问题 解答题
已知数列{an}的前n项和是Sn,且Sn+
1
2
an=1(n∈N+)

(Ⅰ)求数列{an}的通项公式;
(Ⅱ)设bn=log
1
3
(1-Sn+1)(n∈N+)
,令Tn=
1
b1b2
+
1
b2b3
+…+
1
bnbn+1
,求Tn
答案

(Ⅰ)当n=1时,a1=S1,由S1+

1
2
a1=a1+
1
2
a1=1,得:a1=
2
3

当n≥2时,Sn=1-

1
2
anSn-1=1-
1
2
an-1

Sn-Sn-1=

1
2
(an-1-an),即an=
1
2
(an-1-an)

所以an=

1
3
an-1(n≥2).

a1=

2
3
≠0,∴
an
an-1
=
1
3

故数列{an}是以

2
3
为首项,
1
3
为公比的等比数列.

an=a1qn-1=

2
3
•(
1
3
)n-1=2•(
1
3
)n(n∈N*).

(Ⅱ)∵Sn+

1
2
an=1,∴1-Sn=
1
2
an

bn=log

1
3
(1-Sn+1)=log
1
3
(
1
3
)n+1=n+1.

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

所以,Tn=

1
b1b2
+
1
b2b3
+…+
1
bnbn+1
=(
1
2
-
1
3
)+(
1
3
-
1
4
)+…+(
1
n+1
-
1
n+2
)
=
1
2
-
1
n+2
=
n
2(n+2)

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