设数列{an}满足a1=1, an=(4an-1 +2)/(2an-1 +7) ,则通项xn=?

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/03 16:28:08

设数列{an}满足a1=1, an=(4an-1 +2)/(2an-1 +7) ,则通项xn=?
设数列{an}满足a1=1, an=(4an-1 +2)/(2an-1 +7) ,则通项xn=?

设数列{an}满足a1=1, an=(4an-1 +2)/(2an-1 +7) ,则通项xn=?
an=[4a(n-1)+2]/[2a(n-1)+7]
an+2=[4a(n-1)+2+4a(n-1)+14]/[2a(n-1)+7]
=[8a(n-1)+16]/[2a(n-1)+7]
=8[a(n-1)+2]/[2a(n-1)+7]
an -1/2=[4a(n-1)+2-a(n-1)-7/2]/[2a(n-1)+7]
=[3a(n-1)-3/2]/[2a(n-1)+7]
=3[a(n-1)-1/2]/[2a(n-1)+7]
[(an +2)/(an -1/2)]/{[a(n-1)+2]/[a(n-1)-1/2]}=8/3,为定值.
(a1+2)/(a1-1/2)=(1+2)/(1-1/2)=6
数列{(an +2)/(an -1/2)}是以6为首项,8/3为公比的等比数列.
(an +2)/(an -1/2)=6×(8/3)^(n-1)
[6×(8/3)^(n-1) -1]an=3×(8/3)^(n-1)+2
an=[3×(8/3)^(n-1) +2]/[6×(8/3)^(n-1) -1]
=[3×(8/3)^(n-1) -1/2 +5/2]/[6×(8/3)^(n-1) -1]
=(1/2) +5/[12×(8/3)^(n-1) -2]
=(1/2) +5/[2^(3n-1)/3^(n-2) -2]
n=1时,a1=(1/2) +5/[2²/3^(-1) -2]=(1/2)+5/10=1,同样满足.
数列{an的通项公式为an=(1/2) +5/[2^(3n-1)/3^(n-2) -2].

设数列{an}满足a1=1, an=(4an-1 +2)/(2an-1 +7) ,则通项xn=? 设数列{an}满足lg(1+a1+a2+...+an)=n+1,求通项公式an 数列an满足a1=2,an+1=4an+9,则an=? 设数列AN满足A1等于1,3(A1+a2+~+AN)=(n+2)an,求通向公式 关于数列、等差数列的题目设数列an满足an+1=an-2且a1=241)判断an是什么数列2)若an 数列{An}满足a1=1/2,a1+a2+..+an=n方an,求an 已知数列{an}满足a1=1 an+1=an/(3an+1) 则球an 已知数列{an},{bn}满足a1=2,2an=1+2an*an+1,设{bn}=an-1求数列{1n}为等差数列急!!! 设数列{An}满足 A1=6,A2=4 A3=3,且数列{An+1-An}(n属于自然数)是等差数列,球An的通向公式 rt设数列{An}满足 A1=6,A2=4 A3=3,且数列{An+1-An}(n属于自然数)是等比数列,球An的通向公式rt 设函数f(x)=根号下x2+4,若一个正数数列{an}满足:a1=1,an+1=f(an),求数列的通项公 数列{an}满足Sn+Sn+1=5/3an+1,a1=4求an 已知数列{an}满足3an+1+an=4,a1=9,求通项公式. 设数列{an},{bn}满足;a1=4 a2=5/2,an+1=an+bn/2,bn+1=2anbn/an+bn 用数列an表示an+1;并证明;任意n属于设数列{an},{bn}满足;a1=4 a2=5/2,an+1=an+bn/2,bn+1=2anbn/an+bn (1)用数列an表示an+1;并证明;任意n属于N*都 已知数列{an}满足a1=-1,an=[(3n+3)an+4n+6]/n,bn=3^(n-1)/an+2.求数列an的通向公式.设数列bn是的前n项和已知数列{an}满足a1=-1,an=[(3n+3)an+4n+6]/n,bn=3^(n-1)/an+2.(1)求数列an的通向公式.(2)设数列bn是的前n项和为sn, 数列{an}满足a1=3/2,an+1=an2-an+1,求证:1/an=1/(an)-1 - 1/(an+1)-1数列{an}满足a1=3/2,an+1=an2-an+1,求证:1/an=1/(an)-1 - 1/(an+1)-1设Sn=1/a1+1/a2+...+1/an,n>2证明1 设数列an满足a1=2,a(n+1)=3an+2^(n-1),求an2,设数列an满足a1=2,a(n+1)=3an+2n,求an 设{an}是a1=4的单调递增数列,且满足an+1^2+an^2+16=8(an+1+an)+2an+1an,求ann+1均为a的下标 设数列【an】满足a1=1,3(a1+a2+a3+······+an)=(n+2)an,求通项an