题目内容

(2013•门头沟区一模)已知数列{An}的前n项和为Sn,a1=1,满足下列条件
①?n∈N*,an≠0;
②点Pn(an,Sn)在函数f(x)=
x2+x2
的图象上;
(I)求数列{an}的通项an及前n项和Sn
(II)求证:0≤|Pn+1Pn+2|-|PnPn+1|<1.
分析:(I)由题意Sn=
an2+an
2
,当n≥2时an=Sn-Sn-1,由此可得两递推式,分情况可判断数列{an}为等比数列或等差数列,从而可求得通项an,进而求得Sn
(II)分情况讨论:当当an+an-1=0时,Pn((-1)n-1
1-(-1)n
2
)
,计算可得|Pn+1Pn+2|=|PnPn+1|=
5
,从而易得|Pn+1Pn+2|-|PnPn+1|的值;当an-an-1-1=0时,Pn(n,
n2+n
2
)
,利用两点间距离公式可求得|Pn+1Pn+2|,|PnPn+1|,对|Pn+1Pn+2|-|PnPn+1|化简后,再放缩即可证明结论;
解答:(I)解:由题意Sn=
an2+an
2

当n≥2时an=Sn-Sn-1=
an2+an
2
-
an-12+an-1
2

整理,得(an+an-1)(an-an-1-1)=0,
又?n∈N*,an≠0,所以an+an-1=0或an-an-1-1=0,
当an+an-1=0时,a1=1,
an
an-1
=-1

an=(-1)n-1Sn=
1-(-1)n
2

当an-an-1-1=0时,a1=1,an-an-1=1,
得an=n,Sn=
n2+n
2

(II)证明:当an+an-1=0时,Pn((-1)n-1
1-(-1)n
2
)

|Pn+1Pn+2|=|PnPn+1|=
5
,所以|Pn+1Pn+2|-|PnPn+1|=0,
当an-an-1-1=0时,Pn(n,
n2+n
2
)

|Pn+1Pn+2|=
1+(n+2)2
,|PnPn+1|=
1+(n+1)2

|Pn+1Pn+2|-|PnPn+1|=
1+(n+2)2
-
1+(n+1)2

=
1+(n+2)2-1-(n+1)2
1+(n+2)2
+
1+(n+1)2

=
2n+3
1+(n+2)2
+
1+(n+1)2

因为
1+(n+2)2
>n+2,
1+(n+1)2
>n+1,
所以0<
2n+3
1+(n+2)2
+
1+(n+1)2
<1,
综上0≤|Pn+1Pn+2|-|PnPn+1|<1.
点评:本题考查数列与函数的综合,考查分类讨论思想,解决本题的关键是利用an与Sn的关系先求得an
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