设正数数列{an}是个等比数列 且a2=4 a4=16求lim(n趋向于无穷)(lga(n+1)+lga(n+2)+...+lga(2n))/n^2

题目

设正数数列{an}是个等比数列 且a2=4 a4=16求lim(n趋向于无穷)(lga(n+1)+lga(n+2)+...+lga(2n))/n^2

答案

答案：3/2lg2
由a2=4,a4=16,求得a1=2,q=2,即an = 2×2^(n-1)=2^n
所以(lga(n+1)+lga(n+2)+...+lga(2n))/n^2 = lg2((n+1)+(n+2)+...+(2n))/n^2 =
lg2(3n^2+n) /(2n^2)->3/2lg2