作业帮设等差数列an的公差d为2,前n项和为Sn,则lim(an^2-n^2)/Sn=?
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设等差数列an的公差d为2,前n项和为Sn,则lim(an^2-n^2)/Sn=?
设等差数列an的公差d为2,前n项和为Sn,则lim(an^2-n^2)/Sn=?
设等差数列an的公差d为2,前n项和为Sn,则lim(an^2-n^2)/Sn=?
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
an = 2n + (a1-2),Sn = 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 ]
n=∞，lim(an^2-n^2)/Sn=3
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