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=log3(1-Sn+1),求适合方程
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答案
(Ⅰ)当n=1时,a1=S1,由S1+
a1=1,得a1=1 2
.2 3
当n≥2时,
∵Sn=1-
an,Sn-1=1-1 2
an-1,1 2
∴Sn-Sn-1=
(an-1-an),即an=1 2
(an-1-an).1 2
∴an=
an-1.1 3
∴{an}是以
为首项,2 3
为公比的等比数列.1 3
故an=
•(2 3
)n-1=2•(1 3
)n. (7分)1 3
(Ⅱ)1-Sn=
an=(1 2
)n,1 3
bn=log3(1-Sn+1)=log3(
)n+1=-n-1,(9分)1 3
=1 bnbn+1
=1 (n+1)(n+2)
-1 n+1 1 n+2
+1 b1b2
++1 b2b3
=(1 bnbn+1
-1 2
)+(1 3
-1 3
)++(1 4
-1 n+1
)=1 n+2
-1 2
(11分)1 n+2
解方程
-1 2
=1 n+2
,得n=100(14分)25 51