题目内容
已知点(1,
)是函数f(x)=ax(a>0且a≠1)的图象上一点,等比数列{an}的前n项和为f(n)-c,数列{bn}(bn>0)的首项为c,且前n项和Sn满足Sn-Sn-1=
+
(n≥2)
(Ⅰ)求数列{an}和{bn}的通项公式
(Ⅱ)求数列{
}前n项和为Tn.
1 |
3 |
Sn |
Sn-1 |
(Ⅰ)求数列{an}和{bn}的通项公式
(Ⅱ)求数列{
1 |
bnbn+1 |
(Ⅰ)∵f(1)=
,故a=
,
∴f(x)=(
)x,
∵a1=f(1)-c=
-c,a2=[f(2)-c]-[f(1)-c]=-
,a3=[f(3)-c]-[f(2)-c]=-
,
又数列{an}为等比数列,a1=
=
=-
=
-c,
∴c=1,又公比q=
=
,
∴an=-
(
)n-1=-2(
)n,n∈N*;
∵Sn-Sn-1=(
+
)(
-
)=
+
(n≥2),
又bn>0,
>0,
∴
-
=1;
∴数列{
}构成一个首相为1公差为1的等差数列,
∴
=1+(n-1)×1=n,于是Sn=n2;
当n≥2,bn=Sn-Sn-1=n2-(n-1)2=2n-1;
∴bn=2n-1,n∈N*;
(Ⅱ)∵
=
=
(
-
),
∴Tn=
+
+…+
=
[(1-
)+(
-
)+(
-
)+(
-
)+…+(
-
)]
=
(1-
)
=
.
1 |
3 |
1 |
3 |
∴f(x)=(
1 |
3 |
∵a1=f(1)-c=
1 |
3 |
2 |
9 |
2 |
27 |
又数列{an}为等比数列,a1=
a22 |
a3 |
| ||
-
|
2 |
3 |
1 |
3 |
∴c=1,又公比q=
a2 |
a1 |
1 |
3 |
∴an=-
2 |
3 |
1 |
3 |
1 |
3 |
∵Sn-Sn-1=(
Sn |
Sn-1 |
Sn |
Sn-1 |
Sn |
Sn-1 |
又bn>0,
Sn |
∴
Sn |
Sn-1 |
∴数列{
Sn |
∴
Sn |
当n≥2,bn=Sn-Sn-1=n2-(n-1)2=2n-1;
∴bn=2n-1,n∈N*;
(Ⅱ)∵
1 |
bnbn+1 |
1 |
(2n-1)(2n+1) |
1 |
2 |
1 |
2n-1 |
1 |
2n+1 |
∴Tn=
1 |
b1b2 |
1 |
b2b3 |
1 |
bnbn+1 |
=
1 |
2 |
1 |
3 |
1 |
3 |
1 |
5 |
1 |
5 |
1 |
7 |
1 |
7 |
1 |
9 |
1 |
2n-1 |
1 |
2n+1 |
=
1 |
2 |
1 |
2n+1 |
=
n |
2n+1 |
练习册系列答案
相关题目