Question

设等差数列{an}公差d是2,前n项和为Sn,则lim(an^2-n^2)/Sn

Answer 1

S(n)=n[a(1)+a(n)]2=na(1)+n(n-1)d/2=na(1)+n(n-1)

a(n)^2-n^2=[a(1)+(n-1)d]^2-n^2=a(1)^2+4(n-1)a(1)+4(n-1)^2-n^2

所以

lim(a(n)^2-n^2)/S(n)

=lim[na(1)+n(n-1)]/[a(1)^2+4(n-1)a(1)+4(n-1)^2-n^2]

=lim[a(1)/n+(1-1/n)]/{[a(1)/n]^2+4(1-1/n)a(1)/n+4(1-1/n)^2-1}

=(0+1-0)/[0+4×(1-0)×0+4×(1-0)^2-1]

=1/3