已知数列an中,a1=3,an+1=2an-1,设bn=2n/anan+1,求证:数列bn的前n项和Sn<1/3
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/09/05 02:43:37
已知数列an中,a1=3,an+1=2an-1,设bn=2n/anan+1,求证:数列bn的前n项和Sn<1/3
数列an中,a1=3,a=2an-1,
∴a-1=2(an-1),
∴an-1=(a1-1)*2^(n-1)=2^n,
∴an=2^n+1,
∴bn=2n/{(2^n+1)(2^(n+1)+1]},
∴Sn=2/(3*5)+4/(5*9)+……+2n/{(2^n+1)(2^(n+1)+1]}
=1/3-1/5+1/5-1/9+……+[n/2^(n-1)][1/(2^n+1)-1/[2^(n+1)+1],
当n∈N+时n
∴a-1=2(an-1),
∴an-1=(a1-1)*2^(n-1)=2^n,
∴an=2^n+1,
∴bn=2n/{(2^n+1)(2^(n+1)+1]},
∴Sn=2/(3*5)+4/(5*9)+……+2n/{(2^n+1)(2^(n+1)+1]}
=1/3-1/5+1/5-1/9+……+[n/2^(n-1)][1/(2^n+1)-1/[2^(n+1)+1],
当n∈N+时n
已知数列an中,a1=3,an+1=2an-1,设bn=2n/anan+1,求证:数列bn的前n项和Sn<1/3
已知数列{an},{bn}满足a1=2,2an=1+anan+1,bn=an-1,设数列{bn}的前n项和为Sn,令Tn
数列an满足an+1=2an-1且a1=3,bn=an-1/anan+1,数列bn前n项和为Sn.求数列an通项an,
已知数列{an}前n项和Sn=n^2+n,令bn=1/anan+1,求数列{bn}的前n项和Tn
已知数列an的前n项和为Sn,若Sn=2an+n,且bn=An-1/AnAn+1,求证an-1为等比数列;求数列{bn}
在数列an中,已知a1=2,an+1=2an/an +1,令bn=an(an -1).求证bn的前n项和
已知数列{an}的前n项和为Sn,a1=1,a(n+1)=1+2Sn.设bn=n/an,求证:数列{bn}的前n项和Tn
已知数列{an}的前N项和为Sn 且an+1=Sn-n+3,a1=2,设Bn=n/Sn-n+2前N项和为Tn 求证Tn
已知数列{an}的前n项和Sn=-an-(1/2)^(n-1)+2(n为正整数).令bn=2^n*an,求证数列{bn}
设数列{an}满足a1=1,an+1=3an,数列{bn}的前n项和Sn=n2+2n+1.
数列[an]中,前n项和sn=n²+1 (1)求数列[an]通项公式 (2)设bn=1/anan+
在数列an中a1=2,a(n+1)下标=4an-3n+1 1设bn=an-n求证bn是等比数列 2求数列an的前n项和s