已知点(1,13)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=Sn+Sn-1(n

题目简介

已知点(1,13)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=Sn+Sn-1(n

题目详情

已知点(1,
1
3
)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=
Sn
+
Sn-1
(n≥2)
(Ⅰ)求数列{an}和{bn}的通项公式
(Ⅱ)求数列{
1
bnbn+1
}前n项和为Tn
题型:解答题难度:中档来源:不详

答案

(Ⅰ)∵f(1)=class="stub"1
3
,故a=class="stub"1
3

∴f(x)=(class="stub"1
3
)
x

∵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}为等比数列,a1=
a22
a3
=
class="stub"4
81
-class="stub"2
27
=-class="stub"2
3
=class="stub"1
3
-c,
∴c=1,又公比q=
a2
a1
=class="stub"1
3

∴an=-class="stub"2
3
(class="stub"1
3
)
n-1
=-2(class="stub"1
3
)
n
,n∈N*;
∵Sn-Sn-1=(
Sn
+
Sn-1
)(
Sn
-
Sn-1
)=
Sn
+
Sn-1
(n≥2),
又bn>0,
Sn
>0,
Sn
-
Sn-1
=1;
∴数列{
Sn
}构成一个首相为1公差为1的等差数列,
Sn
=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)∵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
bnbn+1

=class="stub"1
2
[(1-class="stub"1
3
)+(class="stub"1
3
-class="stub"1
5
)+(class="stub"1
5
-class="stub"1
7
)+(class="stub"1
7
-class="stub"1
9
)+…+(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

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