题目内容
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{
}的前n项和,求使Tn>
(m2-5m)对所有的n∈N*都成立的最大正整数m的值.
(1)求证:{lgan}是等差数列;
(2)设Tn是数列{
3 |
(lgan)(lgan+1) |
1 |
4 |
(1)依题意,a2=9a1+10=100,故
=10,
当n≥2时,an=9Sn-1+10①又an+1=9Sn+10②
②-①整理得:
=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(
+
++
)
=3(1-
+
-
++
-
)=3-
∴Tn≥
,
依题意有
>
(m2-5m),解得-1<m<6,
故所求最大正整数m的值为5.
a2 |
a1 |
当n≥2时,an=9Sn-1+10①又an+1=9Sn+10②
②-①整理得:
an+1 |
an |
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(
1 |
1•2 |
1 |
2•3 |
1 |
n(n+1) |
=3(1-
1 |
2 |
1 |
2 |
1 |
3 |
1 |
n |
1 |
n+1 |
3 |
n+1 |
3 |
2 |
依题意有
3 |
2 |
1 |
4 |
故所求最大正整数m的值为5.
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