问题 解答题
已知数列{an}的前n项和Sn=-2n2+5n(n∈N*).
(1)求数列{an}的通项公式;
(2)如果两个互不相等的正整数n1,n2满足
n1+n2
2
=q
(q为正整数),试比较
Sn1+Sn2
2
与Sq的大小,并说明理由.
答案

(1)当n=1时,a1=3,--------------1’

当n≥2时,an=Sn-Sn-1=-2n2+5n-[-2(n-1)2+5(n-1)]=-4n+7---------------3’

当n=1时满足通项公式,∴an=-4n+7---------4’

(2)∵n1n2

n1+n2
2
=q,

Sn1+Sn2
2
-Sq=
1
2
(-2
n21
+5n1-2
n22
+5n2)-(-2q2+5q)----6’

=

1
2
[-2(
n21
+
n22
)+10q]+2q2-5q=-(
n21
+
n22
)+2(
n1+n2
2
)2
=-
1
2
[2
n21
+2
n22
-(n1+n2)2]
=
1
2
(n1-n2)2<0
-------10’

Sn1+Sn2
2
Sq-----------12’

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