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,a3;(2)求Sn的表达式.-数学

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,a3;(2)求Sn的表达式.-数学

题目详情

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,a3
(2)求Sn的表达式.
题型:解答题难度:中档来源:贵溪市模拟

答案

(1)当n=1时,由已知得a12-2a1-a12+1=0
∴a1=class="stub"1
2

同理,可解得 a2=class="stub"1
6
,a3=class="stub"1
12
       (5分)
(2)解法一:由题设Sn2-2Sn-anSn+1=0,
当n≥2时,an=Sn-Sn-1
代入上式,得Sn-1Sn-2Sn+1=0,(*) (6分)
由(1)可得S1=a1=class="stub"1
2
,S2=a1+a2=class="stub"1
2
+class="stub"1
6
=class="stub"2
3
由(*)式可得S3=class="stub"3
4

由此猜想:Sn=class="stub"n
n+1
   (8分)
证明:①当n=1时,结论成立.
②假设当n=k时结论成立,
Sk=class="stub"k
k+1
那么,由(*)得Sk+1=class="stub"1
2-Sk

Sk+1=class="stub"1
2-class="stub"k
k+1
=class="stub"k+1
k+2

所以当n=k+1时结论也成立,根据①和②可知,
Sn=class="stub"n
n+1
对所有正整数n都成立.(12分)
解法二:由题设Sn2-2Sn-anSn+1=0,
当n≥2,an=Sn-Sn-1
代入上式,得SnSn-1-2Sn+1=0
Sn=class="stub"1
2-Sn-1

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

class="stub"1
Sn-1
=
2-Sn
Sn-1-1
=-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,以-1为公差的等差数列,
class="stub"1
Sn-1
=-2+(-1)(n-1)
=-n-1
Sn=1-class="stub"1
1+n
=class="stub"n
n+1
 (12分)

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