1.已知数列{an}满足a1=1,a(n+1)(n+1为角标)=an+2,且前n项的和味Sn求(1)求数列{Sn/n}的前n项和Tn(2)求数列{1/Tn}的前n项和2.解不等式a(x-1)/(x-2)>2
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/18 13:22:33
1.已知数列{an}满足a1=1,a(n+1)(n+1为角标)=an+2,且前n项的和味Sn求(1)求数列{Sn/n}的前n项和Tn(2)求数列{1/Tn}的前n项和2.解不等式a(x-1)/(x-2
1.已知数列{an}满足a1=1,a(n+1)(n+1为角标)=an+2,且前n项的和味Sn求(1)求数列{Sn/n}的前n项和Tn(2)求数列{1/Tn}的前n项和2.解不等式a(x-1)/(x-2)>2
1.已知数列{an}满足a1=1,a(n+1)(n+1为角标)=an+2,且前n项的和味Sn
求(1)求数列{Sn/n}的前n项和Tn
(2)求数列{1/Tn}的前n项和
2.解不等式a(x-1)/(x-2)>2
1.已知数列{an}满足a1=1,a(n+1)(n+1为角标)=an+2,且前n项的和味Sn求(1)求数列{Sn/n}的前n项和Tn(2)求数列{1/Tn}的前n项和2.解不等式a(x-1)/(x-2)>2
第一题:
1)
由递推关系 及 a1=1知:数列{an}是以1为首项2为公差的等差数列;
则:Sn=n^2……………………(就是n的平方了,下同);
则Tn=1+2+3+…+n=(1/2)*n*(n+1);
2)
1/Tn =2/[n*(n+1)]=2[(1/n)-(1/n+1)]…………(这是裂项相消);
设:Qn是数列{1/Tn}的前n项和;
则:Qn=2{[1-(1/2)]+[(1/2)-(1/3)]+…+[(1/n)-(1/n+1)]}=2-2/(n+1).
第二题:
原式等价于:ax^2-3ax+2(a-1)>0;
Δ=9a^2-4*a*2(a-1)=a^2+8a;
所以:
a
已知数列{an}满足a(n+1)=an+n,a1=1,则an=
已知数列{an}满足a(n+1)=an+lg2,a1=1,求an
已知数列{an}满足a1=1,3a(n+1)+an-7
已知数列{an}满足a1=1,a(n+1)=2an+1.求证(1)数列a(n+1)是等比数列;(2)求an
已知数列an满足a1=1,a(n+1)=an/(3an+1) 求数列通项公式
已知数列an满足:a1=1,an-a(n-1)=n n大于等于2 求an
已知数列{an}满足a1=33,a(n+1)-an=2n,则an/n的最小值
已知数列{an}满足a1=33,a(n+1)-an=2n,求an/n的最小值
已知数列an满足a1=100,a(n+1)-an=2n,则(an)/n的最小值为
已知数列an满足a1=2,an=a(n-1)+2n,(n≥2),求an
已知数列满足a1=1,an-a(n-1)=n-1,求其通项
已知数列{an}满足a1=1,a(n+1)=nan n+1是角标
根据下列条件,确定数列{An}的通项公式 1.,A1=1,An+1=(n+1)An,求An2已知数列{an}满足a(n+1)=an+n且a1=2,求an
已知数列{an}满足a1=2,a(n+1)-an=a(n+1)*an,则a31=?
已知数列{a}满足a1=1/2,a(n+1)=an+1/(n^2+n),求an已知数列{a}满足a1=1/2,a(n+1)=an+1/(n^2+n),求an
已知数列{an}满足a(n+1)=an+3n+2,且a1=2,求an=?
已知数列{an}满足a1=5,a(n+1)=an+6n+6,则an=
已知数列an满足a(n+1)=an+3n+2,且a1=2,求an