Question

设等差数列{a_n}的公差d是2，前n项的和为S_n，则

lim

n→∞

a 2n -n2

Sn

=______．

Answer 1

由公差d=2，得到a_n=a₁+2（n-1）=2n+a₁-2，S_n=na₁+

n(n-1)

2

×2=n²+n（a₁-1）
则

lim

n→∞

a 2n -n2

Sn

=

lim

n→∞

3n2+4(a1-2)n+(a1-2)2

n2+n(a1-1)

=

lim

n→∞

3+ 4(a1-2) n + (a1-2)2 n2

1+ a1-1 n

=3
故答案为3．