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.-数学

题目简介

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.-数学

题目详情

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{
1
a2n-1a2n+1
}的前n项和.
题型:解答题难度:中档来源:不详

答案

(Ⅰ)设数列{an}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=0
,即
3a1+3d=0
5a1+10d=0
,解得a1=1,d=-1,
故{an}的通项公式为an=a1+(n-1)d=1+(n-1)•(-1)=2-n;
(Ⅱ)由(Ⅰ)知class="stub"1
a2n-1a2n+1
=class="stub"1
(3-2n)(1-2n)
=class="stub"1
2
(class="stub"1
2n-3
-class="stub"1
2n-1
)

从而数列{class="stub"1
a2n-1a2n+1
}的前n项和
S=class="stub"1
2
[(class="stub"1
-1
-class="stub"1
1
)+(class="stub"1
1
-class="stub"1
3
)+…+(class="stub"1
2n-3
-class="stub"1
2n-1
)]

=class="stub"1
2
(-1-class="stub"1
2n-1
)=class="stub"n
1-2n

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