神马作文网设数列{an},Sn,a1=1,Sn=an+1-1,设bn=2^n/(an+1)(an+1+1),Tn=b1+b2+…+
来源：学生作业帮 编辑：神马作文网作业帮 分类：数学作业 时间：2024/11/17 12:36:47
设数列{an},Sn,a1=1,Sn=an+1-1,设bn=2^n/(an+1)(an+1+1),Tn=b1+b2+…+bn,求证:1/3小于等于Tn小于1
Sn=a(n+1)-1
S(n-1)=an-1
an=Sn-S(n-1)=a(n+1)-an
a(n+1)=2an
故{an}是公比为2的等比数列
an=a1*2^(n-1)=2^(n-1)
a(n+1)=2^n
bn=2^n/[a(n+1)][a(n+1)+1]=1/(2^n+1)
因bn
再问： bn=2^n/[(an)+1][a(n+1)+1]
再答： bn=2^n/[(an)+1][a(n+1)+1] =2^n/[2^(n-1)+1][2^n+1] =2{1/[2^(n-1)+1]-1/(2^n+1)} Tn=2{(1/2-1/3)+(1/3-1/4)+...+1/[2^(n-1)+1]-1/(2^n+1)} =2[1/2-1/(2^n+1)] =1-2/(2^n+1) 可见Tn
S(n-1)=an-1
an=Sn-S(n-1)=a(n+1)-an
a(n+1)=2an
故{an}是公比为2的等比数列
an=a1*2^(n-1)=2^(n-1)
a(n+1)=2^n
bn=2^n/[a(n+1)][a(n+1)+1]=1/(2^n+1)
因bn
再问： bn=2^n/[(an)+1][a(n+1)+1]
再答： bn=2^n/[(an)+1][a(n+1)+1] =2^n/[2^(n-1)+1][2^n+1] =2{1/[2^(n-1)+1]-1/(2^n+1)} Tn=2{(1/2-1/3)+(1/3-1/4)+...+1/[2^(n-1)+1]-1/(2^n+1)} =2[1/2-1/(2^n+1)] =1-2/(2^n+1) 可见Tn
已知数列{an}的各项均为正数,前n项和为Sn,且Sn=an(an+1)/2,设bn=1/2Sn,Tn=b1+b2+…+
有两个等差数列an,bn,若Sn/Tn=a1+a2+.an/b1+b2+---+bn=3n-1/2n+3,则a13/b1
数列an前n项和sn,数列bn中b1=a1,bn=an-a(n-1)(n>=2),若an+sn=n(1)设cn=an-1
设数列an的前n项和为Sn=2n∧2,bn为等比数列,且a1=b1,b2(a2-a3)=b1(1)求数列an和bn的通项
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}中a1=1 an+1=3an 数列{bn}的前几项和Sn=n^2+2n,设cn=an*bn,求Tn=C1
an等差数列 bn前n项和sn满足sn=3(bn－1)／2 且a2=b1 a5=b2 ⑴求an bn通项 ⑵设tn为数列
数列{an},a1=1,an=2-2Sn,求an,若bn=n*an,求{bn}的前n项和Tn
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
数列 an=2n-1 设bn=an/3^n 求和tn=b1+..bn?
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
已知数列{an}中a1=1 a[n+1]=3an 数列{bn}的前几项和Sn=n^2+2n,设cn=an*bn,求Tn=