设数列an的前n项和为Sn,a1=1,a2=6,a3=11,且（5n-8)S(n+1)-(5n+2）Sn=An+B,n=1,2,3...,A,B为常数.（1）求A,B的值（2）证明an为等差数列

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设数列an的前n项和为Sn,a1=1,a2=6,a3=11,且（5n-8)S(n+1)-(5n+2）Sn=An+B,n=1,2,3...,A,B为常数.（1）求A,B的值（2）证明an为等差数列
设数列an的前n项和为Sn,a1=1,a2=6,a3=11,且（5n-8)S(n+1)-(5n+2）Sn=An+B,n=1,2,3...,A,B为常数.
（1）求A,B的值（2）证明an为等差数列

设数列an的前n项和为Sn,a1=1,a2=6,a3=11,且（5n-8)S(n+1)-(5n+2）Sn=An+B,n=1,2,3...,A,B为常数.（1）求A,B的值（2）证明an为等差数列
s1=a1=1
s2=a1+a2=7
s3=a1+a2+a3=1+6+11=18
（5n-8)S(n+1)-(5n+2）Sn=An+B
（5*1-8)*7-(5*1+2)*1=A+B
（5*2-8)*18-(5*2+2）*7=2A+B
A=-20 B=-8
（5n-8)S(n+1)-(5n+2）Sn=-20n-8
（5n-8)S(n+1)-(5n-8）Sn-10Sn=-20n-8
（5n-8)a(n+1)-10Sn=-20n-8
（5(n-1)-8)an-10Sn-1=-20(n-1)-8
（5n-8)a(n+1)-（5n+3)an=20
（5(n-1)-8)an-（5(n-1)+3)an-1=20
（5n-8)a(n+1)-（10n-16)an+（5n-8)a(n-1)=0
相减同除（5n-8)得 an+1-an=an-an-1

1
(5*1-8)(6+1) - (5*1+2)*1=A+B
-28=A+B
(5*2-8)*(1+6+11) - (5*2+2)*(1+6)=2A+B
-48=2A+B
A=-20 B=-8
2
(5n-8)Sn+1 -(5n+2)Sn=-20n-8
(5n-8)(Sn+an+1)-(5n+2)Sn=-2...

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1
(5*1-8)(6+1) - (5*1+2)*1=A+B
-28=A+B
(5*2-8)*(1+6+11) - (5*2+2)*(1+6)=2A+B
-48=2A+B
A=-20 B=-8
2
(5n-8)Sn+1 -(5n+2)Sn=-20n-8
(5n-8)(Sn+an+1)-(5n+2)Sn=-20n-8
-10Sn+(5n-8)an+1=-20n-8
10Sn-(8-5n)an+1=20n+8
10Sn-1+(8-5(n-1))an=20(n-1)+8
10(Sn-Sn-1)+(8-5n)an+1-(13-5n)an=20
10an+(8-5n)an+1-(13-5n)an=20
(8-5n)an+1-(3-5n)an=20
(8-5n+5)an-(8-5n)an-1=20
(8-5n)(an+1+an-1)=(16-10n)an
an+1+an-1=2an
an+1-an=an=an-1
an等差数列

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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)S(n 1解得：A=-20，B=-8 （2）证明(5n-8)Sn 1-(5n 2)Sn=-20n-8 则

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