等差数列{an}的前n项和为Sn,公差为2,则lim（n到无穷大）Sn/ an^2=
人气：277 ℃ 时间：2019-11-13 01:25:03
解答
a(n)=a+(n-1)d,[a(n)]^2=[a+(n-1)d]^2=[nd+a-d]^2,s(n)=na+n(n-1)d/2=(d/2)n^2+n(a-d/2),s(n)/[a(n)]^2=[(d/2)n^2+n(a-d/2)]/[nd+a-d]^2=[d/2+(a-d/2)/n]/[d+(a-d)/n]^2,lim(n->无穷大){s(n)/[a(n)]^2}=lim(n->无穷...
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