且(5n-8)S(n+1)-(5n+2)Sn=A*n+B10 - 解决时间:2008-8-20 16:43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+B,n=1,2,3.,其中AB为常数 1)求A,B(2)求证:{an}为等差数列 (第二小题过程要简
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![且(5n-8)S(n+1)-(5n+2)Sn=A*n+B10 - 解决时间:2008-8-20 16:43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+B,n=1,2,3.,其中AB为常数 1)求A,B(2)求证:{an}为等差数列 (第二小题过程要简](/uploads/image/z/4340989-37-9.jpg?t=%E4%B8%94%285n-8%29S%28n%2B1%29-%285n%2B2%29Sn%3DA%2An%2BB10+-+%E8%A7%A3%E5%86%B3%E6%97%B6%E9%97%B4%EF%BC%9A2008-8-20+16%3A43%E8%AE%BE%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%3DA%2An%2BB%2Cn%3D1%2C2%2C3.%2C%E5%85%B6%E4%B8%ADAB%E4%B8%BA%E5%B8%B8%E6%95%B0+1%EF%BC%89%E6%B1%82A%2CB%282%29%E6%B1%82%E8%AF%81%EF%BC%9A%EF%BD%9Ban%EF%BD%9D%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97+%EF%BC%88%E7%AC%AC%E4%BA%8C%E5%B0%8F%E9%A2%98%E8%BF%87%E7%A8%8B%E8%A6%81%E7%AE%80)
且(5n-8)S(n+1)-(5n+2)Sn=A*n+B10 - 解决时间:2008-8-20 16:43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+B,n=1,2,3.,其中AB为常数 1)求A,B(2)求证:{an}为等差数列 (第二小题过程要简
且(5n-8)S(n+1)-(5n+2)Sn=A*n+B
10 - 解决时间:2008-8-20 16:43
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+B,n=1,2,3.,其中AB为常数
1)求A,B(2)求证:{an}为等差数列 (第二小题过程要简单明了)
且(5n-8)S(n+1)-(5n+2)Sn=A*n+B10 - 解决时间:2008-8-20 16:43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+B,n=1,2,3.,其中AB为常数 1)求A,B(2)求证:{an}为等差数列 (第二小题过程要简
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