Question

已知等差数列an的前n项和Sn=-2n的平方-n,求(1)数列an的通项公式;(2)a1+a3+a5+…+a25

Answer 1

1.an=Sn-S(n-1)

=-2n^2-n-[-2(n-1)^2-(n-1)]

=2n^2-4n+2+n-1-(2n^2+n)

=-4n+1

2.a1+a3+a5+…+a25

=S(25)

=-2*25^2-25

=-1250-25

=-1275