Question

设正数数列｛an｝为等比数列,且a2=4,a4=16,求[lga(n+1)+lga(n+2)+…+lga(2n)]/n^

Answer 1

依题意,可得公比q=2,a1=2,通项an=2^n,所以

原式={lg[2^(n+1)*2^(n+2)*...*2^(2n)]}/n^2

={lg2^[(n+1)+(n+2)+...+(2n)]}/n^2

={[(n+1)+2n]*n/2}*lg2/n^2

当n趋向无穷大时,其极限=3/2*lg2