已知函数f(x)=log2x,设f(a1),f(a2),f(a3),…,f(an),…,(n∈N*)是首项和公差都等于1的等差数列.数列{bn}满足bn=an+3n(n∈N*).(1)求数列{an}的

题目简介

已知函数f(x)=log2x,设f(a1),f(a2),f(a3),…,f(an),…,(n∈N*)是首项和公差都等于1的等差数列.数列{bn}满足bn=an+3n(n∈N*).(1)求数列{an}的

题目详情

已知函数f(x)=log2x,设f(a1),f(a2),f(a3),…,f(an),…,(n∈N*)是首项和公差都等于1的等差数列.数列{bn}满足bn=an+3n(n∈N*)
(1)求数列{an}的通项公式,并证明数列{bn}不是等比数列;
(2)令cn=
2n-1
an
,Sn=c1+c2+c3+…+cn,求证:Sn<3.
题型:解答题难度:中档来源:不详

答案

(1)由题意可得 f(an)=n=log2an,∴an=2n,故数列{an}是等比数列.
假设数列{bn}是等比数列,bn=2n+3n,则有 b22=b1b3
由因为 b22=132b1b3=5×35,∴b22b1b3,与假设矛盾,所以假设不成立.
∴数列{bn}不是等比数列.(6分)
(2)∵cn=class="stub"2n-1
an
,Sn=c1+c2+c3+…+cn,
Sn=class="stub"1
2
+class="stub"3
22
+class="stub"5
23
+…+class="stub"2n-3
2n-1
+class="stub"2n-1
2n
,…①
class="stub"1
2
Sn=class="stub"1
22
+class="stub"3
23
+class="stub"5
24
+…+class="stub"2n-3
2n
+class="stub"2n-1
2n+1
,…②,
①-②得
class="stub"1
2
Sn=class="stub"1
2
+class="stub"2
22
+class="stub"2
23
+class="stub"2
24
+…+class="stub"2
2n
-class="stub"2n-1
2n+1
 
=
class="stub"1
2
+(class="stub"1
21
+class="stub"1
22
+class="stub"1
23
+…+class="stub"2
2n-1
)-class="stub"2n-1
2n+1

=
class="stub"1
2
+
class="stub"1
2
[1-(class="stub"1
2
)
n-1
]
1-class="stub"1
2
-class="stub"2n-1
2n+1
=class="stub"1
2
+1-(class="stub"1
2
)
n-1
-class="stub"2n-1
2n+1
=class="stub"3
2
-class="stub"2n+3
2n+1

Sn=3-class="stub"2n+3
2n
<3.(12分)

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