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=?

题目

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=?

答案

an =a1 + 2(n-1) = 2n + (a1-2)Sn = na1 + n(n-1) = n^2 + (a1-1)nlim(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= 3n趋于无穷大时,看最高此项的系数比值即为极限值...