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=
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?!
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?!
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