已知数列{an}的通项为an,前n项和为sn,且an是sn与2的等差中项,数列{bn}中,b1=1,点P(bn,bn+1)在直线x-y+2=0上.(Ⅰ)求数列{an}、{bn}的通项公式an,bn(Ⅱ

题目简介

已知数列{an}的通项为an,前n项和为sn,且an是sn与2的等差中项,数列{bn}中,b1=1,点P(bn,bn+1)在直线x-y+2=0上.(Ⅰ)求数列{an}、{bn}的通项公式an,bn(Ⅱ

题目详情

已知数列{an}的通项为an,前n项和为sn,且an是sn与2的等差中项,数列{bn}中,b1=1,点P(bn,bn+1)在直线x-y+2=0上.
(Ⅰ)求数列{an}、{bn}的通项公式an,bn
(Ⅱ)设{bn}的前n项和为Bn,试比较
1
B1
+
1
B2
+…+
1
Bn
与2的大小.
(Ⅲ)设Tn=
b1
a1
+
b2
a2
+…+
bn
an
,若对一切正整数n,Tn<c(c∈Z)恒成立,求c的最小值.
题型:解答题难度:中档来源:不详

答案

(Ⅰ)由题意可得2an=sn+2,
当n=1时,a1=2,
当n≥2时,有2an-1=sn-1+2,两式相减,整理得an=2an-1即数列{an}是以2为首项,2为公比的等比数列,故an=2n.
点P(bn,bn+1)在直线x-y+2=0上得出bn-bn+1+2=0,即bn+1-bn=2,
即数列{bn}是以1为首项,2为公差的等差数列,
因此bn=2n-1.
(Ⅱ)Bn=1+3+5+…+(2n-1)=n2
class="stub"1
B1
+class="stub"1
B2
+…+class="stub"1
Bn
=class="stub"1
12
+class="stub"1
22
+class="stub"1
32
+…+class="stub"1
n2
<1+class="stub"1
1×2
+class="stub"1
2×3
+..+class="stub"1
(n-1).n
=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∴class="stub"1
B1
+class="stub"1
B2
+…+class="stub"1
Bn
<2

(Ⅲ)Tn=class="stub"1
2
+class="stub"3
22
+class="stub"5
23
+…+
2n-1
2n

class="stub"1
2
Tn=class="stub"1
22
+class="stub"3
23
+class="stub"5
24
+…+class="stub"2n-1
2n+1

①-②得class="stub"1
2
Tn=class="stub"1
2
+class="stub"1
22
+class="stub"1
23
+class="stub"2
23
+…+class="stub"2
2n
-class="stub"2n-1
2n+1

Tn=3-class="stub"1
2n-2
-class="stub"2n-1
2n
<3

T4=class="stub"1
2
+class="stub"3
22
+class="stub"4
23
+class="stub"7
24
=class="stub"37
16
>2

∴满足条件Tn<c的最小值整数c=3.

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