廣東省廣州市廣雅中學(xué)(510160)徐廣華

周期數(shù)列可看作是周期函數(shù)的特例,其定義域?yàn)檎麛?shù)集.靈活利用數(shù)列的周期性,可以巧妙地解決一些特殊數(shù)列的指定項(xiàng)及求和問題.

一、周期數(shù)列的概念

對于數(shù)列{an},如果存在確定的正整數(shù)T及n0,使得對一切n≥n0,恒有an+T=an成立,則稱{an}是從第n0項(xiàng)起的周期為T的周期數(shù)列.當(dāng)n0=1時(shí),稱{an}為純周期數(shù)列;當(dāng)n0≥2時(shí),稱{an}為混周期數(shù)列.

二、周期數(shù)列的重要性質(zhì)

1.周期數(shù)列是無窮數(shù)列,其值域?yàn)橛邢藜?

2.若T是數(shù)列{an}的周期,則對?k∈N?,kT也是數(shù)列{an}的周期.即若an+T=an,則an+kT=an.

3.周期數(shù)列必有最小正周期.

4.若T是周期數(shù)列{an}的最小正周期,T′是數(shù)列{an}的任一周期,則必有T|T′.

5.若{an}是周期為T的周期數(shù)列,Sn是數(shù)列{an}的前n項(xiàng)和,若m=qT+r(0≤r＜T,r∈N?),則am=ar,Sm=qST+Sr.

三、有關(guān)數(shù)列周期性的重要結(jié)論

以下結(jié)論中,n為一切正整數(shù),k為某個(gè)確定的正整數(shù).

1.若數(shù)列{an}滿足an+an+k=C(C為常數(shù)),則{an}是周期為2k的周期數(shù)列.

特別地,若an+k=-an(即C=0時(shí)),則{an}是周期為2k的周期數(shù)列.

推廣:若數(shù)列{an}滿足an+an+1+···+an+k=C(C為常數(shù)),則{an}是周期為k+1的周期數(shù)列.

2.若數(shù)列{an}滿足an·an+k=C(常數(shù)C/=0),則{an}是周期數(shù)列,且2k是它的周期.

推廣:若數(shù)列{an}滿足an·an+1·····an+k=C(常數(shù)C/=0),則{an}是周期為k+1的周期數(shù)列.

3.若非零數(shù)列{an}滿足an+an+1+···+an+k=an·an+1·····an+k,則{an}是周期為k+1的周期數(shù)列.

4.若數(shù)列{an}滿足則{an}是周期為3k的周期數(shù)列.

類似:若數(shù)列{an}滿足則{an}是周期為3k的周期數(shù)列.

類似:若數(shù)列{an}滿足則{an}是周期為3k的周期數(shù)列.

5.若數(shù)列{an}滿足則{an}是周期為3k的周期數(shù)列.

類似:若數(shù)列{an}滿足則{an}是周期為3k的周期數(shù)列.

6.若數(shù)列{an}滿足則{an}是周期為4k的周期數(shù)列.

類似:若數(shù)列{an}滿足則{an}是周期為4k的周期數(shù)列.

7.若數(shù)列{an}滿足則{an}是周期為6k的周期數(shù)列.

類似:若數(shù)列{an}滿足則{an}是周期為6k的周期數(shù)列.

8.若數(shù)列{an}滿足an+2k=an+k-an,則{an}是周期為6k的周期數(shù)列.

特別地,若an+2=an+1-an(即k=1時(shí)),則{an}是周期為6的周期數(shù)列.

9.若數(shù)列{an}滿足則{an}是周期為6k的周期數(shù)列.

10.若數(shù)列{an}滿足則{an}是周期為2k的周期數(shù)列.

上述結(jié)論的簡略證明

1.由an+an+k=C,得an+k+an+2k=C,則an+k+an+2k=an+an+k,故an+2k=an,{an}是周期為2k的數(shù)周期數(shù)列.

推廣:由an+an+1+···+an+k=C,得an+1+an+2+···+an+k+1=C,則an+1+an+2+···+an+k+1=an+an+1+···+an+k,故an+k+1=an,{an}是周期為k+1的周期數(shù)列.

2.an·an+k=C?an+k·an+2k=C,則an+k·an+2k=an·an+k,故an+2k=an,{an}是周期為2k的數(shù)周期數(shù)列.

推廣:由an·an+1·····an+k=C,得an+1·an+2·····an+k+1=C,則an+1·an+2·····an+k+1=an·an+1·····an+k,故an+k+1=an,{an}是周期為k+1的周期數(shù)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}是周期為k+1的周期數(shù)列.

8.由an+2=an+1-an①,得an+3=an+2-an+1②,①+②,得an+3=-an,則an+6=-an+3=an,故{an}是周期為6的周期數(shù)列.

四、周期數(shù)列的簡單應(yīng)用

(2)(2017廣州調(diào)研理16)數(shù)列{an}滿足a1=2,a2=8,an+2+an=an+1,則

(3)(2012福建卷理12)數(shù)列{an}的通項(xiàng)公式前n項(xiàng)和為Sn,則S2012=____.

分析(1)由遞推式得:a7=1,a8=2,故{an}是周期為6的周期數(shù)列,從而a2017=a6×336+1=a1=1.

(2)由遞推式得:an+2=an+1-an,則a3=6,a4=-2,a5=-8,a6=-6,a7=2,a8=8,故{an}是周期為6的周期數(shù)列,從而

例3 已知數(shù)列{an}滿足求數(shù)列{an}的通項(xiàng)公式.

分析(1)由遞推式得:a4=-3,a5=2=a1,故{an}是周期為4的周期數(shù)列,因此,當(dāng)n=4k-3(k∈N?)時(shí),an=2;當(dāng)n=4k-2(k∈N?)時(shí),當(dāng)n=4k-1(k∈N?)時(shí),當(dāng)n=4k(k∈N?)時(shí),an=-3.

例4 設(shè)數(shù)列{an}滿足a1=a2=1,a3=2,且anan+1an+2/=1,anan+1an+2an+3=an+an+1+an+2+an+3.求數(shù)列{an}的前100項(xiàng)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}是周期為4的周期數(shù)列.由a1a2a3a4=a1+a2+a3+a4,a1=a2=1,a3=2,得:a4=4,故S100=25S4=25(1+1+2+4)=200.

由上可見,“周期數(shù)列”這個(gè)概念盡管在目前的高中教材中沒有定義過,但與周期數(shù)列有關(guān)的問題卻在高考和模擬試題中屢見不鮮.記住一些有關(guān)數(shù)列周期性的重要結(jié)論,靈活運(yùn)用周期數(shù)列的重要性質(zhì)解題,可以起到觸類旁通、化繁為簡的效果,在復(fù)習(xí)備考中值得我們重視和研究.