11,14,17,23,41.
综上,存在10个这样的m.

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