已知数列{an}的前n项和为Sn,满足an+Sn=3-82n,设bn=2n•an.(1)求证:数列{bn}是等差数列,并求出数列{an}的通项公式;(2)求数列{an•bn}中最大项;(3)求证:对于

题目简介

已知数列{an}的前n项和为Sn,满足an+Sn=3-82n,设bn=2n•an.(1)求证:数列{bn}是等差数列,并求出数列{an}的通项公式;(2)求数列{an•bn}中最大项;(3)求证:对于

题目详情

已知数列{an}的前n项和为Sn,满足an+Sn=3-
8
2n
,设bn=2nan
(1)求证:数列{bn}是等差数列,并求出数列{an}的通项公式;
(2)求数列{an•bn}中最大项;
(3)求证:对于给定的实数λ,一定存在正整数k,使得当n≥k时,不等式λSn<bn恒成立.
题型:解答题难度:中档来源:江苏一模

答案

(1)证明:∵an+Sn=3-class="stub"8
2n

∴n≥2时,an-1+Sn-1=3-class="stub"8
2n-1

两式相减可得2an-an-1=class="stub"8
2n-1
-class="stub"8
2n

2an-an-1=class="stub"4
2n-1

2nan-2n-1an-1=4
bn=2nan
∴bn-bn-1=4
∵n=1时,a1+S1=3-class="stub"8
21
,∴a1=-class="stub"1
2

b1=21a1=-1
∴数列{bn}是以-1为首项,4为公差的等差数列
bn=4n-5,an=class="stub"4n-5
2n

(2)an•bn=
(4n-5)2
2n

令f(n)=
(4n-5)2
2n
,则
f(n+1)
f(n)
=
(4n-1)2
2(4n-5)2

(4n-1)2
2(4n-5)2
<1,则16n2-72n+49>0
∴n>5时,
f(n+1)
f(n)
<1,n<5时,
f(n+1)
f(n)
>1
∴数列从第一项到第四项,单调递增,从第五项开始,单调递减
所以最大项是第四项class="stub"121
16

(3)证明:∵an=class="stub"4n-5
2n

∴数列{an}的前n项和为Sn=(-1)×class="stub"1
2
+3×class="stub"1
22
+…+(4n-5)×class="stub"1
2n

class="stub"1
2
Sn=(-1)×class="stub"1
22
+…+(4n-9)×class="stub"1
2n
+(4n-5)×class="stub"1
2n+1

两式相减可得class="stub"1
2
Sn=(-1)×class="stub"1
2
+4×class="stub"1
22
+…+4×class="stub"1
2n
-(4n-5)×class="stub"1
2n+1

∴Sn=3-(4n+3)×class="stub"1
2n

∴S1=-class="stub"1
2

∴Sn的值域[-class="stub"1
2
,3),
∵bn=4n-5,∴bn的值域[-1,+∞),
∴对于给定的实数λ,一定存在正整数k,使得当n≥k时,不等式λSn<bn恒成立.

更多内容推荐