题目

设正数数列{an}为一等比数列,且a2=4,a4=16,求lim(lgan+1+lgan+2+...+lga2n)/n^2

提问时间：2020-08-03

答案

因为an>0,a2=4,a4=16
所以q=2,a1=2
所以lim(lgan+1+lgan+2+...+lga2n)/(n^2)
=lim(n/2 * lg(an+1 * a2n))/(n^2)
=lim(lg(a1*q^n * a1*q^(2n-1)))/(2*n)
=lim(lg(a1^2*q^(3n-1))/(2*n)
=lim(lg4 + 3n*lg2 - lg2)/(2*n)
=lim(lg2 + 3n*lg2)/(2*n)
=lim((lg2)/2n)+lim((3lg2)/2)
=0 + 1.5*lg2