题目内容
设数列{an} 对任意n∈N*和实数常数,有an-2an+1 |
anan+1 |
1 |
3 |
(1)若{
1-an |
an |
(2)设{bn}满足bn=(1-an)an,其前n项和Tn,求证:Tn>
2 |
3 |
2n-1 |
2n+1+1 |
分析:(1)由题设知
-1=2(
-1) +t•
-1=2,再由{
}是等比数列,得an=
.
(2)由bn=(1-an)an得bn=(1-
) •
=
<
-
,由此入手能够进行证明.
1 |
an+1 |
1 |
an |
1 |
a1 |
1-an |
an |
1 |
2n+1 |
(2)由bn=(1-an)an得bn=(1-
1 |
2n+1 |
1 |
2n+1 |
2n |
(2n+1)2 |
1 |
2n+1 |
1 |
2n+1+1 |
解答:解:(1)由
=t-2,t∈R,a1=
,
得
-1=2(
-1) +t•
-1=2,
∵{
}是等比数列,
∴
-1=2n,
得an=
.
(2)由bn=(1-an)an得bn=(1-
) •
=
<
-
,
前n项和Tn=b1+b2+…+bn
<
-
=
•
.
an-2an+1 |
anan+1 |
1 |
3 |
得
1 |
an+1 |
1 |
an |
1 |
a1 |
∵{
1-an |
an |
∴
1 |
an |
得an=
1 |
2n+1 |
(2)由bn=(1-an)an得bn=(1-
1 |
2n+1 |
1 |
2n+1 |
2n |
(2n+1)2 |
1 |
2n+1 |
1 |
2n+1+1 |
前n项和Tn=b1+b2+…+bn
<
1 |
3 |
1 |
2n+1+1 |
=
2 |
3 |
2n-1 |
2n+1+1 |
点评:本题考查数列的性质和应用,解题时要认真审题,仔细解答,注意合理地选取公式.
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