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Sn1)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
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![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Sn1)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](/uploads/image/z/10374936-24-6.jpg?t=%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E5%B7%B2%E7%9F%A5a1%3D1%2Ca2%3D6%2Ca3%3D11%2C%E4%B8%94%285n-8%29S%28n%2B1%29-%285n%2B2%29Sn%3DAn%2Bb%2Cn%3D1%2C+%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn1%EF%BC%89%E6%B1%82A%E4%B8%8EB%E7%9A%84%E5%80%BC%EF%BC%9B%EF%BC%882%EF%BC%89%E8%AF%81%E6%98%8E%EF%BC%9A%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%EF%BC%9B%EF%BC%883%EF%BC%89%E8%AF%81%E6%98%8E%EF%BC%9A%E4%B8%8D%E7%AD%89%E5%BC%8F%E6%A0%B9%E5%8F%B75amn-%E6%A0%B9%E5%8F%B7aman%3E1%E5%AF%B9%E4%BB%BB%E4%BD%95%E6%AD%A3%E6%95%B4%E6%95%B0m%2Cn)
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Sn1)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
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
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Sn1)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
(1)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
1)S1=1,S2=7,S3=18
当n=1时,-3S2-7S1=A+B
当n=2时,2S3-12S2=2A+B
得A=-20,B=-8
第三问用数学归纳法