问题
解答题
设数列{an},{bn}满足:a1=4,a2=
(1)用an表示an+1;并证明:∀n∈N+,an>2; (2)证明:{ln
(3)设Sn是数列{an}的前n项和,当n≥2时,Sn与2(n+
|
答案
(1)由已知得a1=4,a2=
,所以b1=1故an+1bn+1=anbn═a1b1=4;5 2
由已知:an>0,a1>2,a2>2,bn=
∴an+1=4 an
+an 2
,2 an
由均值不等式得an+1>2
故∀n∈N+,an>2
(2)
=(an+1+2 an+1-2
)2,an+1+2=an+2 an-2
,(an+2)2 2an
an+1-2=(an-2)2 2an
所以ln
=2lnan+1+2 an+1-2
,所以{lnan+2 an-2
}是等比数列an+2 an-2
(3)由(2)可知ln
=(ln3)×2n-1=ln32n-1∴an=an+2 an-2 32n-1+1 32n-1-1
设Cn=
=4 32n-1
<4 (32n-2)(32n-2)
Cn-1,(n≥2)1 4
Cn<
Cn-1<(1 4
)2Cn-2<<(1 4
)n-1C1=2(1 4
)n-11 4
∴当n≥2时,an<2+2(
)n-11 4
Sn=a1+a2++an<4+2(n-1)+2[
+(1 4
)2++(1 4
)n-1]1 4
=2n+2+2×
(1-1 4
)1 4n-1 1- 1 4
=2n+2+
(1-2 3
)<2n+1 4n-1
.8 3