已知数列{an}满足a1=1,a2=12,且[3+(-1)n]an+2=2an-2[(-1)n-1](n=1,2,3,…)(1)求a3,a4,a5,a6的值及数列{an}的通项公式;(2)令bn=a2

题目简介

已知数列{an}满足a1=1,a2=12,且[3+(-1)n]an+2=2an-2[(-1)n-1](n=1,2,3,…)(1)求a3,a4,a5,a6的值及数列{an}的通项公式;(2)令bn=a2

题目详情

已知数列{an}满足a1=1,a2=
1
2
,且[3+(-1)n]an+2=2an-2[(-1)n-1]
(n=1,2,3,…)
(1)求a3,a4,a5,a6的值及数列{an}的通项公式;
(2)令bn=a2n-1•a2n,记数列{bn}的前n项和为Tn,求证Tn<3.
题型:解答题难度:中档来源:不详

答案

(1)分别令n=1,2,3,4
可求得:a3=3,a4=class="stub"1
4
a5=5,a6=class="stub"1
8

当n为奇数时,不妨设n=2m-1,m∈N*,则a2m+1-a2m-1=2.
∴{a2m-1}为等差数列,∴a2m-1=1+(m-1)•2=2m-1即am=n.
当n为偶数时,设n=2m,m∈N*,则2a2m+2-a2m=0.
∴{a2m}为等比数列,a2m=class="stub"1
2
•(class="stub"1
2
)m-1=class="stub"1
2m
,故an=(class="stub"1
2
)class="stub"n
2

综上所述,an=
n,(n为奇)
(class="stub"1
2
)class="stub"n
2
,(n为偶数)

(2)bn=a2n-1a2n=(2n-1)•class="stub"1
2n

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

class="stub"1
2
Tn=1×class="stub"1
22
+3×class="stub"1
22
++(2n-3)•class="stub"1
2n
+(2n-1)•class="stub"1
2n

两式相减:class="stub"1
2
Tn=class="stub"1
2
+2[class="stub"1
22
+class="stub"1
23
++class="stub"1
2n
]-(2n-1)•class="stub"1
2n+1

=class="stub"1
2
+2•
class="stub"1
4
(1-class="stub"1
2n-1
)
1-class="stub"1
2
-(2n-1)•class="stub"1
2n+1

Tn=3-class="stub"2n+3
2n
,故Tn<3.

更多内容推荐