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

来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/08 07:52:14
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

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...

全部展开

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

收起

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

设数列{an}的前n项和为Sn,a1=10,a(n+1)=9Sn+10 数列{an},中,a1=1/3,设Sn为数列{an}的前n项和,Sn=n(2n-1)an 求Sn 设数列an的首项a1等于1,前n项和为sn,sn+1=2n设数列an的首项a1等于1,前n项和为sn,sn+1=2n 设数列an的前n项和为Sn,已知a1=1,(2Sn)/n=a(n+1)-1/3n^2-n-2/3 设数列an的前n项和为sn,已知a1=a,a不等于3,a(n+1)=sn+3^n 设数列{an}的前n项和为Sn,已知a1=a,an+1=Sn+3^n,n∈N+.设bn=Sn+3n,求数列{bn}的通项公式 设数列{an}的前n项和为Sn,已知a1=a,An+1 =Sn+3^n (n∈N+),设bn=Sn-3^n,求数列{bn}的通项公式. 设数列An的前n项和为Sn,已知a1=1,An+1=Sn+3n+1求证数列{An+3}是等比数列 设数列{An}的前n项和为Sn,已知A1=a,A(n+1)=Sn+3∧n,n是正整数,设Bn=Sn-3∧n,求数列{Bn}的通项 设数列{an}的前n项和为sn,已知a1=a,an+1=Sn+3^n,n属于N*.)设数列{an}的前n项和为sn,已知a1=a,an+1=Sn+3^n,n属于N*.(1)设bn=Sn-3^n,求数列{bn}通项公式;(2)若an+1>=an,n属于N*,求a的取值范围....Thanks.... 设数列【An】的前n项和为Sn,A1=10,An+1=9Sn+10.设Bn=lgAn,求证数列【Bn】为等差数列 设数列An的前n项和为Sn,已知a1=a,an+1=Sn+3n设Bn=Sn-3n次方,求数列Bn的通项公式 已知数列{an}的前n项和Sn,且a1=a(a为非零常数),当n>=2时,an=2Sn^2/2已知数列{an}的前n项和Sn,且a1=a(a为非零常数),当n>=2时,an=2Sn^2/2Sn-11)求证:数列{1/Sn}是等差数列2)设bn=Sn/an,数列bn的前n项和为Tn.已 设数列{an}的前n项和为Sn,若a1=1,Sn=2an+Sn+(n∈N+),则a6= 【高考】若数列{an}满足,a1=1,且a(n+1)=an/(1+an),设数列{bn}的前n项和为Sn,且Sn=2-bn,求{bn/an}的前...【高考】若数列{an}满足,a1=1,且a(n+1)=an/(1+an),设数列{bn}的前n项和为Sn,且Sn=2-bn,求{bn/an}的前n项和Tn 设数列{An}的前n项的和为Sn已知A1=a A(n+1)=Sn+3^n (1)设Bn=Sn-3^n 求数列{Bn}的通项公式? 设数列{an}的前n项和为Sn,且a1=1,Sn=nan-2n(n-1) (1)求a2,a3,a4,并求出数列{an}的通项公式.(2)设数列{1/a 一道数学题:设数列{an}的前n项和为Sn.已知a1=a,a(n+1)=Sn+3^n,n属于N*.设数列{an}的前n项和为Sn.已知a1=a,a(n+1)=Sn+3^n,n属于N*.(1)设bn=Sn-3^n,求数列{bn}的通项公式; (2)若a(n+1)≥an,n属于N*,求a的取值