问题
解答题
已知数列{an}的前n项和Sn与通项an满足Sn=
(1)求数列{an}的通项公式; (2)设函数f(x)=log
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答案
(1)n≥2时,an=
(1-an) -1 2
(1-an-1) =-1 2
an+1 2
an-1,1 2
2an=-an+an-1
=an an-1
,---------------------------------------------------------------------------(3分)1 3
S1=a1=
(1-a1)得a1=1 2
,1 3
∴数an是以首a1=
,公比1 3
的等比数列,1 3
∴an=(
)n------(5分)1 3
(2)∵f(x)=log
x,bn=f(a1)+f(a2)+…+f(an),1 3
∴bn=log
a1+log1 3
a2 +…+log1 3
an=log1 3
(a1•a2…•an)-----------(10分)1 3
即log
(1 3
)1+2+…+n=1+2+…+n=1 3
-------------------(12分)n(n+1) 2
∴
=1 bn
=2(2 n(n+1)
-1 n
),1 n+1
∴Tn=
+1 b1
+…+1 b2
=2[(1-1 bn
)+(1 2
-1 2
)+…+(1 3
-1 n
)]=1 n+1
--------(14分)2n n+1