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已知等比数列{An}满足An>0,n=1,2,…,A5?A2n%5=...

由a5?a2n-5=an2=22n,且an>0,解得an=2n,则log2a1+log2a2+log2a3+…+log2a2n-1=log(a1a2n?1)?(a2a2n?2) …an 2=log2n22=n2.故答案为:n2

由题意等比数列{an}a>0,n=1,2,…,当n>1时,log2a1+log2a3+log2a5+…+log2a2n+1=log2a1a3a5…a2n+1.又a5?a2n-5=22n(n≥3)∴a1a3a5…a2n+1=2(n+1)2∴log2a1+log2a3+log2a5+…+log2a2n+1=log22(n+1)2=(n+1)2故选:C.

解:∵{An}为等比数列,∴A5×A2n-5=A1×A2n-1=A2×A2n-2=……=2^2n。而㏒2A1+㏒2A2+㏒2A3+……+㏒2A2n-1=㏒2(A1×A2×A3×……×A2n-2×A2n-1)=㏒2{2^2n}^(n/2)=㏒2(2^n^2)=n^2。考查等比数列的性质及对数运算,如有问题,敬请追问,万望采纳!祝您学习...

(1)a1=a,a2=2S1-21-12=2a-3,a3-5=2(a1+a2)-22-22-5=6a-19,∵a1,a2,a3-5成等比数列,∴(2a-3)2=a(6a-19),解得a=-1或a=92.(2)∵an+1=2Sn-2n-n2(n∈N*),①∴an=2Sn-1-2n-1-(n-1)2(n≥2,n∈N*),②∴当n≥2时,①-②得an+1-an=2an-2n-1-...

17(2)解:由 (1-λ)Sn=-λan+2·4^n/3+1/3 ① 得: (1-λ)S(n-1)=-λa(n-1)+2·4^(n-1)/3+1/3 ② ①-②,得: (1-λ)an=-λ[an-a(n-1)]+2·4^(n-1)(4-1)/3 即 an=λa(n-1)+2·4^(n-1) ③ ③式一定可以写成这样的形式: an-k·2·4^n=λ[a(n-1)-k·2·4^(n-1)] ④ 其中,...

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