Question

数列的题目，看看

Answer 1

（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
�B8�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