已知各项都不为零的数列{an}的前n项和为Sn,且Sn=12anan+1(n∈N*),a1=1.(1)求数列{an}的通项公式;(2)求证:1a12+1a22+1a32+…+1an2<74.-数学

题目简介

已知各项都不为零的数列{an}的前n项和为Sn,且Sn=12anan+1(n∈N*),a1=1.(1)求数列{an}的通项公式;(2)求证:1a12+1a22+1a32+…+1an2<74.-数学

题目详情

已知各项都不为零的数列{an}的前n项和为Sn,且Sn=
1
2
anan+1(n∈N*)
,a1=1.
(1)求数列{an}的通项公式;
(2)求证:
1
a12
+
1
a22
+
1
a32
+…+
1
an2
7
4
题型:解答题难度:中档来源:潮州二模

答案

(1)∵Sn=class="stub"1
2
anan+1
,①
Sn-1=class="stub"1
2
an-1an(n≥2)
,②
①-②得an=Sn-Sn-1=class="stub"1
2
(an+1-an-1)an

∵an≠0,∴an+1-an-1=2.
数列{an}的奇数项组成首项为a1=1,公差为2的等差数列;偶数项组成首项为a2,公差为2的等差数列.
∵a1=1,∴a2=
S1
class="stub"1
2
a1
=2

∴a2n-1=1+(n-1)×2=2n-1,a2n=2+(n-1)×2=2n.
∴数列{an}的通项公式为an=n.(n∈N*);
(2)证明:当n≥3时,class="stub"1
an2
=class="stub"1
n2
<class="stub"1
(n-1)n
=class="stub"1
(n-1)
-class="stub"1
n
,则
class="stub"1
a12
+class="stub"1
a22
+class="stub"1
a32
+…+class="stub"1
an2
=class="stub"1
12
+class="stub"1
22
+class="stub"1
32
+…+class="stub"1
n2
<1+class="stub"1
4
+(class="stub"1
2
-class="stub"1
3
)+(class="stub"1
3
-class="stub"1
4
)+…+class="stub"1
(n-1)
-class="stub"1
n
=class="stub"7
4
-class="stub"1
n
<class="stub"7
4

当n=1时,class="stub"1
a12
=1<class="stub"7
4
;  当n=2时,class="stub"1
a12
+class="stub"1
a22
=class="stub"5
4
<class="stub"7
4

class="stub"1
a12
+class="stub"1
a22
+class="stub"1
a32
+…+class="stub"1
an2
<class="stub"7
4

更多内容推荐