已知数列{an}的各项均是正数,其前n项和为Sn,满足Sn=4-an.(1)求数列{an}的通项公式;(2)设bn=12-log2an(n∈N*),数列{bnbn+2}的前n项和为Tn,求证:Tn<3

题目简介

已知数列{an}的各项均是正数,其前n项和为Sn,满足Sn=4-an.(1)求数列{an}的通项公式;(2)设bn=12-log2an(n∈N*),数列{bnbn+2}的前n项和为Tn,求证:Tn<3

题目详情

已知数列{an}的各项均是正数,其前n项和为Sn,满足Sn=4-an
(1)求数列{an}的通项公式;
(2)设bn=
1
2-log2an
(n∈N*),数列{bnbn+2}的前n项和为Tn,求证:Tn
3
4
题型:解答题难度:中档来源:不详

答案

(1)由Sn=4-an.得S1=4-a1,解得a1=2,
而an+1=Sn+1-Sn=(4-an+1)-(4-an)=an-an+1,即2an+1=an,
an+1
an
=class="stub"1
2

可见,数列{an}是首项为2,公比为class="stub"1
2
的等比数列.
∴an=2•(class="stub"1
2
)n-1
=(class="stub"1
2
)n-2

(2)证明:∵bn=class="stub"1
2-log2an
=class="stub"1
2-(2-n)
=class="stub"1
n

∴bnbn+2=class="stub"1
n(n+2)
=class="stub"1
2
(class="stub"1
n
-class="stub"1
n+2
)

∴数列{bnbn+2}的前n项和
Tn=class="stub"1
2
[(1-class="stub"1
3
)+(class="stub"1
2
-class="stub"1
4
)+(class="stub"1
3
-class="stub"1
5
)+(class="stub"1
4
-class="stub"1
6
)+…+(class="stub"1
n-2
-class="stub"1
n
)+(class="stub"1
n-1
-class="stub"1
n+1
)+(class="stub"1
n
-class="stub"1
n+2
)]
=class="stub"1
2
(1+class="stub"1
2
-class="stub"1
n+1
-class="stub"1
n+2

=class="stub"1
2
class="stub"3
2
-class="stub"1
n+1
-class="stub"1
n+2
)=class="stub"3
4
-class="stub"1
2
class="stub"1
n+1
+class="stub"1
n+2
<class="stub"3
4

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