已知数列{an}的前n项和为Sn,若a1=12且an+2Sn•Sn-1=0(n≥2).(Ⅰ)求证{1Sn}是等差数列,并求出an的表达式;(Ⅱ)若bn=2(1-n)an(n≥2),求证b22+b32+

题目简介

已知数列{an}的前n项和为Sn,若a1=12且an+2Sn•Sn-1=0(n≥2).(Ⅰ)求证{1Sn}是等差数列,并求出an的表达式;(Ⅱ)若bn=2(1-n)an(n≥2),求证b22+b32+

题目详情

已知数列{an}的前n项和为Sn,若a1=
1
2
且an+2Sn•Sn-1=0(n≥2).
(Ⅰ)求证{
1
Sn
}
是等差数列,并求出an的表达式;
(Ⅱ) 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

(I)证明:当n≥2时,an=Sn-Sn-1
又an+2SnSn-1=0
∴Sn-Sn-1+2SnSn-1=0(n≥2),
若Sn=0,则an=0,
∴a1=0与a1=class="stub"1
2
矛盾
∴Sn≠0,Sn-1≠0.
class="stub"1
Sn-1
-class="stub"1
Sn
+2=0即class="stub"1
Sn
-class="stub"1
Sn-1
=2,
class="stub"1
S2
-class="stub"1
S1
=2.
∴{class="stub"1
Sn
}是首项为2,公差为2的等差数列
由(I)知数列{class="stub"1
Sn
}是等差数列.
class="stub"1
Sn
=2+(n-1)•2=2n即Sn=class="stub"1
2n

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

又当n=1时,S1=a1=class="stub"1
2

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

(Ⅱ)证明:由(I)知bn=2(1-n)•class="stub"1
2n(1-n)
=class="stub"1
n
(n≥2)
∴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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