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等比数列{An}An>0,且A5乘以A2n%5=2^2n(n≥3),则当n≥...

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...

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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

在等比数列{an}中,a5?a2n-5=22n(n≥3),即a5?a2n-5=a2n=(2n)2,(n≥3),∴an=2n,(n≥3),∴a1=2,a2=4.∴log2an=log22n=n,∵n≥1时,log2a1+log2a2+log2a3+…log2an,∴log2a1+log2a2+log2a3+…log2an=1+2+…+n=n(n+1)2,故选:B.

回答:(1)n大于等于1的作用是:log2a(2n--1)有意义。 (2)题中的log2a(1)+log2a(2)+.....+log2a(2n--1)=log2(a1)*a(2)......*a(2n--1) 是运用了对数的性质“对数的和等于积的对数。”如“logaM+logaN=log a(M*N).” (3) A5*A2n--5=A1*A2n--1. 这...

(I)a4=a23+1a2=9+12=5a5=a24?1a3=25?13=8(II)假设存在实数λ,使得数列{an+1-λan}(n∈N*)是等差数列,则2(a3-λa2)=(a2-λa1)+(a4-λa3),解得λ=1由a3=3,a4=5,a5=8,a6=13得2(a5-a4)≠(a4-a3)-(a3-a2)与数列{an+1-an}(n∈N*)是等...

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