设数列{an}的前n项和为Sn且Sn2-2Sn-anSn+1=0,n=1,2,3…(1)求a1,a2(2)求Sn与Sn-1(n≥2)的关系式,并证明数列{1Sn-1}是等差数列.(3)求S1•S2•S

题目简介

设数列{an}的前n项和为Sn且Sn2-2Sn-anSn+1=0,n=1,2,3…(1)求a1,a2(2)求Sn与Sn-1(n≥2)的关系式,并证明数列{1Sn-1}是等差数列.(3)求S1•S2•S

题目详情

设数列{an}的前n项和为SnSn2-2Sn-anSn+1=0,n=1,2,3…
(1)求a1,a2
(2)求Sn与Sn-1(n≥2)的关系式,并证明数列{
1
Sn-1
}是等差数列.
(3)求S1•S2•S3…S2010•S2011的值.
题型:解答题难度:中档来源:不详

答案

(1)∵Sn2-2Sn-anSn+1=0,
∴取n=1,得S12-2S1-a1S1+1=0,即a12-2a1-a12+1=0,解之得a1=class="stub"1
2

取n=2,得S22-2S2-a2S2+1=0,即(class="stub"1
2
+a2)2-2(class="stub"1
2
+a2)-a2(class="stub"1
2
+a2)+1=0,解之得a2=class="stub"1
6

(2)由题设Sn2-2Sn-anSn+1=0,
当n≥2时,an=Sn-Sn-1,代入上式,化简得SnSn-1-2Sn+1=0
∴Sn=class="stub"1
2-Sn-1
,可得Sn-1-1=class="stub"1
2-Sn-1
-1=
Sn-1-1
2-Sn-1

class="stub"1
Sn-1
=
2-Sn
Sn-1
=-1+class="stub"1
Sn-1-1

∴数列{class="stub"1
Sn-1
}是以class="stub"1
S1-1
=-2为首项,公差d=-1的等差数列.
(3)由(2)得class="stub"1
Sn-1
=-2+(n-1)×(-1)=-n-1,
可得Sn=1-class="stub"1
n+1
=class="stub"n
n+1

∴S1•S2•S3•…•S2010•S2011=class="stub"1
2
×class="stub"2
3
×class="stub"3
4
×…×class="stub"2010
2011
×class="stub"2011
2012
=class="stub"1
2012

即S1•S2•S3•…•S2010•S2011的值为class="stub"1
2012

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