已知数列{an}的前n项和为Sn,且满足an+2Sn•Sn-1=0(n≥2),a1=12.(1)求证:{1Sn}是等差数列;(2)求an表达式;(3)若bn=2(1-n)an(n≥2),求证:b22+

题目简介

已知数列{an}的前n项和为Sn,且满足an+2Sn•Sn-1=0(n≥2),a1=12.(1)求证:{1Sn}是等差数列;(2)求an表达式;(3)若bn=2(1-n)an(n≥2),求证:b22+

题目详情

已知数列{an}的前n项和为Sn,且满足
an+2Sn•Sn-1=0(n≥2),a1=
1
2

(1)求证:{
1
Sn
}是等差数列;
(2)求an表达式;
(3)若bn=2(1-n)an(n≥2),求证:b22+b32+…+bn2<1.
题型:解答题难度:中档来源:不详

答案

解(1)∵-an=2SnSn-1,
∴-Sn+Sn-1=2SnSn-1(n≥2)
Sn≠0,∴class="stub"1
Sn
-class="stub"1
Sn-1
=2,又class="stub"1
S1
=class="stub"1
a1
=2,
∴{class="stub"1
Sn
}是以2为首项,公差为2的等差数列.

(2)由(1)class="stub"1
Sn
=2+(n-1)2=2n,
∴Sn=class="stub"1
2n

当n≥2时,an=Sn-Sn-1=-class="stub"1
2n(n-1)

n=1时,a1=S1=class="stub"1
2

∴an=
class="stub"1
2
(n=1)
-class="stub"1
2n(n-1)
(n≥2)


(3)由(2)知bn=2(1-n)an=class="stub"1
n

∴b22+b32+…+bn2=class="stub"1
22
+class="stub"1
32
+…+class="stub"1
n2
class="stub"1
1×2
+class="stub"1
2×3
+…+class="stub"1
(n-1)n

=(1-class="stub"1
2
)+(class="stub"1
2
-class="stub"1
3
)+…+(class="stub"1
n-1
-class="stub"1
n
)=1-class="stub"1
n
<1.

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