已知等比数列{an}的前n项和An=(13)n-c.数列{bn}(bn>0)的首项为c,且前n项和Sn满足Sn-Sn-1=1(n≥2).(1)求数列{an}的通项公式;(2)求数列{bn}的通项公式;

题目简介

已知等比数列{an}的前n项和An=(13)n-c.数列{bn}(bn>0)的首项为c,且前n项和Sn满足Sn-Sn-1=1(n≥2).(1)求数列{an}的通项公式;(2)求数列{bn}的通项公式;

题目详情

已知等比数列{an}的前n项和An=(
1
3
)n-c
.数列{bn}(bn>0)的首项为c,且前n项和Sn满足
Sn
-
Sn-1
=1(n≥2).
(1)求数列{an}的通项公式;
(2)求数列{bn}的通项公式;
(3)若数列{
1
bnbn+1
}前n项和为Tn,问Tn
1001
2010
的最小正整数n是多少?.
题型:解答题难度:中档来源:不详

答案

(1)a1=A1=class="stub"1
3
-c,  a2=A2-A1=(class="stub"1
9
-c)-(class="stub"1
3
-c)=-class="stub"2
9
a3=A3-A2=(class="stub"1
27
-c)-(class="stub"1
9
-c)=-class="stub"2
27

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

所以 c=1;
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
,n∈N*.
(2)∵
Sn
-
Sn-1
=1(n≥2),  S1=b1=1

∴数列{
Sn
}是首项为1公差为1的等差数列.
Sn
=1+(n-1)×1.
∴Sn=n2.
当n≥2,bn=Sn-Sn-1=n2-(n-1)2=2n-1.
∴bn=2n-1(n∈N*);                
(3)Tn=class="stub"1
b1b2
+class="stub"1
b2b3
+class="stub"1
b3b4
+…+class="stub"1
bnbn+1

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

=class="stub"1
2
(1-class="stub"1
2n+1
)

=class="stub"n
2n+1

Tn=class="stub"n
2n+1
>class="stub"1001
2010
n>class="stub"1001
8

故满足Tn>class="stub"1001
2010
的最小正整数为126.

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