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

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/14 11:28:53

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,n=1, 数列{an}的前n项和为Sn
1）求A与B的值；（2）证明：数列{an}为等差数列；（3）证明：不等式根号5amn-根号aman>1对任何正整数m,n都成立。主要回答第三问

由已知得：S1=1,S2=7,S3=18
令n=1,n=2,得：-3*7-7*1=A*1+B,2*18-12*7=2A+B
解得：A=-20,B=-8
（2）证明(5n-8)Sn+1-(5n+2)Sn=-20n-8
则 (5n-3)Sn+2-(5n+7)Sn+1=-20n-28
两式相减,得：(5n-3)Sn+2-(10n-1)Sn+1+(5n+2)Sn=-20
(5n-3)Sn+2-(5n-3)Sn+1-(5n+2)Sn+1+(5n+2)Sn=-20
(5n-3)an+2-(5n+2)an+1=20
则 (5n+2)an+3-(5n+7)an+2=20
两式相减,得：(5n+2)an+3-(10n+4)an+2+(5n+2)an+1=0
an+3-2an+2+an+1=0
又已知a1=1,a2=6,a3=11,
综上,an+2-2an+1+an=0即2an+1=an+an+2
证得{an}为等差数列

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=A*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=-20n- 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=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+ 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 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,满足2Sn=an－2∧n+1 +1 ,且a1,a2+5,a3成等差设数列{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a1=1,a2=6,a3=11,且（5n