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=

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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=
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=

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=
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->无穷大）{[d/2+(a-d/2)/n]/[d+(a-d)/n]^2}=(d/2)/d^2=1/(2d)
d=2
lim(n->无穷大){s(n)/[a(n)]^2}=0.25
这是什么考试题!老有人问?!

Sn=na1+n(n-1)
an=a1+2(n-1)
Sn/an^2~n^2/(2n)^2-->1/4

an=a1+2(n-1)
Sn=(a1+a1+2(n-1))n/2=a1n+(n-1)n
lim（n到无穷大） Sn/ an^2=[a1n+(n-1)n]/[a1+2(n-1)]^2=1/4

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