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=
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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
(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
再问: (5(n-1)-8)an-10Sn-1=-20(n-1)-8怎么到(5n-8)a(n+1)-(5n+3)an=20
再答: (5n-8)a(n+1)-10Sn=-20n-8 (5(n-1)-8)an-10Sn-1=-20(n-1)-8 两式相减右边=-20n-8+20n-20+8=-20 (5n-8)a(n+1)-(5n+3)an=-20 (5(n-1)-8)an-(5(n-1)+3)an-1=-20 两式相减(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
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
再问: (5(n-1)-8)an-10Sn-1=-20(n-1)-8怎么到(5n-8)a(n+1)-(5n+3)an=20
再答: (5n-8)a(n+1)-10Sn=-20n-8 (5(n-1)-8)an-10Sn-1=-20(n-1)-8 两式相减右边=-20n-8+20n-20+8=-20 (5n-8)a(n+1)-(5n+3)an=-20 (5(n-1)-8)an-(5(n-1)+3)an-1=-20 两式相减(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
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