在等比数列an中

Question

在等比数列an中已知对于大于1的正整数n都有a2+a3+...+an=2^n-2，求lim{[lgan+lga(n+1)+lga(n+2)+...+lga(2n)]/n^2}

Answer 1

令公比为q,则由a2+a3+...+an=2^n-2得a2=2^2-2=2,
a2+a3+...+an
=a2*[1-q^(n-1)]/(1-q)
=2[1-q^(n-1)]/(1-q)
=2^n-2
解得q=2
所以an=a2*q^(n-2)=2^(n-1)
所以lim{[lgan+lga(n+1)+lga(n+2)+...+lga(2n)]/n^2}
=lim{lg[an*a(n+1)*a(n+2)+...+a(2n)]/n^2}
=lim{[lg2^[(n-1+2n-1)(n+1)/2]/n^2}
=lim[(3n-2)(n+1)/(2n^2)]lg2
=3/2*lg2