问题 解答题
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
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
=
1
2

可见,数列{an}是首项为2,公比为

1
2
的等比数列.

∴an=2•(

1
2
)n-1=(
1
2
)n-2

(2)证明:∵bn=

1
2-log2an
=
1
2-(2-n)
=
1
n

∴bnbn+2=

1
n(n+2)
=
1
2
(
1
n
-
1
n+2
)

∴数列{bnbn+2}的前n项和

Tn=

1
2
[(1-
1
3
)+(
1
2
-
1
4
)+(
1
3
-
1
5
)+(
1
4
-
1
6
)+…+(
1
n-2
-
1
n
)+(
1
n-1
-
1
n+1
)+(
1
n
-
1
n+2
)]

=

1
2
(1+
1
2
-
1
n+1
-
1
n+2

=

1
2
3
2
-
1
n+1
-
1
n+2
)=
3
4
-
1
2
1
n+1
+
1
n+2
3
4

单项选择题
名词解释