an为等比,a1 ...an=2^n-1

n≥2时,Sn=4a(n－1)＋2,与S(n＋1)=4an＋2相减,得：a(n＋1)=4an－4a(n－1),即：a(n＋1)－2an=2[an－a(n－1)],则：bn=2b(n－1),其中n≥2.

a(n)>0.s(n)=[a(n)+1]/4,a(1)=s(1)=[a(1)+1]/4,a(1)=1/3.s(n+1)=[a(n+1)+1]/4,a(n+1)=s(n+1)-s(n)=[a(n+1)+

a3=a1+2d=a1+4a4=a1+3d=a1+6因为a1,a3,a4成等比数列,则a4/a3=a3/a1(a1+4)^2=a1(a1+6)解之,a1=-8则a2=a1+d=-8+2=-6

(n+1)=[a(n+1)-2]/[a(n+1)+1]=[(3an+2)/(an+2)-2]/[(3an+2)/(an+2)+1]=an-2/4an+4bn=an-2/an+1故bn+1/bn=1/4

(1)a(n+1)=2an+2^(n+1)等式两边同除以2^(n+1)a(n+1)/2^(n+1)=an/2ⁿ+1a(n+1)/2^(n+1)-an/2ⁿ=1,为定值a1/2=

楼上都解对了.在百度文库中搜“数列求算技巧“,我自己总结的,看了你就会这一类的题了!

a2+(a1)/2=2.5A(n+2)=(An+A(n+1))/2a(n+2)+[a(n+1)]/2=a(n+1)+(an)/2所以数列{a(n+1)+(an)/2}是首项为2.5,公比为1的等比数列

(1)an=2a+3,∴an+3=2[a+3],∴数列{an+3}是等比数列.（2）an+3=(a1+3)*2^(n-1),an=(a1+3)*2^(n-1)-3=（6)*2^(n-1)-3.再问：2

(an*an+1)/(an-1*an)=3=>an+1/an-1=3=>a2n=3^n,a2n-1=2*3^(n-1)=>bn=5*3^(n-1)

易得ana(n+1)=a1a2q^(n-1)=2q^(n-1)故2q^(n-1)+2q^n>2q^(n+1)即1+q>q^2解得(1-√5)/2再问：q>0时，求an的前2n项和sn再答：ana(n+

a(n+1)=2an/1+an,1/a(n+1)=1/2an+1/2,1/a(n+1)-1=1/2*(1/an-1),[a(n+1)-1]/a(n+1)=1/2*(an-1)/an所以(an-1)/a

Dn=(C1×C2×C3×……×Cn)^(1/n)成等比数列Bn=Sn/n=(nA1+(1/2)n(n-1)d)/n=A1+(n-1)(d/2)Bn是以A1为首项,d/2为公差的等差数列.类比Dn=(

（Ⅰ）设等差数列{an}的公差为d（d≠0），由题意得a22＝a1a4，即(a1+d)2＝a1(a1+3d)，∴（2+d）2=2（2+3d），解得 d=2，或d=0（舍），∴an=a1+（n

a(n+1)=an^2+6an+6=(an+3)^2-3,即a(n+1)+3=(an+3)^2,从而log5[a(n+1)+3]=2log5(an+3)而cn=log5(an+3),则结合上式即得c(

(n+1)=a(n+1)+1=[2an+1]+1=2an+2=2(an+1)=2bn,所以{bn}是公比为2的等比数列.b1=a1+1=2,所以bn=b1*q^(n-1)=2*2^(n-1)=2^n.

证明：由题设a(n+1)=3an/(1+2an)变形得1/a(n+1)=(1+2an)/(3an)1/a(n+1)=(1/3)(1/an)+(2/3)[1/a(n+1)]-1=(1/3)[(1/an)

首项a1=2,公差d=2ak=a1+(k-1)d=2kS(k+2)=(k+2)(a1+a(k+2))/2=(k+2)(a1+a1+(k+2-1)d)/2=(k+2)(a1+k+1)=(k+2)(k+3

因为{an}为等比数列所以an=a1*q^(n-1)a1*a5=a1*a1*q^4=16a1^2*q^4=16a1*q^2=±4所以a1=4/q^2①或a1=-4/q^2②a2+a4=a1*q+a1*

an=n

Sn为9?再问：嗯是S3再问：写错再问：S3等于9再答：S3=a1+a2+a3=3a1+3d=9因为a1.a3.a7成等比数列所以a1a7=(a3)^2a1^2+6a1d=a1^2+4a1d+4d^2