题目内容
已知函数f(x)=(
)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=
+
(n≥2).
(1)求数列{an}的通项公式;
(2)证明数列{
}是等差数列,并求Sn;
(3)若数列{
}前n项和为Tn,问Tn>
的最小正整数n是多少?
(4)设cn=
,求数列{cn}的前n项和Pn.
1 |
3 |
Sn |
Sn-1 |
(1)求数列{an}的通项公式;
(2)证明数列{
Sn |
(3)若数列{
1 |
bnbn+1 |
1000 |
2009 |
(4)设cn=
2bn |
an |
(1)因为a1=f(1)-c=
-c,
a2=[f(2)-c]-[f(1)-c]=-
,
a3=[f(3)-c]-[f(2)-c]=-
.
又数列{an}成等比数列,
所以a1=
=
=-
=
-c,
解得c=1.…(2分)
又公比q=
=
,
所以an=-
•(
)n-1=-2•(
)n-1,n∈N*.…(3分)
(2)∵Sn-Sn-1=
+
,n≥2,
即(
-
)(
+
)=
+
,n≥2
∴
-
=1,(n≥2)…(5分)
又
=
=
=1
∴数列{
}构成一个首项为1,公差为1的等差数列,
∴
=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)
∵
=
=
(
-
),
∴Tn=
+
+
+…+
=
(1-
)+
(
-
)+
(
-
)+…+
(
-
)
=
(1-
)=
,…(10分)
由Tn=
>
,得n>
,
故满足Tn>
的最小正整数为112.…(11分)
(4)cn=
=(1-2n)•3n.…(12分)
∴Pn=(-1)×3+(-3)×32+(-5)×33+…+(1-2n)×3n①3Pn=(-1)×32+(-3)×33+(-5)×34+…+(3-2n)×3n+(1-2n)×3n+1②
②-①得2Pn=3+2×32+2×33+…+2×3n+(1-2n)×3n+1
∴Pn=(1-n)•3n+1-3.…(14分)
1 |
3 |
a2=[f(2)-c]-[f(1)-c]=-
2 |
9 |
a3=[f(3)-c]-[f(2)-c]=-
2 |
27 |
又数列{an}成等比数列,
所以a1=
a22 |
a3 |
| ||
-
|
2 |
3 |
1 |
3 |
解得c=1.…(2分)
又公比q=
a2 |
a1 |
1 |
3 |
所以an=-
2 |
3 |
1 |
3 |
1 |
3 |
(2)∵Sn-Sn-1=
Sn |
Sn-1 |
即(
Sn |
Sn-1 |
Sn |
Sn-1 |
Sn |
Sn-1 |
∴
Sn |
Sn-1 |
又
S1 |
b1 |
c |
∴数列{
Sn |
∴
Sn |
(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)
∵
1 |
bnbn+1 |
1 |
(2n-1)(2n+1) |
1 |
2 |
1 |
2n-1 |
1 |
2n+1 |
∴Tn=
1 |
b1b2 |
1 |
b2b3 |
1 |
b3b4 |
1 |
bnbn+1 |
=
1 |
2 |
1 |
3 |
1 |
2 |
1 |
3 |
1 |
5 |
1 |
2 |
1 |
5 |
1 |
7 |
1 |
2 |
1 |
2n-1 |
1 |
2n+1 |
=
1 |
2 |
1 |
2n-1 |
n |
2n+1 |
由Tn=
n |
2n+1 |
1000 |
2009 |
1000 |
9 |
故满足Tn>
1000 |
2009 |
(4)cn=
2bn |
an |
∴Pn=(-1)×3+(-3)×32+(-5)×33+…+(1-2n)×3n①3Pn=(-1)×32+(-3)×33+(-5)×34+…+(3-2n)×3n+(1-2n)×3n+1②
②-①得2Pn=3+2×32+2×33+…+2×3n+(1-2n)×3n+1
|
∴Pn=(1-n)•3n+1-3.…(14分)
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