设数列an的前n项和为Sn,Sn-tSn-1=n,a1=1(1)t=2,求a2,a3(2){an+1}是等比数列,求t的值(3)求sn
来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/23 23:44:31
设数列an的前n项和为Sn,Sn-tSn-1=n,a1=1(1)t=2,求a2,a3(2){an+1}是等比数列,求t的值(3)求sn设数列an的前n项和为Sn,Sn-tSn-1=n,a1=1(1)t
设数列an的前n项和为Sn,Sn-tSn-1=n,a1=1(1)t=2,求a2,a3(2){an+1}是等比数列,求t的值(3)求sn
设数列an的前n项和为Sn,Sn-tSn-1=n,a1=1
(1)t=2,求a2,a3
(2){an+1}是等比数列,求t的值
(3)求sn
设数列an的前n项和为Sn,Sn-tSn-1=n,a1=1(1)t=2,求a2,a3(2){an+1}是等比数列,求t的值(3)求sn
(1)A1=1,S1=1.根据递推式,S2=4,S3=11,所以A2=3,A3=7
(2)S2=t+2,S3=t^2+2t+3,所以A2=t+1,A3=t^2+t+1,根据题意,2,t+2,t^2+t+2成等比数列,解得t=0或2
(3)对于Sn-t(Sn-1)=n,两边同除t^n,得Sn/(t^n)-S(n-1)/t^(n-1)=n/(t^n)
可用累加法得到Sn/(t^n)=S1/t+2/(t^2)+3/(t^3)+...+n/(t^n)
右边等式是典型的错位相减法,
t=1时,Sn=n(n+1)/2
t=0时,Sn=n
其他情况下,求得Sn/(t^n)后即可求得Sn=[t^(n+1)-t]/(t-1)-n
设数列an的前n项和为Sn,Sn-tSn-1=n,a1=1(1)t=2,求a2,a3(2){an+1}是等比数列,求t的值(3)求sn
设数列{an}满足a1=t,a2=t^2,前n项和为Sn,且Sn+2-(t+1)Sn+1+tSn=0(n属于N*)(1)证明:数列{an}为等比数列,并求出数列{an}的通项公式(2)当1/2
设数列{an}满足a1=t,a2=t^2,前n项和为Sn,且Sn+2-(t+1)Sn+1+tSn=0(n属于N*)(1)证明:数列{an}为等比数列,并求出数列{an}的通项公式(2)当1/2
合情推理设正数列{An}前n项和为Sn 且存在正数t 使得对所有自然数n 有√(tSn)=(t+An)/2 则通过归纳猜想可得到Sn=
1.设数列{an}的前n项和为Sn,其中an≠0,a1为常数,且-a1,Sn,an+1(n+1是a的下标啊,我打不出来)成等差数列,求{an}的通项公式2.数列{an}中,a1=1,Sn为其前n项的和,当t>0时,有3tSn-(2t+3)Sn-1=3t(n-1是S的下
设数列{an}满足a1=t,a2=t的平方,前n项和为Sn,且Sn+2-(t+1)Sn+1+tSn=0设数列{an}满足a1=t,a2=t^2(其中t为非零常数),前n项和为Sn且Sn+2-(t+1)Sn+1+tSn=0(n属于N*)求证1/b1+1/b2+…+1/bn若1/2<t<2,bn=2an/(
数列{an},中,a1=1/3,设Sn为数列{an}的前n项和,Sn=n(2n-1)an 求Sn
设数列an的首项a1=1,前n项和Sn=满足关系式tSn-(t+1)S(n-1)=t (t大于0,n属于N* n大于等于2) 求证:数列an是等比数列
设数列an的前n项和为Sn,若Sn=1-2an/3,则an=
设数列{an}的前N项和为Sn,已知1/Sn+1/S2+1/S3+.+1/Sn=n/(n+1),求Sn
求:设数列 {an}的前n项和为Sn,数列{Sn}的前n项和为Tn,满足Tn=2Sn一n²,n∈求:设数列 {an}的前n项和为Sn,数列{Sn}的前n项和为Tn,满足Tn=2Sn一n²,n∈N
设数列an的首项a1等于1,前n项和为sn,sn+1=2n设数列an的首项a1等于1,前n项和为sn,sn+1=2n
设数列{an}的前n项和为Sn,满足Sn-tSn-1=n(n大于等于2,n属于N),t为常数,且a1=1.(1)当t=2时,求a2和a3;(2)若{an +1}是等比数列,求t的值;
设数列(an)de 首项a1=1,前n项和Sn满足3tSn-(2t+3).Sn-1=3t(t>0,n=1,2,3,4,...)证明数列an是等比数列
设数列{an}的前n项和为Sn,a1=10,a(n+1)=9Sn+10
设数列{an}的前n项和为Sn,已知ban-2n=(b-1)Sn
设等比数列an的公比为q前n项和为Sn若Sn+1,Sn,Sn+2成等数列,求q
设等比数列{an}的公比为q,前项和为sn,求数列{sn}的前n项和un