已知数列{an}满足a1=2,an+1=2(1+1n)2an,n∈N*.(1)求数列{an}的通项;(2)设bn=ann,求ni=1bi;(3)设cn=nan,求证ni=1Ci<1724.-数学

题目简介

已知数列{an}满足a1=2,an+1=2(1+1n)2an,n∈N*.(1)求数列{an}的通项;(2)设bn=ann,求ni=1bi;(3)设cn=nan,求证ni=1Ci<1724.-数学

题目详情

已知数列{an}满足a1=2,an+1=2(1+
1
n
2an,n∈N*
(1)求数列{an}的通项;
(2)设bn=
an
n
,求
n




i=1
bi

(3)设cn=
n
an
,求证
n




i=1
Ci
17
24
题型:解答题难度:中档来源:不详

答案

(1)由已知得,
an+1
(n+1)2
=  2•
an
n2

{
an
n2
}
是公比为2的等比数列,首项a1=2,
an
n2
=2n,
∴an=n22n;
(2)bn=
an
n
=n2n,
n








i=1
bi
=1•2+2•22+3•23+…+n2n,
2
n








i=1
bi
=1•22+2•23+3•24+…+(n-1)2n+n2n+1,
∴-
n








i=1
bi
=2+22+23+…+2n-n2n+1=
2(1-2n)
1-2
-n2n+1
n








i=1
bi
=(n-1)2n+1+2;

(3)cn=class="stub"n
an
=class="stub"1
n2n

当n≥2时,class="stub"1
n2n
=class="stub"n-1
n(n-1)2n
< class="stub"n+1
n(n-1)2n
=class="stub"1
(n-1)2n-1
- class="stub"1
n2n

∴c1+c2+c3+…+cn=class="stub"1
2
+class="stub"1
8
+class="stub"1
24
+…+class="stub"1
n2n

class="stub"1
2
+class="stub"1
8
+class="stub"1
24
+class="stub"1
3•23
-class="stub"1
4•24
…+class="stub"1
(n-1)2n-1
-class="stub"1
n2n
class="stub"17
24

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