在数列an中a1=5分之3,an 1=2-an分之1

a(n+1)=2an/(an+1)∴1/a(n+1)=(an+1)/2an=1/2an+1/2∴1/a(n+1)-1=1/2an+1/2-1=1/2an-1/2=(1/2)(1/an-1),1/a1-

a(n+1)=-2an+3a(n+1)+k=-2an+3+k=-2(an-3/2-k/2)则令k=-3/2-k/2k=-1则两边同时加-1a(n+1)-1=-2(an-1)[a(n+1)-3]/(an

这是一道构造等比数列的题,而且要用到两次构造.具体方法如下：因为a(n+1)-3a(n)=3n所以a(n+1)=3a(n)+3n设a(n+1)+k=3(a(n)+k)则a(n+1)=3a(n)+2k所

由条件得a1=2,a2=5.且有：a2-a1=3*1,a3-a2=3*2,a4-a3=3*3,...an-a(n-1)=3*(n-1),累加得,an-a1=3*(1+2+3+...+n-1)=3n(n

a(n+1)=3a(n)+2n+1=3a(n)+3n-(n+1)+2=3a(n)+3n-(n+1)+3-1,a(n+1)+(n+1)+1=3[a(n)+n+1],{a(n)+n+1}是首项为a(1)+

a(n+2)-3a(n+1)+2an=0a(n+2)-a(n+1)=2a(n+1)-2ana(n+2)-a(n+1)=2[a(n+1)-an][a(n+2)-a(n+1)]/[a(n+1)-an]=2

a(n+1)=(2/3)an+1∴a(n+1)-3=(2/3)[(an)-3]∴bn=(an)-3是公比为2/3的等比数列.首项b1=a1-3=-2∴bn=-2×(2/3)^(n-1)∴an=3-2×

a(n+1)=2an+3n-4a(n)=2a(n-1)+3(n-1)-4上面两式相减得a(n+1)-3a(n)+2a(n-1)=3a(n)-3a(n-1)+2a(n-2)=3两式相减得a(n+1)-4

a(n)=3a(n-1)-2a(n)-1=3a(n-1)-3(a(n)-1)/(a(n-1)-1)=3所以an-1为等比数列an=3^(n-1)+1请选为最佳答案,谢谢!

移向得a(n+1)an+2a(n+1)=3an即[a(n+1)an+2a(n+1)]/3an=1化解得a(n+1)/3+2/3=1得a(n+1)=1又a1=1综上an=1你们老师什么恶趣味额.不懂再问

/>由题意可得：A(n+1)-An=2n则有：An=（An-A(n-1))+(A(n-1)-A（n-2))+(A(n-2)-A(n-3))+……+(A3-A2)+(A2-A1)+A1=2(n-1)+2

a2=(1/2+3)/3*1/2=3/7a3=(7/3+3)/3*7/3=16/21a4=（16/21+3)/3*16/21=79/48

根据an+1＝2an2+an，得2an+1+an+1an=2an，两边同时除以an+1an，得到2an+1−2an＝1，所以数列{2an}是公差为1的等差数列，且2a1＝2，所以2an＝n+1，所以a

an+1=4an-3n+1an+1-(n+1)=4[an-n][an+1-(n+1)]/[an-n]=4等比a1-1=3an-n=3*4^(n-1)an=3*4^(n-1)+n2\sn=[3*4^(n

a(n+1)-(n+1)=4an-3n+1-(n+1)=4an-4n得证

1an+1=an+6nan=an-1+6(n-1)..a2=a1+6*2a1=6-5Sn=Sn-1+6*(1+2+..+n)-5an=6*(1+n)n/2-5=3n(n+1)-52an*an+1=3^

两边加n+1,得a"+n+1=3(a'+n)+1;令bn=an+n,得b"=3b'+1,得(b"+1/2)=3(b'+1/2)数列{bn+1/2}是等比数列,得bn=5/2*3^(n-1)-1/2,故

an+a(n+1)=6/5^(n+1)=(5+1)/5^(n+1)=1/5^n+1/5^(n+1)a(n+1)-1/5^(n+1)=-(an-1/5^n)a1-1/5^1=1/5-1/5=0an-1/