设数列{an}的前n项和为Sn,已知a1=1,a2=6,a3=11,且(5n-8)Sn+1-(5n+2)Sn=An+B,n=1,2,3,…,其中A、B为常数.(
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设数列{an}的前n项和为Sn,已知a1=1,a2=6,a3=11,且(5n-8)Sn+1-(5n+2)Sn=An+B,n=1,2,3,…,其中A、B为常数. (1)求A与B的值. (2)证明数列{an}为等差数列. |
答案
(1)由已知得s1=a1=1,s2=a1+a2=7,s3=a1+a2+a3=18
| 由(5n-8)sn+1-(5n+2)sn=An+B知 | -3s2-7s1=A+B | 2s3-12s2=2A+B |
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| ⇒⇒A=-20,B=-8 |
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(2)证明:由(1)知(5n-8)sn+1-(5n+2)sn=-20n-8① 所以(5n-3)sn+2-(5n+7)sn+1=-20n-28② ②-①得(5n-3)sn+2-(10n-1)sn+1+(5n+2)sn=-20③ 所以(5n+2)sn+3-(10n+9)sn+2+(5n+7)sn+1=-20④ ④-③得(5n+2)sn+3-(15n+6)sn+2+(15n+6)sn+1-(5n+2)sn=0 因为an+1=sn+1-sn,所以(5n+2)an+3-(10n+4)an+2+(5n+2)an+1=0 又因为5n+2≠0,所以an+3-2an+2+an+1=0,即an+3-an+2=an+2-an+1 n≥1, 又a3-a2=a2-a1=5.∴数列{an}为等差数列. |
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