设数列{an}的前n项和为Sn,已知a1=1,a2=6,a3=11

来源:百度知道 编辑:UC知道 时间:2024/06/14 16:20:30
且(5n-8)S(n+1)-(5n+2)Sn=A*n+B
悬赏分:10 - 解决时间:2008-8-20 16:43
设数列{an}的前n项和为Sn,已知a1=1,a2=6,a3=11,且(5n-8)S(n+1)-(5n+2)Sn=A*n+B, n=1,2,3...., 其中AB为常数
1)求A,B(2)求证:{an}为等差数列 (第二小题过程要简单明了)

(1)解:由已知得:S1=1,S2=7,S3=18
令n=1,n=2,得:-3*7-7*1=A*1+B,2*18-12*7=2A+B
解得:A=-20,B=-8
(2)证明(5n-8)Sn+1-(5n+2)Sn=-20n-8
则 (5n-3)Sn+2-(5n+7)Sn+1=-20n-28
两式相减,得:(5n-3)Sn+2-(10n-1)Sn+1+(5n+2)Sn=-20
(5n-3)Sn+2-(5n-3)Sn+1-(5n+2)Sn+1+(5n+2)Sn=-20
(5n-3)an+2-(5n+2)an+1=20
则 (5n+2)an+3-(5n+7)an+2=20
两式相减,得:(5n+2)an+3-(10n+4)an+2+(5n+2)an+1=0
an+3-2an+2+an+1=0
又已知a1=1,a2=6,a3=11,
综上,an+2-2an+1+an=0即2an+1=an+an+2
证得{an}为等差数列