a(n+1)-1 = (an-1)/(an+1)
1/[a(n+1)-1] = (an+1)/(an-1)
= 1 + 2/(an-1)
1/[a(n+1)-1] +1 = 2( 1/(an -1) + 1 )
[1/[a(n+1)-1] +1]/( 1/(an -1) + 1 ) =2
( 1/(an -1) + 1 )/( 1/(a1 -1) + 1 ) =2^(n-1)
1/(an -1) + 1 = 2^n
an -1 =1/( -1+2^n)
an = 1+1/( -1+2^n)
已知数列{An}中,已知a1=2,an+1=2an∕an+1,求证,数列{1∕an-1}是等差...
( 1\/(an -1) + 1 )\/( 1\/(a1 -1) + 1 ) =2^(n-1)1\/(an -1) + 1 = 2^n an -1 =1\/( -1+2^n)an = 1+1\/( -1+2^n)
在数列{an}中,a1=1,an+1=(1+1\/n)an+(n+1)∕2n 设bn=an\/n,求证bn+1...
an+1=(n+1)\/n* an+(n+1)\/2^n,两边同除以(n+1)可得：a(n+1)\/(n+1)-an\/n=1\/2^n 因为bn=an\/n 所以b(n+1)-bn=1\/2^n 则有：b(n+1)-bn=1\/2^n bn-b(n-1)=1\/2^(n-1)………b2-b1=1\/2 等式两边累加可得：b(n+1)-b1=1\/2+...+1\/2^(n-1)+1\/2^n，...
在线等!!!在数列{an}中,a1=1,an+1=(1+1\/n)an+(n+1)∕2n
1\/2^(n-2)……a2\/2=a1\/1 1\/2^1上式相加，相同项消去an\/n=a1\/1 (1\/2^1 1\/2^2 ……1\/2^(n-1))=1 1\/2×(1-(1\/2)^(n-1))\/(1-1\/2)=2-1\/2^nbn=2-1\/2^n 2.an\/n=2-1\/2^nan=2n-n\/2^n{an}分为两部分，2n是等差数列，n\/2^n是等差数列与等比数列相乘第...
已知数列an满足a1﹦a,a(n﹢1)﹦[(4n﹢6)an﹢4n﹢10]∕(2n﹢1),判断{...
直接构造[a(n+1)+2]\/(2n+3)=[(4n+6)an+8n+12]\/(2n+1)(2n+3)=2(an+2)\/(2n+1)，因此新数列{(an+2)\/(2n+1)}是首项为a+2\/3，公比为2的等比数列
数列an满足an 1=1∕1-an,a8=2,则a4等于多少
an=2n-1,则{... 4 2015-06-06 在等比数列an中,a1=1\/8,q=2,则a4与a8的等比中... 1 2016-02-25 已知各项均为正的数列{an},满足(an+1)∧2=2(an... 1 2015-02-09 己知数列{an}满足an+1+(-1)nan=n,(n∈N*... 1 2013-07-01 数列{an}满足an+1+(-1)^n an=2n-1,则{...
已知数列{an}的前n项和为Sn,且Sn+(1∕2)an=1。求数列的通项公式。
即a[n]是公比为1\/3的等比数列 当n=1时，S1=a1 S1+1\/2a1=1，解得 a1=2\/3 所以 {an}=a[1]*(1\/3)^(n-1)=2\/3^n 这样， S[n]=1-1\/2*a[n]=1-1\/3^n b[n]=㏒3(1-S[n＋1])=㏒3(1\/3^(n＋1))=-(n+1)下面的方程，还缺一半，题目不完整 1\/(b[n]b[n+1]...
等差数列{an}中,a3=8,a7=20,若数列{1∕an*an+1}的前n项和为4∕25,则n...
a3=a1+2d=8a7=a1+6d=20解得a1=2d=31∕an*an+1=1\/d(1\/an-1\/an+1)=1\/3(1\/an-1\/an+1)(an+1-an=d)前n项和=1\/3(1\/a1-1\/a2)+1\/3(1\/a2-1\/a3)+……+1\/3（1\/an-1-1\/an)+1\/3(1\/an-1\/an+1)=1\/3(1\/a1-1\/an+1）=4\/25an+1=50a1+nd=50n=16 ...
数学数列高手进
2。 （1）因为an+1=4an-3n+1 所以a(n+1)-(n+1)=4(an-n)即 [a(n+1)-(n+1)]\/(an-n)=4 因此数列{an-n}是公比为4的等比数列。（2）由（1）知：an-n=4^(n-1)*(a1-1)=4^(n-1)an=4^(n-1)+n Sn=[1+4+4^2+4^3+...+4^(n-1)]+(1+2+3+...+n)=...
已知an=2-1╱a(n-1).求1╱an是等差数列
an-a(n-1)=2-1∕a(n-1)-2+1∕a(n-2) 即1∕an-1∕a(n-1)=1∕a(n-1)-1∕a(n-2) 为等差已证
设Sn是等差数列﹛An﹜的前n项,若A2n∕An﹦4n-1\/2n-1,则S2n\/Sn=?
设等等差数列{An}的首项为A1，公差为X，于是有A2n=A1+(2n-1)X,An=A1+(n-1)X，又有A2n\/An﹦4n-1\/2n-1，代入得2A1=X S2n=A1+A2+...+A2n=2nA1+2n^2X=(2n^2+n)X Sn =A1+A2+...+An =nA1+(1\/2)n^2X=(1\/2)(n^2+n)X 因此S2n\/Sn=4n+2\/n+1 --- 方法是这样...
