设数列{an}满足a1+2a2+22a3+…+2n-1an=n2,n∈N*.(1)求数列{an}的通项公式;(2)设bn=1log12an,cn=bnbn+1,记Sn=c1+c2+…+cn,证明:Sn

题目简介

设数列{an}满足a1+2a2+22a3+…+2n-1an=n2,n∈N*.(1)求数列{an}的通项公式;(2)设bn=1log12an,cn=bnbn+1,记Sn=c1+c2+…+cn,证明:Sn

题目详情

设数列{an}满足a1+2a2+22a3+…+2n-1an=
n
2
,n∈N*

(1)求数列{an}的通项公式;
(2)设bn=
1
log
1
2
an
cn=bnbn+1
,记Sn=c1+c2+…+cn,证明:Sn<1.
题型:解答题难度:中档来源:不详

答案

(1)由题意,a1+2a2+22a3+…+2n-2an-1+2n-1an=class="stub"n
2

当n≥2时,a1+2a2+22a3+…+2n-2an-1=class="stub"n-1
2

两式相减,得2n-1an=class="stub"n
2
-class="stub"n-1
2
=class="stub"1
2

所以,当n≥2时,an=class="stub"1
2 n
,…(4分)
当n=1时,a1=class="stub"1
2
也满足上式,
所求通项公式an=class="stub"1
2 n
,n∈N*.…(6分)
(2)∵bn=class="stub"1
logclass="stub"1
2
an
=class="stub"1
logclass="stub"1
2
(class="stub"1
2
)n
=class="stub"1
n
.…(8分)
cn=
n+1
-
n
n(n+1)
=class="stub"1
n
-class="stub"1
n+1
,…(10分)
∴Sn=c1+c2+…+cn
=(1-class="stub"1
2
)+(class="stub"1
2
-class="stub"1
3
)+(class="stub"1
3
-class="stub"1
4
)
+…+(class="stub"1
n
-class="stub"1
n+1
)

=1-class="stub"1
n+1
<1.…(12分)

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