已知等差数列{an}的前n项和Sn满足S3=0，S5=-5．（Ⅰ）求{an}的通项公式；（Ⅱ）求数列{1a2n−1a2n+1}的前n项和．

Question

已知等差数列{a_n}的前n项和S_n满足S₃=0，S₅=-5．
（Ⅰ）求{a_n}的通项公式；
（Ⅱ）求数列{(1)/(a2n−1a2n+1)}的前n项和．

Answer 1

（Ⅰ）设数列{a_n}的首项为a₁，公差为d，则Sn＝na1+(n(n−1)d)/(2)．
由已知可得

3a1+3d＝0

5a1+(5(5−1))/(2)d＝−5

，即

a1+d＝0

a1+2d＝−1

，解得a₁=1，d=-1，
故{a_n}的通项公式为a_n=a₁+（n-1）d=1+（n-1）•（-1）=2-n；
（Ⅱ）由（Ⅰ）知(1)/(a2n−1a2n+1)＝(1)/((3−2n)(1−2n))＝(1)/(2)((1)/(2n−3)−(1)/(2n−1))．
从而数列{(1)/(a2n−1a2n+1)}的前n项和
S_n=(1)/(2)[((1)/(−1)−(1)/(1))+((1)/(1)−(1)/(3))+…+((1)/(2n−3)−(1)/(2n−1))]
=(1)/(2)(−1−(1)/(2n−1))＝(n)/(1−2n)．