已知数列{an}满足前N项和sn=n平方+1数列{bn}满足bn=2/an +1且前n项和为Tn 设T 2n+1 -Tn
来源:学生作业帮 编辑:搜狗做题网作业帮 分类:数学作业 时间:2024/07/20 07:54:33
已知数列{an}满足前N项和sn=n平方+1数列{bn}满足bn=2/an +1且前n项和为Tn 设T 2n+1 -Tn
1/ 求{bn}的通向公式
2/ 判断{cn}得单调性
3/ n大于等于2时T 2n+1 -Tn小于1/5-7/12loga (a-1)恒成立求a的范围
cn=T 2n+1 -Tn
1/ 求{bn}的通向公式
2/ 判断{cn}得单调性
3/ n大于等于2时T 2n+1 -Tn小于1/5-7/12loga (a-1)恒成立求a的范围
cn=T 2n+1 -Tn
![已知数列{an}满足前N项和sn=n平方+1数列{bn}满足bn=2/an +1且前n项和为Tn 设T 2n+1 -Tn](/uploads/image/z/17760458-2-8.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3%E5%89%8DN%E9%A1%B9%E5%92%8Csn%3Dn%E5%B9%B3%E6%96%B9%2B1%E6%95%B0%E5%88%97%7Bbn%7D%E6%BB%A1%E8%B6%B3bn%3D2%2Fan+%2B1%E4%B8%94%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BATn+%E8%AE%BET+2n%2B1+-Tn)
(1)
∵数列{an}满足前N项和sn=n平方+1
∴Sn=n^2+1
S(n-1)=(n-1)^2+1
An=Sn-S(n-1)
=n^2+1-[(n-1)^2+1]
=2n-1
A1=S1=2
Bn=2/An +1=2/(2n-1)+1=(2n+1)/(2n-1)
B1=2/A1+1=2
Bn是一个首项为2,通项为(2n+1)/(2n-1) 的数列
(2)
Cn=T(2n+1)-Tn
要判断Cn的单调性只要判断Cn-C(n-1)是大于0还是小于0即可
Cn-C(n-1)=T(2n+1)-Tn-[T(2n-1)-T(n-1)]
=[T(2n+1)-T(2n-1)]-[Tn-T(n-1)]
=B(2n+1)+B(2n)-Bn
=[2(2n+1)+1]/[2(2n+1)-1]+[2(2n)+1]/[2(2n)-1]-[(2n+1)/(2n-1)]
=1+2[1/(4n+1)+1/(4n-1)-1/(2n-1)]
∵1/(4n+1)+1/(4n-1)-1/(2n-1)
= (1-8n)/[(4n+1)*(4n-1)*(2n-1)]
又∵1-8n0,4n-1>0,2n-1>0
∴(1-8n)/[(4n+1)*(4n-1)*(2n-1)]
∵数列{an}满足前N项和sn=n平方+1
∴Sn=n^2+1
S(n-1)=(n-1)^2+1
An=Sn-S(n-1)
=n^2+1-[(n-1)^2+1]
=2n-1
A1=S1=2
Bn=2/An +1=2/(2n-1)+1=(2n+1)/(2n-1)
B1=2/A1+1=2
Bn是一个首项为2,通项为(2n+1)/(2n-1) 的数列
(2)
Cn=T(2n+1)-Tn
要判断Cn的单调性只要判断Cn-C(n-1)是大于0还是小于0即可
Cn-C(n-1)=T(2n+1)-Tn-[T(2n-1)-T(n-1)]
=[T(2n+1)-T(2n-1)]-[Tn-T(n-1)]
=B(2n+1)+B(2n)-Bn
=[2(2n+1)+1]/[2(2n+1)-1]+[2(2n)+1]/[2(2n)-1]-[(2n+1)/(2n-1)]
=1+2[1/(4n+1)+1/(4n-1)-1/(2n-1)]
∵1/(4n+1)+1/(4n-1)-1/(2n-1)
= (1-8n)/[(4n+1)*(4n-1)*(2n-1)]
又∵1-8n0,4n-1>0,2n-1>0
∴(1-8n)/[(4n+1)*(4n-1)*(2n-1)]
已知数列an满足前n项和Sn=n平方+1.数列bn满足bn=2\an+1,且前n项和为Tn,设Cn=T的2n+1个数—T
已知数列{an}的前n项和为Sn=n^2+1,数列{bn}满足:bn=2/(an+1),且前n项和为Tn,设Cn=T(2
已知数列{an}的前n项和为Tn,且满足Tn=1-an,数列{bn}的前n项和Sn,Sn=1-bn,设Cn=1/Tn,证
已知数列{an}的前n项和为Sn=n^2+1,数列{bn}满足bn=2/(an)+1,前n项和为Tn,设Cn=T(2n+
已知数列{an}的前n项和为Sn=n^2+1,数列{bn}满足:bn=2/(an+1),且前n项和为Tn,设Cn
已知数列an的前n项和为sn=2n^2+5n+1,数列bn的前n项和tn满足Tn=(3/2)bn-3/2 求数列an的通
已知数列an的通项公式为an=2n-1,数列bn的前n项和为tn且满足tn=1- b
数列an,满足Sn=n^2+2n+1,设bn=an*2^n,求bn的前n项和Tn
数列题.已知数列{An}的前n项和为Sn,且Sn=n^2 +n,数列{bn}满足bn=1/AnA(n+1) ,Tn是数列
已知数列bn满足bn=b^2n,其前n项和为Tn,求(1-bn)/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
数列an的前n项和为Sn,Sn=2an-1,数列bn满足b1=2,bn+1=an+bn.求数列bn的前n项和Tn