a1=1,an=an/1an

数列an中,a1=3,an+1=an/2an+1,则an=?a(n+1)=an/(2an+1)1/a(n+1)=(2an+1)/an=1/an+21/a(n+1)-1/an=2,为定值.1/a1=1/3,数列{1/an}是以1/3为首项,2

已知数列{an}满足a1=1an+1=an/(3an+1)则球an求通项公式倒数变形经常可以尝试一下:

已知等差数列{an}中,a1=2.an+1=an+3分之an求anan=3n-1由an+1=an+3得知公差d=3所以an=a1+（n-1）d=3n-1

数列{An}满足a1=1/2,a1+a2+..+an=n方an,求ana1+a2+..+an=Sn=n²*an所以n>=2时S(n-1)=(n-1)²*a(n-1)相减an=Sn-S(n-1)=n²*an-(n

数列{an}中,2an*an+1+2an+1-an=0,a1=1,an=?2anXa(n+1)+2a(n+1)-an=02+2/an-1/a(n+1)=0(同除ana(n+1))即1/(a(n+1)-2/an=2令数列Cn=1/an所以C(

已知数列{an}a1=3an+1=(3an+2)/(an+2)bn=(an-2)/(an+1)求证bn是等比数列b(n+1)=[a(n+1)-2]/[a(n+1)+1]=[(3an+2)/(an+2)-2]/[(3an+2)/(an+2)+

已知数列{an}中满足(An+1-An)(An+1+An)=16,且a1=1,an1.[a(n+1)-an][a(n+1)+an]=16a(n+1)²-an²=16,为定值.a1=1a1²=1数列{an

数列｛an｝中,a1=1/23an*an-1+an-an-1=0,通项公式an3ana(n-1)+an-a(n-1)=0同除以a(n-1)an3+1/a(n-1)-1/an=01/an-1/a(n-1)=3bn=1/an是以1/a1=2为首

已知a1=2,an不等于0,且an+1-an=2an+1an,求ana[n+1]-a[n]=2a[n+1]a[n]1/a[n]-1/a[n+1]=21/a[n+1]=(1/a[n])-21/a[n]为等差数列,公差为-2,首项1/a[1]=

A1=1,An+1=3An/(3An+6)证(1/An)+1等比a(n+1)=3an/(3an+6)1/a(n+1)=(3an+6)/(3an)=2/an+11/a(n+1)+1=2/an+2=2(1/an+1)[1/a(n+1)+1]/(

在数列an中,a1=1,an=3an-1+2则an=an=3an-2+2=>an+1=3(an-1+1)设bn=an+1则bn=3bn-1b1=a1+1=1+1=2所以bn=b1q∧(n-1)=2*3∧(n-1)所以an=bn-1=2*3∧

a1=1,an+1=an/(3+an),求an=?（n为下标）an+1=an/(3+an),取倒数1/a(n+1)=(3+an)/an=1+3/an1/a(n+1)+1/2=3(1/an+1/2)所以数列{1/an+1/2}是以3为公比的等

已知数列{an}中,a1=1/2,an+1+3an=0,an=()a(n+1)+3an=0a(n+1)=-3ana(n+1)/an=-3又a1=1/2所以数列{an}是以首项为1/2,公比为-3的等比数列所以an=1/2×（-3）^(n-1

数列{An}中,A1=1,An=0.5An-1-0.5,则An=_________.A(n)+1=0.5A(n-1)+0.5=0.5[A(n-1)+1],A(n)是首项为2,公比为1/2的等比数列,所以A(n)+1=2*(1/2)^(n-1

设数列{an}中,a1=2,an+1=an+n+1,则通项an=?an1里的n1是下标吗

数列{an}满足a1=1an+1=2n+1an/an+2n(1)a(n+1)/2^(n+1)=an/(an+2^n)2^(n+1)/a(n+1)=(an+2^n)/an=1+2^n/an2^(n+1)/a(n+1)-2^n/an=1所以{2

在数列{an}中,a1=1,an+1=an/1+nan,求anAn+1=An/(1+nAn)An+1(1+nAn)=AnAn+1+nAn+1*An=An两边除以An+1An1/An+n=1/An+11/An+1-1/An=n1/A2-1/A

已知｛an｝,a1=1,an+1=2/3an+1,求anA(n+1)+x=2/3An+1+x=(2/3)(An+3/2+3x/2)令x=3/2+3x/2x=-3所以A(n+1)-3=2/3An+1+x=(2/3)(An-3)[A(n+1)-

an+1,n+1是下表,an+1=an^3a1=2,求anan=2^3*（n-1）（n>=1）

数列an中,a1=3,an+1=3an,则an=,sn=a(n+1)=3ana(n+1)/an=3所以{an}是以a1=3,q=3为公比的等比数列an=3*3^(n-1)=3^nSn=a1+a2+...+an=3+3^2+...+3^n=3