Question

已知等差数列{an}的前n项和Sn满足S3=0,S5=-5

已知等差数列{an}的前n项和Sn满足S3=0,S5=-51)求{an}的通项公式(2)求数列{1/a2n-1a2n+1}的前n项和

Answer 1

解：
(1)
S3=3a2=0，a2=0
S5=5a3=-5，a3=-1
d=a3-a2=-1-0=-1
a1=a2-d=0-(-1)=1
an=a1+(n-1)d=1+(-1)(n-1)=-n+2
数列{an}的通项公式为an=-n+2
(2)
1/[a(2n-1)a(2n+1)]
=1/[(-2n+1+2)(-2n-1+2)]
=1/[(2n-3)(2n-1)]
=½[1/(2n-3) -1/(2n-1)]
=½[1/(2n-3) -1/(2(n+1)-3)]
Tn=½[1/(2·1-3)- 1/(2·2-3)+1/(2·2-3)- 1/(2·3-3)+...+1/(2n-3) -1/(2(n+1)-3)]
=½[-1 -1/(2n-1)]
=n/(1-2n)