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=?](/uploads/image/z/14835551-23-1.jpg?t=%E8%AE%BE%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97an%E7%9A%84%E5%85%AC%E5%B7%AEd%E4%B8%BA2%2C%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E5%88%99lim%28an%5E2-n%5E2%29%2FSn%3D%3F)
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
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
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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