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在等比数列{An}中,An>0(n∈N*),且A1A3=4,A3+1...

(1)a3与a5的等比中项a4=2, ∴a1a5+2a3a5+a2a8 =a4^2(1/q^2+2+q^2) =25, ∴1/q^2+2+q^2=25/4, ∴q^4-(17/4)q^2+1=0, q∈(0,1), ∴q^2=1/4,q=1/2, ∴an=a4*q^(n-4)=1/2^(n-5). (2)bn=logan=5-n, ∴Sn=n(9-n)/2. (3)S1/1+S2/2+...+Sn/n =(1/4)n(17-n)(...

a1=4,a1+a3=2a^2+1可得 原式为4+4q^2=32q^2+1 化简得,28q^2=3 q^2=3/28 开平方,舍去负值

(I)因为a3a5+2a4a6+a3a9=100,即a42+2a4a6+a62=100,∴(a4+a6)2=100,又∵an>0,∴a4+a6=10,…(2分)又∵4为a4与a6的等比中项,∴a4?a6=16,…(3分)∴a4,a6是方程x2-10x+16=0的两个根,而q∈(0,1),∴a4>a6,∴a4=8,a6=2,…(4分)∴a1q3=8...

a1+a2=1、a3+a4=9 又: (a3+a4)/(a1+a2)=q² 得: q²=9,即:q=3 而: (a4+a5)=(a3+a4)q=27

∵a1=1,a1、a3、a13 成等比数列,∴(1+2d)2=1+12d.得d=2或d=0(舍去),∴an =2n-1,∴Sn=n(1+2n?1)2=n2,∴2Sn+14an+3=2n2+142n+2.令t=n+1,则2Sn+14an+3=t+8t-2t=2时,t+8t-2=4,t=3时,t+8t-2=113,∴2Sn+14an+3的最小值为113.故选:D.

由等比数列的性质可知:a4a6=a3a7=16,又a4+a6=10由a4a6=16a4+a6=10解得a4=8a6=2 ①或a4=2a6=8 ②由①得q2=a6a4=14与公比q>1矛盾,故应舍去;由②得q2=a6a4=4,解得q=2,或q=-2(舍去),故a3=a4q=22=1.故选D

(Ⅰ)依题意b1=2,b3=23=8,…(2分)设数列{bn}的公比为q,由bn=2an+1>0,可知q>0,…(3分)由b3=b1?q2=2?q2=8,得q2=4,又q>0,则q=2,…(4分)故bn=b1qn?1=2?2n?1=2n,…(5分)又由2an+1=2n,得an=n-1.…(6分)(Ⅱ)依题意cn=(...

(1)∵等差数列{an}中,a1=1,公差d>0,且a2,a3+1,a4+4成等比数列,∴(2+2d)2=(1+d)(5+3d),解得d=-1(舍)或d=1,∴an=n,又b1=2,b2=4,∴bn=2n.(2)n=1时,c1a1=b1,解得c1=2,n≥2时,(c1a1+c2a2+…+cnan)-(c1a1+c2a2+…+cn?1an?1...

(1)∵等差数列{an}的前n项和Sn,公差d≠0,且a3+S5=42,a1,a4,a13成等比数列,∴a1+2d+5a1+5×42d=42(a1+3d)2=a1(a1+12d)d≠0,解得a1=3,d=2,∴an=3+(n-1)×2=2n+1.(2)∵{bnan}是首项为1,公比为3的等比数列,∴bn2n+1=3n-1,即bn=(2n+1)?...

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