等差数列{an}的各项为正数,a1=1前n项和为sn
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因为lga1,lga2,lga4成等差数列lga1+lga4=2lga2,lga1*a4=lg(a2)^2所以a1*a4=(a2)^2a1(a1+3d)=(a1+d)^2得a1=dan=ndBn=1/
因为a3+b5=21,a5+b3=13,{an}是等差数列,{bn}是等比数列所以a1+2d+b1*q^4=21,a1+4d+b1*q^2=13因为a1=b1=1所以2d+q^4=20,4d+q^2=
(1)根据题意,设公差为d则a3=a1+2d=2d+1a9=a1+8d=8d+1有(2d+1)^2=8d+1d=1故通项:an=n(2)根据题意,设公比为q则b2=qb3=q^2有q-0.5q^2=0
a(n+1)=√[bn*b(n+1)]2bn=an+an+12bn=√[bn*b(n-1)]+√[bn*b(n+1)]2√bn=√b(n-1)+√b(n+1)所以数列{√bn}为等差数列√b1=√2(
公差d,公比q代入得:d=q=2an=2n-1;bn=2^(n-1)an/bn=(2n-1)/2^(n-1)Sn=1+3/2+5/2^2+……+(2n-1)/2^(n-1)Sn/2=1/2+3/2^2
首先你的条件a5+a3=13恐怕打错了吧,应该为a5+b3=13吧,这才好算点设an=1+(n-1)d,bn=q^(n-1)根据已知:1+2d+q^4=211+4d+q^2=13解得d=2,q=2或-
设等差数列的公差为d等比数列的公比为q由题意得1+2d+q^4=21(1)1+4d+q^2=13(2)(1)*2-(2)得2q^4-q^2-28=0解得q^2=4又由题意,知{bn}各项为正,所以q=
1.n=1时,2a1=2S1=a1²+1-4a1²-2a1-3=0(a1+1)(a1-3)=0a1=-1(数列各项均为正,舍去)或a1=3n≥2时,2an=2Sn-2S(n-1)=
可用递推法:2Sn=An+An*An递推2Sn-1=An-1+An-1*An-1两市相减,得:An+An-1=An*An-An-1*An-1因为An为正数,所以An-An-1=1之后求An,然后用求和
sn=(1/8)(an+2)²S(n-1)=(1/8)[a(n-1)+2]²an=Sn-S(n-1)=(1/8){(an+2)²-[a(n-1)+2]²}=(1
a1+a10=a2+a9=.a3+a8=10>=2更号下A3*A8A3*A8=25
4Sn=2an+an^24S(n-1)=2a(n-1)+a(n-1)^2相减得4an=2an-2an(n-1)+[an+a(n-1)][an-a(n-1)]2[an+a(n-1)]=[an+a(n-1
∵lga1,lga2,lga3成等差数列∴lga1+lga3=2lga2=lg(a2)^2即lg(a1*a3)=lg(a2)^2∴a1*a3=(a2)^2而a2=a1+d,a3=a1+2d∴a1(a1
首先需要计算出a(n)的通项loga1、loga2、loga3成等差数列,所以a(2)*a(2)=a(1)*a(3)(a(1)+d)^2=a(1)*(a(1)+2d)得:d=0所以a(n)=a(1)常
根号Sn的通项公式是nSn=n^2an=Sn-Sn-1=n^2-(n-1)^2=2n-1
log2A(n+1)=log2An+1=log2[2An],则:A(n+1)=2An,则[A(n+1)]/[An]=2=常数,则数列{An}是以A1=1为首项、以q=2为公比的等比数列,得:An=2^
Sn、an、1成等差,则2an=Sn+1(n=1时,得a1=1),当n≥2时,有2a(n-1)=S(n-1)+1,则2an-2a(n-1)=an,即an/[a(n-1)]=2=常数,所以{an}是等比
由题意知2an=Sn+1/2,an>0,当n=1时,2a1=a1+1/2,解得a1=1/2,当n≥2时,Sn=2an-1/2,S(n-1)=2a(n-1)-1/2,两式相减得an=Sn-S(n-1)=
(1)由Sn,an,12成等差数列,可得2an=Sn+12,∴a1=12,a2=1(2)由2an=Sn+12可得,2Sn=4an-1(n≥1),∴2Sn-1=4an-1-1(n≥2)∴两式相减得2an
由题意2an=Sn+1/2Sn=2an-1/2n=1时,S1=a1a1=2a1-1/2a1=1/2S(n+1)-Sn=a(n+1)2a(n+1)-1/2-[2an-1/2]=a(n+1)a(n+1)=