已知a>0,b<0,且a+b≠0,令a1=a,b1=b,且对任意的正整数k,当ak+bk≥0时,ak+1=12ak-14bk,bk+1=34bk;当ak+bk<0时,bk+1=-14ak+12bk,a

题目简介

已知a>0,b<0,且a+b≠0,令a1=a,b1=b,且对任意的正整数k,当ak+bk≥0时,ak+1=12ak-14bk,bk+1=34bk;当ak+bk<0时,bk+1=-14ak+12bk,a

题目详情

已知a>0,b<0,且a+b≠0,令a1=a,b1=b,且对任意的正整数k,当ak+bk≥0时,ak+1=
1
2
ak-
1
4
bk
bk+1=
3
4
bk
;当ak+bk<0时,bk+1=-
1
4
ak+
1
2
bk
ak+1=
3
4
ak

(1)求数列{an+bn}的通项公式;
(2)若对任意的正整数n,an+bn<0恒成立,问是否存在a,b使得{bn}为等比数列?若存在,求出a,b满足的条件;若不存在,说明理由;
(3)若对任意的正整数n,an+bn<0,且b2n=
3
4
b2n+1
,求数列{bn}的通项公式.
题型:解答题难度:中档来源:徐州一模

答案

(1)当ak+bk≥0时,ak+1=class="stub"1
2
ak-class="stub"1
4
bk
bk+1=class="stub"3
4
bk

∴ak+1+bk+1=class="stub"1
2
ak-class="stub"1
4
bk+class="stub"3
4
bk
=class="stub"1
2
(ak+bk)

当ak+bk<0时,bk+1=-class="stub"1
4
ak+class="stub"1
2
bk
ak+1=class="stub"3
4
ak

∴ak+1+bk+1=-class="stub"1
4
ak+class="stub"1
2
bk+class="stub"3
4
ak
=class="stub"1
2
(ak+bk)

∴总有ak+1+bk+1=class="stub"1
2
(ak+bk)

∵a1=a,b1=b,
∴a1+b1=b+a
∴数列{an+bn}是以a+b为首项,以class="stub"1
2
为公比的等比数列
∴bn+an=(b+a)(class="stub"1
2
)n-1.
(2)∵an+bn<0恒成立
∴(b+a)(class="stub"1
2
)n-1
<0恒成立
∴b+a<0
∵当ak+bk<0时,bk+1=-class="stub"1
4
ak+class="stub"1
2
bk
ak+1=class="stub"3
4
ak

an=a•(class="stub"3
4
)n-1

bn=(a+b)•(class="stub"1
2
)n-1-a•(class="stub"3
4
)n-1
不可能是个等比数列
故{bn}不是等比数列
(3)∵an+bn<0,bk+1=-class="stub"1
4
ak+class="stub"1
2
bk
ak+1=class="stub"3
4
ak

b2n+1=-class="stub"1
4
a2n+class="stub"1
2
b2n
a2n+1=class="stub"3
4
a2n

b2n=class="stub"3
4
b2n+1

b2n+1=class="stub"4
3
b2n
=-class="stub"1
4
a2n+class="stub"1
2
b2n

b2n=-class="stub"3
10
a2n
=-class="stub"3
10
a•(class="stub"3
4
)2n-1

∴bn=-class="stub"3a
10
•(class="stub"3
4
)n-1

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