已知数列{an}的每一项都是正数,满足a1=2,且an+12-anan+1-2an2=0;等差数列{bn}的前n项和为Tn,b2=3,T5=25.(1)求数列{an}、{bn}的通项公式;(2)比较1

题目简介

已知数列{an}的每一项都是正数,满足a1=2,且an+12-anan+1-2an2=0;等差数列{bn}的前n项和为Tn,b2=3,T5=25.(1)求数列{an}、{bn}的通项公式;(2)比较1

题目详情

已知数列{an}的每一项都是正数,满足a1=2,且an+12-anan+1-2an2=0;等差数列{bn}的前n项和为Tn,b2=3,T5=25.
(1)求数列{an}、{bn}的通项公式;
(2)比较
1
T1
+
1
T2
+…+
1
Tn
与2的大小;
(3)若
b1
a1
+
b2
a2
+…+
bn
an
<c恒成立,求整数c的最小值.
题型:解答题难度:中档来源:不详

答案

(1)an+12-anan+1-2an2=0
得(an+1-2an)(an+1+an)=0,
由于数列{an}的每一项都是正数,∴an+1=2an,∴an=2n.
设bn=b1+(n-1)d,由已知有b1+d=3,5b1+class="stub"5×4
2
d=25,
解得b1=1,d=2,∴bn=2n-1.
(2)由(1)得Tn=n2,∴class="stub"1
Tn
=class="stub"1
n2

当n=1时,class="stub"1
T1
=1<2.
当n≥2时,class="stub"1
n2
class="stub"1
(n-1)n
=class="stub"1
n-1
-class="stub"1
n

class="stub"1
T1
+class="stub"1
T2
+…+class="stub"1
Tn
<1+class="stub"1
1
-class="stub"1
2
+class="stub"1
2
-class="stub"1
3
++class="stub"1
n-1
-class="stub"1
n
=2-class="stub"1
n
<2.
(3)记Pn=
b1
a1
+
b2
a2
+…+
bn
an
=class="stub"1
2
+class="stub"3
22
+class="stub"5
23
+…+class="stub"2n-1
2n

class="stub"1
2
Pn=class="stub"1
22
+class="stub"3
23
++class="stub"2n-3
2n
+
2n-1
2n+1

两式相减得Pn=3-class="stub"2n+3
2n

∵Pn递增,∴class="stub"1
2
≤Pn<3,P4=class="stub"37
16
>2,
∴最小的整数c=3.

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