设数列{an}的前n项和为sn,a1=10,an+1=9sn+10.设Tn是数列（3/（lgan)(lgan+1)}的前

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设数列{an}的前n项和为sn,a1=10,an+1=9sn+10.设Tn是数列（3/（lgan)(lgan+1)}的前n项和,求使Tn>1/4(m2-5m
对所有n∈N都成立的最大正整数m的值

a1=10
an+1=9sn+10
an=9sn-1+10
an+1-an=9an
an+1=10an
a1=10
an=10^n
bn=3/[lg(an)lg(an+1)]=3/[(n)(n+1)]=3*[1/n-1/(n+1)]
Tn=b1+...+bn=3[1-1/(n+1)]
Tn>(m^2-5m)/4
3-3/(n+1)>(m^2-5m)/4
n=1时
3/2>(m^-5m)/4
2m^2-10m

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