Question

等差数列{a_n}满足

lim

n→∞

Sn

2n2

=1，试写出满足上述条件的{a_n}的一个通项公式______．

Answer 1

设等差数列的公差为d，首项为a₁
∴Sn=na1+

1

2

n(n-1)d
∴

lim

n→∞

Sn

2n2

=

lim

n→∞

na1+ 1 2 n(n-1)d

2n2

=

lim

n→∞

a1 n + 1 2 d(1- 1 n )

2

=

d

4

=1
∴d=4
故答案为：a_n=4n