求{an}的通项

Question

设数列{an}的前n项和为Sn，点（n，Sn/n）（n属于N*）均在函数y=3x-2的图象上。（1）求数列{an}的通项公司；（2）设bn=3/anan+1，Tn是数列{bn}的前n项和，求使得Tn〈m/20对所有n属于N*都成立的最小正整数m。
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Answer 1

1.sn/n=3n-2
sn=3n^2-2n

sn-1=3(n-1)^2-2(n-1)
an=sn-sn-1
=3n^2-2n-3n^2+6n-3+2n-2
=6n-5

bn=3/ana(n+1)=3/(6n-5)(6(n+1)-5)
=3/(6n-5)(6n+1)
=（1/2)*(1/(6n-5)-1/(6n+1))
Tn=(1/2)*(1-1/7+1/7-1/13+........+1/(6n-5)-1/(6n+1))
=(1/2)*(1-1/(6n+1))<m/20
m>10(1-1/(6n+1))
对所有n属于N正，则当n→∞时，10(1-1/(6n+1))最大
mmin=10