怎么能证明(5n-8)S(n+1)-(5n+2)Sn=-28,是等差数列?知道 a1=1,a2=6,a3=11
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怎么能证明(5n-8)S(n+1)-(5n+2)Sn=-28,是等差数列?知道 a1=1,a2=6,a3=11
如题
如题
由a1=1,a2=6,a3=11,得S1=1,S2=7,S3=18.
把n=1,2分别代入(5n-8)S(n+1)-(5n+2)Sn=An+B,
得A+B=-28, 2A+B=-48,解得A=-20, B=-8
则5n[S(n+1)-Sn]-8S(n+1)-2Sn=-20n-8.
即5na(n+1)-8S(n+1)-2Sn=-20-8n……(1)
又5(n+1)a(n+2)-8S(n+2)-2S(n+1)=-20(n+1)-8……(2)
(2)-(1),得5(n+1)a(n+2)-5na(n+1)-8a(n+2)-2a(n+1)=-20
即(5n-3)a(n+2)-(5n+2)a(n+1)=-20……(3)
又(5n+2)a(n+3)-(5n+7)a(n+2)=-20……(4)
(4)-(3),得(5n+2)[a(n+3)-2a(n+2)+a(n+1)]=0,因为5n+2≠0
所以a(n+3)-2a(n+2)+a(n+1)=0
所以a(n+3)-a(n+2)=a(n+2)-a(n+1),n≥1.又a3-a2=a2-a1=5,
因此,数列{an}是首项为1,公差为5的等差数列.
把n=1,2分别代入(5n-8)S(n+1)-(5n+2)Sn=An+B,
得A+B=-28, 2A+B=-48,解得A=-20, B=-8
则5n[S(n+1)-Sn]-8S(n+1)-2Sn=-20n-8.
即5na(n+1)-8S(n+1)-2Sn=-20-8n……(1)
又5(n+1)a(n+2)-8S(n+2)-2S(n+1)=-20(n+1)-8……(2)
(2)-(1),得5(n+1)a(n+2)-5na(n+1)-8a(n+2)-2a(n+1)=-20
即(5n-3)a(n+2)-(5n+2)a(n+1)=-20……(3)
又(5n+2)a(n+3)-(5n+7)a(n+2)=-20……(4)
(4)-(3),得(5n+2)[a(n+3)-2a(n+2)+a(n+1)]=0,因为5n+2≠0
所以a(n+3)-2a(n+2)+a(n+1)=0
所以a(n+3)-a(n+2)=a(n+2)-a(n+1),n≥1.又a3-a2=a2-a1=5,
因此,数列{an}是首项为1,公差为5的等差数列.
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