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)-(5n+2)Sn=-20n-
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/07/13 21:44:32
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)-(5n+2)Sn=-20n-8,求证An为等差数列
证明:
(5n-8)S(n+1)-(5n+2)Sn=-20n-8
(5n-8)an+(5n-8)Sn-(5n+2)Sn=-20n-8
(5n-8)an-10Sn=-20n-8 ①
令n=n-1,得
(5n-13)a(n-1)-10S(n-1)=-20n+12 ②
①-②,得
(5n-18)an-(5n-13)a(n-1)=-20 ③
再令n=n-1,得
(5n-23)a(n-1)-(5n-18)a(n-2)=-20 ④
③-④,得
(5n-18)an+(5n-18)a(n-2)=(5n-13)a(n-1)+(5n-23)a(n-1)=2(5n-18)a(n-1)
得
an+a(n-2)=2a(n-1)
an-a(n-1)=a(n-1)-a(n-2)
所以an是等差数列!
(5n-8)S(n+1)-(5n+2)Sn=-20n-8
(5n-8)an+(5n-8)Sn-(5n+2)Sn=-20n-8
(5n-8)an-10Sn=-20n-8 ①
令n=n-1,得
(5n-13)a(n-1)-10S(n-1)=-20n+12 ②
①-②,得
(5n-18)an-(5n-13)a(n-1)=-20 ③
再令n=n-1,得
(5n-23)a(n-1)-(5n-18)a(n-2)=-20 ④
③-④,得
(5n-18)an+(5n-18)a(n-2)=(5n-13)a(n-1)+(5n-23)a(n-1)=2(5n-18)a(n-1)
得
an+a(n-2)=2a(n-1)
an-a(n-1)=a(n-1)-a(n-2)
所以an是等差数列!
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)-(5n+2)Sn=-20n-
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)-(5n+2)Sn=A*n
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=
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)—(5n+2)Sn=Pn+
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 =A*n+
数列{an}的前n项和为Sn,已知a1=1,a2=6,a3=11,且(5n-8)S(n+1)-(5n+2)Sn=An+b
设{an}的前n项和为S,已知a1=1,a2=6,a3=11,(5n-8)Sn+1-(5n+2)Sn= -20n-8.n
已知数列an的前n项和为sn,且sn+an=1/2(n2+5n+2)(2属于n*) 计算a1 a2 a3 a4
设数列{an}的前n项和为sn,已知a1+2a2+3a3+…+nan=(n-1)Sn+2n(n∈N*)
设数列an的前n项和为Sn,已知S1=1,Sn+1/Sn=n+c/n,且a1,a2,a3成等差数列
设数列{an}的前n项和为Sn,已知a1=1,a2=6,a3=11,且(5n
设{an}的前n项和为S,已知a1=1,a2=6,a3=11,(5n-8)Sn+1-(5n+2)Sn=a*n+b,n=1