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

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

人气：469 ℃　时间：2020-01-27 15:25:23

解答

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/...