设等差数列的公差d为2,前n项和为Sn,则lim(n→∞)(an^2-n^2)/Sn_数学解答

设等差数列的公差d为2,前n项和为Sn,则lim(n→∞)(an^2-n^2)/Sn

人气：124 ℃　时间：2019-10-17 07:56:23

解答

an=a1+2(n-1)=a1+2n-2；Sn=na1+n(n-1)2/2=na1+n^2-n
(an^2-n^2)/Sn=[(a1+2n-2)^2-n^2]/(na1+n^2-n)=[(a1-2)^2+4(a1-2)n+3n^2]/(na1+n^2-n)
上下同除n^2得[(a1-2)^2/n^2+4(a1-2)/n+3]/(a1/n+1-1/n),所以lim=3/1=3