作业帮设等差数列的公差d为2,前n项和为Sn,则lim(n→∞)(an^2-n^2)/Sn
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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
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
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
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
an =a1 + 2(n-1) = 2n + (a1-2)
Sn = na1 + n(n-1) = n^2 + (a1-1)n
lim(an^2-n^2)/Sn=lim[3*n^2 +4(a1-1)n + (a1-2)^2 ]/[n^2 + (a1-1)n ]
= lim（3*n^2）/n^2
= 3
n趋于无穷大时，看最高此项的系数比值即为极限值
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