已知数列,且a1=3,a3=9 求数列an 的通向公式

a1a1da12d=2π,3a13d=2π,a1d=2π/3.==>a2=2π/3.a1a1da12d=a1d2a2=3a2=2π.a3a5=a12da14d=2a16d.=2(a13d)=2a4.…

a1+a2+a3=12a1+a1+d+a1+2d=126+3d=12d=2an=a1+d(n-1)=2+2n-2=2nsn=b1+b2+b3+b4+b5+.+bn=3^2+3^4+3^6+.3^2n=

（2）an+1-an=2^(n+1)-2^n=2^n,1/（an+1-an）=1/2^n；则1/（a2-a1）+1/（a3-a2）+.+1/（an+1-an）=1/2+1/4+.+1/2^n=1/2[

设等差数列{log2（an-1）}的公差为d．由a1=3，a3=9得2（log22+d）=log22+log28，即d=1．所以log2（an-1）=1+（n-1）×1=n，即an=2n+1．

假设公比为q,则a2=a1*q,a3=a1*q^2,a4=a3+(a3-a2)=a1(2q^2-q)所以a1+a1(2q^2-q)=16,a1*q+a1*q^2=12解得a1=1,q=3,或者a1=1

0,4,8,16

a(n+1)-an=2^(n+1)+1-2^n-1=2^n,所以1/a2-a1+1/a3-a2.+.+1/a（n+1）-an=1/2+1/4+...+1/2^n=1/2*(1-1/2^n)/(1-1/

a1=2a1+a2+a3=12a2=4d=2an=2nbn=3^an=3^2n=9^n数列bn是以9为首项,公比=9的等比数列Sn=9(1-9^n)/(1-9)=(9^[n+1]-9)/8

a1=2,a1+a2+a3=12a2=4d=2an=2n2.Sn=2*3+4*3^2+6*3^3+……+2n*3^n3Sn=2*3^2+4*3^3+……+(2n-2)*3^n+2n*3^[n+1]相减

{an}是等差数列,且a1=2,a1+a2+a3=12而2a2=a1+a3所以a2=4所以公差d=a2-a1=2所以an=a1+(n-1)d=2nbn=(1/2)^n*2n和Tn=b1+b2+……+b

设公差值为ca1+a2+a3=a1+（a1+c）+(a1+c+c)=3a1+3c=12c=2an=a1+c(n-1)=2nbn=3^(2n)b(n+1)/bn=3^(2n+2)/3^2n=9所以bn是

An=6Sn/(An+3)6Sn=(An)^2+3Ann>=26S(n-1)=(A(n-1))^2+3A(n-1)6An=(An)^2+3An-(A(n-1))^2-3A(n-1)(An)^2-(A(

(1)log2（a1-1）-log2(a3-1)=-2dlog2(8)-log2(2)=2dd=1log2(an-1)=nan=2^n+1（n属于N*）(2)1/（an-a（n-1）)=1/(2^(n

已知数列{bn}={log2(an-1)}为等差数列,且a1=3a3=9→b1=log2(3-1)=log2(2)=1,b2=log2(9-1)=log2(8)=3,公差d=3-1=2,∴bn=1+(

log2(an+1-an/3)-log2(an-an-1/3)=1log2[(a(n+1)-an/3)/(an-a(n-1)/3)]=log2(2)(a(n+1)-an/3)/(an-a(n-1)/3

因为数列{2^an}(n属于正自然数)为等比数列所以q^2=（2^a3）/（2^a1）=2^6所以q=8所以2^an=（2^a1）q^（n-1）=2^（3n）所以an=3n希望楼主采纳我的解法,因为我

1.设an=3+(n-1)d15=a1+a2+a3=3+3+d+3+2d=9+3dd=2an=3+2(n-1)=2n+1;2.1/[ana(n+1)]=1/[(2n+1)(2n+3)]=(1/2)[1

(!)由题意可知log2(a1-1)+2d=log(a3-1)所以log2(2)+2d=log2(8)1+2d=3d=1故an=a1+(n-1)d=log2(2)+(n-1)*1=1+n-1=n(2)

（I）设等差数列{log2（an-1）}的公差为d．由a1=3，a3=9得2（log22+d）=log22+log28，即d=1．所以log2（an-1）=1+（n-1）×1=n，即an=2n+1．（