设数列{an}的前n项和为Sn,a1=10,an+1=9Sn+10.(1)求证:{lgan}是等差数列;(2)设Tn是数列{3(lgan)(lgan+1)}的前n项和,求使Tn>14(m2-5m)对所

题目简介

设数列{an}的前n项和为Sn,a1=10,an+1=9Sn+10.(1)求证:{lgan}是等差数列;(2)设Tn是数列{3(lgan)(lgan+1)}的前n项和,求使Tn>14(m2-5m)对所

题目详情

设数列{an}的前n项和为Sn,a1=10,an+1=9Sn+10.
(1)求证:{lgan}是等差数列;
(2)设Tn是数列{
3
(lgan)(lgan+1)
}的前n项和,求使Tn
1
4
(m2-5m)
对所有的n∈N*都成立的最大正整数m的值.
题型:解答题难度:中档来源:不详

答案

(1)依题意,a2=9a1+10=100,故
a2
a1
=10

当n≥2时,an=9Sn-1+10①又an+1=9Sn+10②
②-①整理得:
an+1
an
=10,故{an}
为等比数列,
且an=a1qn-1=10n,∴lgan=n∴lgan+1-lgan=(n+1)-n=1,
即{lgan}n∈N*是等差数列.
(2)由(1)知,Tn=3(class="stub"1
1•2
+class="stub"1
2•3
++class="stub"1
n(n+1)
)

=3(1-class="stub"1
2
+class="stub"1
2
-class="stub"1
3
++class="stub"1
n
-class="stub"1
n+1
)=3-class="stub"3
n+1
Tn≥class="stub"3
2

依题意有class="stub"3
2
>class="stub"1
4
(m2-5m),解得-1<m<6

故所求最大正整数m的值为5.

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