已知等差数列{an}的前n项和Sn满足S3=0,S5=-5,
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已知等差数列{an}的前n项和Sn满足S3=0,S5=-5,
(1)求{an}的通项公式;
(2)求数列{
(1)求{an}的通项公式;
(2)求数列{
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a
(1)设等差数列{an}的公差为d,
∵前n项和Sn满足S3=0,S5=-5,∴ 3a1+3d=0 5a1+10d=−5, 解得a1=1,d=-1. ∴an=1-(n-1)=2-n. (2) 1 a2n−1a2n+1= 1 (3−2n)(1−2n)= 1 2( 1 2n−3− 1 2n−1), ∴数列{ 1 a2n−1a2n+1}的前n项和= 1 2[(−1−1)+(1− 1 3)+( 1 3− 1 5)+…+( 1 2n−3− 1 2n−1)] = 1 2(−1− 1 2n−1) = n 1−2n.
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