关于周期数列的重要性质与结论的探究

广东省广州市广雅中学(510160)徐广华

周期数列可看作是周期函数的特例,其定义域为正整数集.灵活利用数列的周期性,可以巧妙地解决一些特殊数列的指定项及求和问题.

一、周期数列的概念

对于数列{an},如果存在确定的正整数T及n0,使得对一切n≥n0,恒有an+T=an成立,则称{an}是从第n0项起的周期为T的周期数列.当n0=1时,称{an}为纯周期数列;当n0≥2时,称{an}为混周期数列.

二、周期数列的重要性质

1.周期数列是无穷数列,其值域为有限集.

2.若T是数列{an}的周期,则对∀k∈N∗,kT也是数列{an}的周期.即若an+T=an,则an+kT=an.

3.周期数列必有最小正周期.

4.若T是周期数列{an}的最小正周期,T′是数列{an}的任一周期,则必有T|T′.

5.若{an}是周期为T的周期数列,Sn是数列{an}的前n项和,若m=qT+r(0≤r＜T,r∈N∗),则am=ar,Sm=qST+Sr.

三、有关数列周期性的重要结论

以下结论中,n为一切正整数,k为某个确定的正整数.

1.若数列{an}满足an+an+k=C(C为常数),则{an}是周期为2k的周期数列.

特别地,若an+k=-an(即C=0时),则{an}是周期为2k的周期数列.

推广:若数列{an}满足an+an+1+···+an+k=C(C为常数),则{an}是周期为k+1的周期数列.

2.若数列{an}满足an·an+k=C(常数C/=0),则{an}是周期数列,且2k是它的周期.

推广:若数列{an}满足an·an+1·····an+k=C(常数C/=0),则{an}是周期为k+1的周期数列.

3.若非零数列{an}满足an+an+1+···+an+k=an·an+1·····an+k,则{an}是周期为k+1的周期数列.

4.若数列{an}满足则{an}是周期为3k的周期数列.

类似:若数列{an}满足则{an}是周期为3k的周期数列.

类似:若数列{an}满足则{an}是周期为3k的周期数列.

5.若数列{an}满足则{an}是周期为3k的周期数列.

类似:若数列{an}满足则{an}是周期为3k的周期数列.

6.若数列{an}满足则{an}是周期为4k的周期数列.

类似:若数列{an}满足则{an}是周期为4k的周期数列.

7.若数列{an}满足则{an}是周期为6k的周期数列.

类似:若数列{an}满足则{an}是周期为6k的周期数列.

8.若数列{an}满足an+2k=an+k-an,则{an}是周期为6k的周期数列.

特别地,若an+2=an+1-an(即k=1时),则{an}是周期为6的周期数列.

9.若数列{an}满足则{an}是周期为6k的周期数列.

10.若数列{an}满足则{an}是周期为2k的周期数列.

上述结论的简略证明

1.由an+an+k=C,得an+k+an+2k=C,则an+k+an+2k=an+an+k,故an+2k=an,{an}是周期为2k的数周期数列.

推广:由an+an+1+···+an+k=C,得an+1+an+2+···+an+k+1=C,则an+1+an+2+···+an+k+1=an+an+1+···+an+k,故an+k+1=an,{an}是周期为k+1的周期数列.

2.an·an+k=C⇒an+k·an+2k=C,则an+k·an+2k=an·an+k,故an+2k=an,{an}是周期为2k的数周期数列.

推广:由an·an+1·····an+k=C,得an+1·an+2·····an+k+1=C,则an+1·an+2·····an+k+1=an·an+1·····an+k,故an+k+1=an,{an}是周期为k+1的周期数列.

3.由an+an+1+· +an+k=an·an+1·····an+k①,得an+1+an+2+· +an+k+1=an+1·an+2·····an+k+1②,②-①,得an+k+1-an=(an+k+1-an)an+1·an+2·····an+k,由an/=0,得an+k+1=an,故{an}是周期为k+1的周期数列.

8.由an+2=an+1-an①,得an+3=an+2-an+1②,①+②,得an+3=-an,则an+6=-an+3=an,故{an}是周期为6的周期数列.

四、周期数列的简单应用

(2)(2017广州调研理16)数列{an}满足a1=2,a2=8,an+2+an=an+1,则

(3)(2012福建卷理12)数列{an}的通项公式前n项和为Sn,则S2012=____.

分析(1)由递推式得:a7=1,a8=2,故{an}是周期为6的周期数列,从而a2017=a6×336+1=a1=1.

(2)由递推式得:an+2=an+1-an,则a3=6,a4=-2,a5=-8,a6=-6,a7=2,a8=8,故{an}是周期为6的周期数列,从而

例3 已知数列{an}满足求数列{an}的通项公式.

分析(1)由递推式得:a4=-3,a5=2=a1,故{an}是周期为4的周期数列,因此,当n=4k-3(k∈N∗)时,an=2;当n=4k-2(k∈N∗)时,当n=4k-1(k∈N∗)时,当n=4k(k∈N∗)时,an=-3.

例4 设数列{an}满足a1=a2=1,a3=2,且anan+1an+2/=1,anan+1an+2an+3=an+an+1+an+2+an+3.求数列{an}的前100项和S100.

分析由anan+1an+2an+3=an+an+1+an+2+an+3①得an+1an+2an+3an+4=an+1+an+2+an+3+an+4②②-①,得(an+4-an)an+1an+2an+3=an+4-an,由anan+1an+2/=1,得an+4=an,故{an}是周期为4的周期数列.由a1a2a3a4=a1+a2+a3+a4,a1=a2=1,a3=2,得:a4=4,故S100=25S4=25(1+1+2+4)=200.

由上可见,“周期数列”这个概念尽管在目前的高中教材中没有定义过,但与周期数列有关的问题却在高考和模拟试题中屡见不鲜.记住一些有关数列周期性的重要结论,灵活运用周期数列的重要性质解题,可以起到触类旁通、化繁为简的效果,在复习备考中值得我们重视和研究.