已知函数f(x)=(13)x,等比数列{an}的前n项和为f(n)-c,正项数列{bn}的首项为c,且前n项和Sn满足Sn-Sn-1=Sn+Sn-1(n≥2).(1)求数列{an}的通项公式;(2)证

题目简介

已知函数f(x)=(13)x,等比数列{an}的前n项和为f(n)-c,正项数列{bn}的首项为c,且前n项和Sn满足Sn-Sn-1=Sn+Sn-1(n≥2).(1)求数列{an}的通项公式;(2)证

题目详情

已知函数f(x)=(
1
3
)x
,等比数列{an}的前n项和为f(n)-c,正项数列{bn}的首项为c,且前n项和Sn满足Sn-Sn-1=
Sn
+
Sn-1
(n≥2).
(1)求数列{an}的通项公式;
(2)证明数列{
Sn
}是等差数列,并求Sn
(3)若数列{
1
bnbn+1
}前n项和为Tn,问Tn
1000
2009
的最小正整数n是多少?
(4)设cn=
2bn
an
,求数列{cn}的前n项和Pn
题型:解答题难度:中档来源:不详

答案

(1)因为a1=f(1)-c=class="stub"1
3
-c

a2=[f(2)-c]-[f(1)-c]=-class="stub"2
9

a3=[f(3)-c]-[f(2)-c]=-class="stub"2
27

又数列{an}成等比数列,
所以a1=
a22
a3
=
class="stub"4
81
-class="stub"2
27
=-class="stub"2
3
=class="stub"1
3
-c

解得c=1.…(2分)
又公比q=
a2
a1
=class="stub"1
3

所以an=-class="stub"2
3
•(class="stub"1
3
)n-1
=-2•(class="stub"1
3
)n-1,n∈N*.…(3分)
(2)∵Sn-Sn-1=
Sn
+
Sn-1
,n≥2,
(
Sn
-
Sn-1
)(
Sn
+
Sn-1
)=
Sn
+
Sn-1
,n≥2
Sn
-
Sn-1
=1
,(n≥2)…(5分)
S1
=
b1
=
c
=1

∴数列{
Sn
}构成一个首项为1,公差为1的等差数列,
Sn
=1+(n-1)×1=n,∴Sn=n2.…(6分)
(3)由(2)得Sn=n2
当n≥2时,bn=Sn-Sn-1=n2-(n-1)2=2n-1,(*)
又b1=S1=1,适合(*)式
∴bn=2n-1,(n∈N*)…(8分)
class="stub"1
bnbn+1
=class="stub"1
(2n-1)(2n+1)
=class="stub"1
2
(class="stub"1
2n-1
-class="stub"1
2n+1
)

Tn=class="stub"1
b1b2
+class="stub"1
b2b3
+class="stub"1
b3b4
+…+class="stub"1
bnbn+1

=class="stub"1
2
(1-class="stub"1
3
)+class="stub"1
2
(class="stub"1
3
-class="stub"1
5
)+class="stub"1
2
(class="stub"1
5
-class="stub"1
7
)+…+class="stub"1
2
(class="stub"1
2n-1
-class="stub"1
2n+1
)

=class="stub"1
2
(1-class="stub"1
2n-1
)=class="stub"n
2n+1
,…(10分)
由Tn=class="stub"n
2n+1
class="stub"1000
2009
,得n>class="stub"1000
9

故满足Tn>class="stub"1000
2009
的最小正整数为112.…(11分)
(4)cn=
2bn
an
=(1-2n)•3n
.…(12分)
Pn=(-1)×3+(-3)×32+(-5)×33+…+(1-2n)×3n3Pn=(-1)×32+(-3)×33+(-5)×34+…+(3-2n)×3n+(1-2n)×3n+1
②-①得2Pn=3+2×32+2×33+…+2×3n+(1-2n)×3n+1
=3+2×
32(1-3n-1)
1-3
+(1-2n)×3n+1
=(2-2n)•3n+1-6.

Pn=(1-n)•3n+1-3.…(14分)

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