强大的数学题：设数列{An}的前N项和为Sn已知A1=.

题目

强大的数学题：设数列{An}的前N项和为Sn已知A1=.
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 = -20n-8 (n=1,2,3,4,.)
请证明数列{An}为等差数列

答案

因为：(5n-8)Sn+1-(5n+2)Sn = -20n-8...(1)
所以：(5(n+1)-8)Sn+2 -（5(n+1)+2)Sn+1 = -20(n+1)-8
即：(5n-3)Sn+2 -(5n+7)Sn+1 = -20n-28...(2)
(2)-(1)
(5n-3)Sn+2 -(10n-1)Sn+1 +(5n+2)Sn = -20...(3)
所以,
(5(n+1)-3)Sn+3 -(10(n+1)-1)Sn+2 +(5(n+1)+2)Sn+1 = -20
即：(5n+2)Sn+3 -(10n+9)Sn+2 +(5n+7)Sn+1 = -20...(4)
(4)-(3)
(5n+2)Sn+3 -(15n+6)Sn+2 +(15n+6)Sn+1 -(5n+2)Sn =0...(5)
因为：An+1 =Sn+1 -Sn,带入（5）
所以：
(5n+2)An+3 -(10n+4)An+2 +(5n+2)An+1 =0

因为：(5n+2)不等于0.
所以：
An+3 -2An+2 +An+1 =0
所以：
An+3 -An+2 = An+2 -An+1...(当n≥1时)
又因为：
A3 -A2 =A2 -A1 =5
所以,数列{An}为等差数列