已知数列an是等比数列满足a1=3a4=24数列bn满足b1=4

a(1)=1,a(n+1)-a(n)=(1/3)^n,3^na(n+1)=3*3^(n-1)a(n)+1,3^na(n+1)+1/2=3[3^(n-1)a(n)+1/2]{3^(n-1)a(n)+1/

An=A1+(A2-A1)++(An-An-1)=1(1-(1/3)^n)/(1-1/3)=2/3-2(1/3)^(n+1)

1、证：a(n+1)=3an+2a(n+1)+1=3an+3[a(n+1)+1]/(an+1)=3,为定值.a1+1=1+1=2数列{an+1}是以2为首项,3为公比的等比数列.2.an+1=2×3^

(1)∵a（n+1）=2an+1∴a[n+1]+1=2a[n]+2=2(a[n]+1)∴a[n]+1为等比数列,等比=2(2)a[n]+1=（a[1]+1）*2^(n-1)=2^n∴a[n]=-1+2

a(n+1)=2an-n+1a(n+1)=2an-2n+(n+1)a(n+1)-(n+1)=2(an-n)∴{an-n}是公比为2,首项为2-1=1的等比数列an-n=1×2^(n-1)=2^(n-1

(1)∵a1,a2-a1,a3-a2,…an-an-1,…是首相为1,公比为三分之一的等比数列∴an－an-1＝(1/3)^(n－1)an-1－an-2＝(1/3)^(n－2)……a2－a1＝1/3∴

a1*p=a2a1*p^3=a4,a1*p-a1=a1*p^3-a1*Pp-1=p^(p^2-1);(p-1)(p*(p+1)-1)=0,p=1,或p^2+p-1=0,p=(-1+√5)/2,p=(-

a1,a2,a4成等差数列2a2=a1+a4即2a1*q=a1+a1q^3a1不为0所以：2q=1+q^3q^3-2q+1=0q^3-q^2+q^2-2q+1=0q^2*(q-1)+(q-1)^2=0

a1,a2,a4成等差数列所以2a2=a1+a4{an}是等比数列a2=a1qa4=a1q^3所以2×a1q=a1+a1q^3即：q^3-2q+1=0(q-1)(q^2+q-1)=0q=1或q=(-1

an=2^n步骤：等比数列{an},=>an=a1*q^(n-1),(a1、q不为0)=>a2=a1q,a3=a1q^2,a4=a1q^3,2a1+a3=3a2=>2a1+a1q^2=3a1q,=>q

设公比为q,数列是递增数列,q>1数列是等比数列,a1a5=a2a4=729,又a1+a5=246,a1、a5是方程x²-246x+729=0的两根.(x-3)(x-243)=0x=3或x=

（1）应该是数列{an+1}证：a(n+1)=2an+1a(n+1)+1=2an+2[a(n+1)+1]/(an+1)=2,为定值.a1+1=1+1=2数列{an+1}是以2为首项,2为公比的等比数列

电脑打字太麻烦思路应该是对的~

由题意可知,an+2-an+1=2(an+1-2an)且a2-a1=2,所以是公比为2,首项为2的等比数列.求出an+1-an的通向为an+1=2^n+an求和2^n,Sn=2^n-2所以,an=a1

a(n+1)+1=2an+2=2(an+1)[a(n+1)+1]/(an+1)=2所以an+1是等比数列[a(n+1)+1]/(an+1)=2则q=2所以an+1=(a1+1)*2^(n-1)=2^n

设[an+1+p(n+1)+q]/[an+pn+q]=m得an+1+p（n+1）+q=man+mpn+mq．又an+1=2an+n+1,则2an+n+1+pn+p+q=man+mpn+mq,即（2-m

a1²=S-a1S=a1/(1-q)其中|q|

把通项表示为a[n],比较清楚a[n+1]=a[n]+6a[n-1]两边同时加上2a[n]变成：a[n+1]+2a[n]=3a[n]+6a[n-1]=3(a[n]+2a[n-1])两边同时减去3a[n

1=a1a2=r,故bn=r*q^(n-1)又b(n+1)/bn=a(n+1)*a(n+2)/(an*a(n+1))=a(n+2)/an、b(n+1)/bn=q可得当n为奇数时an=a1*q^((n+

由a3=8,s3=14易求出a1=2,a2=4,所以可知公比为2即得数列an=2^n（b1/a1）+（b2/a2）+…+（bn/an）=n/（2^n）,（1）则（b1/a1）+（b2/a2）+…+[b