问题
解答题
已知数列{an}的前n项和是Sn,且Sn+
(Ⅰ)求数列{an}的通项公式; (Ⅱ)设bn=log
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答案
(Ⅰ)当n=1时,a1=S1,由S1+
a1=a1+1 2
a1=1,得:a1=1 2
.2 3
当n≥2时,Sn=1-
an,Sn-1=1-1 2
an-1.1 2
则Sn-Sn-1=
(an-1-an),即an=1 2
(an-1-an),1 2
所以an=
an-1(n≥2).1 3
∵a1=
≠0,∴2 3
=an an-1
.1 3
故数列{an}是以
为首项,2 3
为公比的等比数列.1 3
故an=a1qn-1=
•(2 3
)n-1=2•(1 3
)n(n∈N*).1 3
(Ⅱ)∵Sn+
an=1,∴1-Sn=1 2
an.1 2
∴bn=log
(1-Sn+1)=log1 3
(1 3
)n+1=n+1.1 3
∴
=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)