数列{an}满足a1=1,an+3=an+3,an+2≥an+2(n∈N*).(1)求a7,a5,a3,a6;(2)求数列{an}的通项公式an;(3)求证:1a12+1a22+1a32+…+1an2

题目简介

数列{an}满足a1=1,an+3=an+3,an+2≥an+2(n∈N*).(1)求a7,a5,a3,a6;(2)求数列{an}的通项公式an;(3)求证:1a12+1a22+1a32+…+1an2

题目详情

数列{an}满足a1=1,an+3=an+3,an+2≥an+2(n∈N*).
(1)求a7,a5,a3,a6;        
(2)求数列{an}的通项公式an
(3)求证:
1
a12
+
1
a22
+
1
a32
+…+
1
an2
<2
题型:解答题难度:中档来源:不详

答案

(1)∵a1=1,an+3=an+3,
∴a4=4,a7=7
∵an+2≥an+2
∴a3≥3,a5≥a3+2,a7≥a5+2,
∴a5=5,a3=3,a6=a3+3=6
(2)∵an+3=an+3,an+2≥an+2(n∈N*)
∴an+3≤an+2+1(n∈N*)
∴an+1≤an+1,an+2≤an+1+1
∴an+1+an+2+an+3≤an+an+1+an+2+3,即an+3≤an+3
∴an+1=an+1,an+2=an+1+1,an+3=an+2+1
∴{an}为等差数列,公差d=1.
∴an=n
(3)证明:n=1时,class="stub"1
a12
=1<2成立n>1时,
class="stub"1
an2
=class="stub"1
n2
<class="stub"1
n(n-1)
=class="stub"1
n-1
-class="stub"1
n
(n>1)
class="stub"1
a12
+class="stub"1
a22
+class="stub"1
a32
+…+class="stub"1
an2

1+(1-class="stub"1
2
)+(class="stub"1
2
-class="stub"1
3
)+…+(class="stub"1
n-1
-class="stub"1
n
)
=2-class="stub"1
n
<2
class="stub"1
a12
+class="stub"1
a22
+class="stub"1
a32
+…+class="stub"1
an2
<2

更多内容推荐