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.
问题描述:
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
答
s1=a1=1s2=a1+a2=7s3=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+BA=-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...(5(n-1)-8)an-10Sn-1=-20(n-1)-8怎么到(5n-8)a(n+1)-(5n+3)an=20(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