Question

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

Answer 1

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/n)]/{[a(1)/n]^2+4(1-1/n)a(1)/n+4(1-1/n)^2-1}
=(0+1-0)/[0+4×(1-0)×0+4×(1-0)^2-1]
=1/3

Answer 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),
当d=0时，a(n)=a,s(n)=na.
当d=0,a=0时,s(n)/[a(n)]^2没有意义。
当d=0,a不等于0时，s(n)/[a(n)]^2=n/a，lim(n->无穷大){s(n)/[a(n)]^2}=(n->无穷大){n/a}=正无穷大。
d不等于0时，
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)