已知数列{an}是首项为1的等差数列,其公差d>0,且a3,a7+2,3a9成等比数列.(Ⅰ)求数列{an}的通项公式;(Ⅱ)求证:a1+a22+a322+…+an2n-1<4(n∈N*).-高二数学

题目简介

已知数列{an}是首项为1的等差数列,其公差d>0,且a3,a7+2,3a9成等比数列.(Ⅰ)求数列{an}的通项公式;(Ⅱ)求证:a1+a22+a322+…+an2n-1<4(n∈N*).-高二数学

题目详情

已知数列{an}是首项为1的等差数列,其公差d>0,且a3,a7+2,3a9成等比数列.
(Ⅰ)求数列{an}的通项公式;
(Ⅱ)求证:a1+
a2
2
+
a3
22
+…+
an
2n-1
<4
(n∈N*).
题型:解答题难度:中档来源:不详

答案

(Ⅰ)因为an=1+(n-1)d,则a3=1+2d,a7=1+6d,a9=1+8d.(3分)
由已知,(a7+2)2=a3•3a9,则(3+6d)2=3(1+2d)(1+8d),即2d2-d-1=0.(5分)
所以(2d+1)(d-1)=0.
因为d>0,则d=1,
故an=n.(6分)
(Ⅱ)设Sn=a1+
a2
2
+
a3
22
+…+
an
2n-1
,则Sn=1+class="stub"2
2
+class="stub"3
22
+…+class="stub"n
2n-1

class="stub"1
2
Sn=class="stub"1
2
+class="stub"2
22
+…+class="stub"n
2n
.(8分)
两式相减得,class="stub"1
2
Sn=1+class="stub"1
2
+class="stub"1
22
+class="stub"1
23
+…+class="stub"1
2n-1
-class="stub"n
2n
=
1-class="stub"1
2n
1-class="stub"1
2
-class="stub"n
2n
=2-class="stub"n+2
2n

所以Sn=4-class="stub"n+2
2n-1
.(12分)
因为class="stub"n+2
2n-1
>0,则4-class="stub"n+2
2n-1
<4,故a1+
a2
2
+
a3
22
+…+
an
2n-1
<4.(13分)

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