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在等比数列An中,An大于0,且A5A6=9,求log3A2+log...

解答: 这个是对数的运算法则 同底对数相加,底数不变,真数相乘, 即 loga(M)+loga(N)=loga(MN) 可以推广到有限多个。

∵等比数列{an}中,an>0,a5a6=9,∴a1?a2…a10=(a5?a6)5=310,∴log3a1+log3a2+…+log3a10=log3310=10.故选B.

a5a6=a4a7=a3a8=a2a9=a1a10=9(正数的等比数列有这性质)好好想想就知道 后面为log3a1a2a3a4a5a6a7a8a9a10=~~~不用我算了吧

等比数列{an}的各项均为正数,且a4a6=9=a52,即a5=3,则log3a1+log3a2+…log3a9 =(log3a1+log3a9 )+(log3a2+log3a8)+(log3a3+log3a7 )+(log3a4+log3a6)+log3a5 =4(log3a4+log3a6)+log3a5=4log3(a4a6)+log3a5=4×2+1=9, 故答案为9.

既然你说到了1006次方,那你应该求出了原式=log3[(a1a2012)(a2a2011)...(a1006a1007)] 因为等比数列具有an*am=a(n+m)的性质, 所以a1*a2012=a2*a2011=…a1006*a1007 那么原式=log2(a1006a1007*a1006a1007*a1006a1007*…a1006a1007) 又因为一共有10...

a5=a1q^4 q^4=2^4 q=2 sn=36(2^n-1) log3a1+log3a2+…+log3a10=20 log3a1a2a3……a10=20 log3(a5a6)^5=20 log3(a5a6)=4 a5a6=3^4=81

等比数列{an}中a4*a5*a6=a5*(a4*a6)=a5*a5^2=a5^3=3 [^表示指数,a5^2=a4*a6]于是log3(a5^3)=1=3log3(a5)于是log3(a5)=1/3而a1*a2*a8*a9=a1*a9*a2*a8=a5^2*a5^2=a5^4[a5是a1和a9的中项,又是a2和a8的中项]=根据对数性质log3a1+log3a2+log3a8+log...

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...

上面那个- - 明明是n+1项好不好orz 原式=log3{[a(2)*a(2n)]^n/2 * a(n+1)} 因为a(n+1)^2=[a(2)*a(2n)=9,且每一项大于零 所以a(n+1)=3 所以 原式=log3(9^n/2 * 3) =log3 (3^n*3) =n+1 求项数判断法有很多╮(╯▽╰)╭。。。我是把具体数字带入最后一...

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

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