6.5構(gòu)造法求通項(xiàng)答案

6.5構(gòu)造法求通項(xiàng)課標(biāo)要求精細(xì)考點(diǎn)素養(yǎng)達(dá)成1.會(huì)利用數(shù)列的遞推公式求通項(xiàng)公式可化為等差數(shù)列求通項(xiàng)之待定系數(shù)法了解遞推公式是給出數(shù)列的一種方法,培養(yǎng)學(xué)生的數(shù)學(xué)抽象素養(yǎng)2.掌握由遞推公式構(gòu)造新數(shù)列相鄰項(xiàng)的差與商的一般方法可化為等比數(shù)列求通項(xiàng)之待定系數(shù)法經(jīng)歷遞推公式構(gòu)造相鄰項(xiàng)的差與商的待定系數(shù)法,培養(yǎng)學(xué)生的邏輯推理素養(yǎng)3.感悟轉(zhuǎn)化與化歸思想的應(yīng)用其他形式求通項(xiàng)通過遞推公式探究數(shù)列通項(xiàng)公式的求解,培養(yǎng)學(xué)生的數(shù)學(xué)建模素養(yǎng)1.(對(duì)接教材)在數(shù)列{an}中,a1=2,nan+1=(n+1)an+2,則an=.

答案4n2解析由題得an+1n+1=則ann=an-1n-1+2(n-1)an-1n-1=…,a22=a1累加可得ann=a11+2(即ann=a1+21n-11n+1n-21n-1+…+112(則ann=2+21?1n(n≥2,即an=4n2(n≥2,n∈Z),又a1=2適合此式,所以an=4n2.2.(對(duì)接教材)若數(shù)列{an}的首項(xiàng)a1=1,且滿足an+1=2an+1,則an=.

答案2n1解析由an+1=2an+1得an+1+1=2(an+1),由a1=1,得a1+1=2.因此{(lán)an+1}是以2為首項(xiàng),2為公比的等比數(shù)列,于是an+1=2n,所以an=2n1.3.(對(duì)接教材)已知數(shù)列{an}的首項(xiàng)a1=35,且滿足an+1=3an2an+1答案1解析由an+1=3an2an+1,得1an+1=13an+23,故1an+11=13·1an-1,由a1=35,得1a11=234.(易錯(cuò)自糾)已知數(shù)列{an}的前n項(xiàng)和Sn=3n+k,求數(shù)列{an}的通項(xiàng)公式.解析當(dāng)n=1時(shí),a1=S1=3+k;當(dāng)n≥2時(shí),an=SnSn1=2·3n1.所以an=3+5.(真題演練)(2022·新高考Ⅰ卷)記Sn為數(shù)列{an}的前n項(xiàng)和,已知a1=1,Snan是公差為(1)求{an}的通項(xiàng)公式.(2)證明:1a1+1a2+…+解析(1)因?yàn)閍1=1,所以S1=a1=1,所以S1a1又因?yàn)镾nan是公差為所以Snan=1+13(n1)=n+23,所以當(dāng)n≥2時(shí),Sn1=(n所以an=SnSn1=(n+2)整理得(n1)an=(n+1)an1,即anan所以an=a1×a2a1×a3a2=1×31×42×53×…×nn-2顯然對(duì)于n=1也成立,所以{an}的通項(xiàng)公式an=n((2)由(1)得1an=2n(所以1a1+1a2+…+1an=21?12+12-13an+1=Aan+B(A≠0,1,B≠0)形式典例1已知數(shù)列{an}中,a1=4,an+1=4an6,則an等于().A.22n+1+2 B.22n+12C.22n1+2 D.22n12答案C解析因?yàn)閍n+1=4an6,所以an+12=4(an2),所以an+1-2an-2=4,所以數(shù)列{an2}是以2為首項(xiàng),4為公比的等比數(shù)列,所以an2=2×4n1,所以an=2當(dāng)遇到an+1=pan+q(p≠0,1,q≠0)的形式時(shí),第一步構(gòu)造出an+1+t=p(an+t)的形式;第二步利用待定系數(shù)求出t的值.則數(shù)列{an+t}是公比為p的等比數(shù)列.訓(xùn)練1在數(shù)列{an}中,a1=45,4an+1=3an15,求數(shù)列{an}解析因?yàn)樵跀?shù)列{an}中,a1=45,4an+1=3an1所以an+1=34an1等式兩邊同加上15,得an+1+15=又a1=45所以數(shù)列an+15是首項(xiàng)為a1+15=1所以an=34nan+1=pan+qn+c(p≠0,1,q≠0)形式典例2已知a1=1,當(dāng)n≥2時(shí),an=12an1+2n1,求{an}的通項(xiàng)公式解析設(shè)an+An+B=12[an1+A(n1)+B所以an=12an112An12A所以-12又a14+6=3,所以{an4n+6}是以3為首項(xiàng),12為公比的等比數(shù)列所以an4n+6=312n-1,所以an=32n先構(gòu)造出an+An+B=p[an1+A(n1)+B],然后利用待定系數(shù)法求出A和B的值,即可判斷出數(shù)列{an+An+B}是公比為p的等比數(shù)列.訓(xùn)練2已知數(shù)列{an}滿足a1=1,an+1=2an+n1,求數(shù)列{an}的通項(xiàng)公式.解析因?yàn)閍n+1=2an+n1,所以an+1+(n+1)=2an+2n,即an+1+(n+1)又a1+1=2,所以數(shù)列{an+n}是以2為首項(xiàng),2為公比的等比數(shù)列,則an+n=2·2n1=2n,故an=2nn.an+1=pan+rqn形式1.類型一:p=q典例3已知數(shù)列{an}的首項(xiàng)a1=12,滿足an+1=12an+12n(n∈N*),求數(shù)列{a解析依題意2n+1an+1=2nan+2,又2a1=1,所以數(shù)列{2nan}是以1為首項(xiàng),2為公差的等差數(shù)列,所以2nan=1+2(n1)=2n1,所以an=(2n1)12訓(xùn)練3已知數(shù)列{an}中,a1=1,an=2an1+2n(n∈N且n≥2),求數(shù)列{an}的通項(xiàng)公式.解析由an=2an1+2n,得an2nan-12n-1=1,又a12=12,所以an2n是首項(xiàng)為12,公差為1的等差數(shù)列,于是an2n=12+(n2.類型:p≠q典例4已知數(shù)列{an}滿足an+1=2an+4×3n1,a1=1,求數(shù)列{an}的通項(xiàng)公式.解析(法一)因?yàn)閍n+1=2an+4×3n1,設(shè)an+1+λ·3n=2(an+λ·3n1),所以an+1=2anλ·3n1,則λ=4,即an+14×3n=2(an4×3n1),則數(shù)列{an4×3n1}是首項(xiàng)為a14×311=3,公比為2的等比數(shù)列,所以an4×3n1=3×2n1,即an=4×3n13×2n1.(法二)因?yàn)閍n+1=2an+4×3n1,兩邊同時(shí)除以3n+1得an+13n+1=2所以an+13n+143=23·所以an3n-43是以所以an3n43=23n-1,則an3n=4323n-1,所以當(dāng)p=q時(shí),等式兩邊同時(shí)除以qn,即可構(gòu)造出一個(gè)等差數(shù)列.當(dāng)p≠q時(shí),可設(shè)an+1+λ·3n=p(an+λ·3n1),利用待定系數(shù)法求出參數(shù)的值,即可構(gòu)造出等比數(shù)列.訓(xùn)練4若數(shù)列{an}滿足a1=2,an+12an=3n1.證明:{an+13an}是等比數(shù)列.解析因?yàn)閍n+12an=3n1,所以an+22an+1=3n,故an+22an+1=3(an+12an),則an+23an+1=2an+16an=2(an+13an).又a1=2,則a2=5,a23a1=1,故{an+13an}是以1為首項(xiàng),2為公比的等比數(shù)列.an+2=pan+1+qan形式典例5已知數(shù)列{an}中,a1=1,a2=2,an+2=23an+1+13an,求數(shù)列{an}解析an+2=23an+1+13an化為an+2san+1=t(an+1san),即an+2=(s+t)an+1sta所以s+t=23,st不妨令an+2an+1=13(an+1an),a2a1=1所以{an+1an}是以1為首項(xiàng),13為公比的等比數(shù)列,則an+1an=-累加法可得ana1=-130+-131+…+-13n-2=1?an=7434·-13n又a1=1符合上式,故an=7434·設(shè)出an+2san+1=t(an+1san),利用待定系數(shù)法求出s和t的值,則可得到數(shù)列{an+1san}是公比為t的等比數(shù)列.訓(xùn)練5在數(shù)列{an}中,a1=1,a2=13,且滿足2anan+1=an1(3an+1an)(n≥2),則an=答案1解析因?yàn)閍1=1,a2=13,2anan+1=an1(3an+1an),顯然an≠0,所以2anan+1=3an+1an1an1an,同除an1anan+1得2an-1=3an1an+1,所以21an-1an-1=1an+11an,所以1an+1-1an1an-1an-1=2,所以1an+1-1an是以2為首項(xiàng),2為公比的等比數(shù)列,所以1an+11an=2×2倒數(shù)法、對(duì)數(shù)法1.形如an+1=an+panan+1的遞推形式兩邊同除以anan+1構(gòu)造等差數(shù)列;2.形如an+1=ankan+b(b3.形如an+1=panr典例6(1)在數(shù)列{an}中,a1=1,an+1=an1+an(n∈N*),則這個(gè)數(shù)列的通項(xiàng)a(2)已知數(shù)列{an}滿足a1=3,an+1=an22an+2,則這個(gè)數(shù)列的通項(xiàng)an=(3)已知數(shù)列{an}滿足nan+1(n+1)an=1(n∈N*),a5=4,則a2025=.

答案(1)1n(2)an=22n-1+1(3解析(1)因?yàn)閍n+1=an1+an,等式兩邊同時(shí)取倒數(shù)得1an+1=1an+1,則1所以1an是首項(xiàng)1a1=1,所以1an=1+(n1)·1=n,所以an=當(dāng)n=1時(shí),a1=11=1也成立綜上所述,an=1n(n∈N*)(2)因?yàn)閍n+1=an22an+2,所以an+11=(an1)則ln(an+11)=ln(an1)2=2ln(an1),又ln(a11)=ln2,所以數(shù)列{ln(an1)}是以ln2為首項(xiàng),2為公比的等比數(shù)列,則ln(an1)=2n1·ln2=ln22所以an=22n-1(3)由nan+1(n+1)an=1(n∈N*)得an+1n+1ann則ann=ann-an-1n-1+an-1n-1-an-2n-2+…+a22-a1由于a5=4,故a55=45=45+a1,所以a故ann=n-1n,所以an=n1,故a2025=特征根法求通項(xiàng)公式典例設(shè)數(shù)列{an}滿足a1=2,an+1=5an-1an+3,求數(shù)列解析考慮方程x=5x-1x+3?(x1)2=0,故1是數(shù)列{an}的不動(dòng)點(diǎn),所以可以嘗試在遞推式兩邊同時(shí)減去1,得到an+1注意到左、右兩邊分別出現(xiàn)了an+11和an1這樣相似的結(jié)構(gòu),并且都是在分子上,我們可以嘗試構(gòu)造新數(shù)列{an1},當(dāng)然也可以直接變形得1an+1-1=an+34(an-1)?1an+1-1=1an-1+14.因此數(shù)列1an-1是首項(xiàng)為1,公差為形如an+1=aan+bcan+d的遞推數(shù)列,處理時(shí)分兩種情況:(1)若其只有一個(gè)不動(dòng)點(diǎn)x0,則1an-x0是等差數(shù)列訓(xùn)練設(shè)數(shù)列{an}滿足a1=3,an+1=4an-2an+1,求數(shù)列解析考慮方程x=4x-2x+1?(x1)(x2)=0,故數(shù)列{an}的不動(dòng)點(diǎn)為1和2,分別在遞推式兩邊同時(shí)減去1和2,得an+11=3(an-1)an+1,an+12=2(an-2)an+1.兩式相除可得an+1一、單選題1.在數(shù)列{an}中,a1=1,an+1=2an2+an,n∈N*,則anA.2n+1 B.2nn+1答案A解析在{an}中,a1=1,由an+1=2an2+an可得1an所以1an是首項(xiàng)為1,公差為1所以1an=1+(n1)·12=n+12,所以2.已知數(shù)列{an}的前n項(xiàng)和為Sn,滿足Sn+2n=2an,則a2022=().A.220222 B.220232C.220242 D.220212答案B解析當(dāng)n=1時(shí),S1+2=2a1,a1=2;當(dāng)n≥2時(shí),Sn1+2(n1)=2an1,Sn+2nSn12(n1)=2an2an1,即an=2an1+2,所以an+2=2(an1+2),an+2a所以{an+2}是以a1+2=4為首項(xiàng),以2為公比的等比數(shù)列.所以an+2=4·2n1,an=2n+12,a2022=220232.3.已知數(shù)列{an}滿足a1=a2=2,an=3an1+4an2(n≥3),則a9+a10=().A.47 B.48C.49 D.410答案C解析由題意得a1+a2=4,由an=3an1+4an2(n≥3),得an+an1=4(an1+an2),即an+an-1an所以數(shù)列{an+an+1}是等比數(shù)列,公比為4,首項(xiàng)為4,an+an+1=4n,所以a9+a10=49.4.已知在數(shù)列{an}中,a1=56,an+1=13an+12n+1,則anA.32n23n B.23n32n答案A解析因?yàn)閍1=56,an+1=13an+所以2n+1·an+1=23·2nan+1,整理得2n+1·an+13=23(2nan所以數(shù)列{2nan3}是以2a13=43為首項(xiàng),23所以2nan3=4323n-1,解得a二、多選題5.(2023·江蘇鎮(zhèn)江考前模擬)已知數(shù)列{an}滿足a1=1,an+1=an2+3an,則下列結(jié)論正確的有A.1an+3為等比數(shù)列 B.{an}的通項(xiàng)公式為aC.{an}為遞增數(shù)列 D.1an的前n項(xiàng)和Tn=2n+23答案ABD解析因?yàn)閍1=1,an+1=an所以1an+1=2+3anan=2an+3又因?yàn)?a1+3=4,所以數(shù)列1an+3是以4為首項(xiàng),2為公比的等比數(shù)列1an+3=4×2n1=2n+1,即an=12n+1因?yàn)閍n+1an=12n+2-312因?yàn)閚≥1,所以2n+23>0,2n+13>0,2n+1>0,所以an+1an<0,所以{an}為遞減數(shù)列,故C錯(cuò)誤;1an=2n+13,則Tn=(22+23+24+…+2n+1)3n=4(1?2n)1?23n=2n+23n6.(2023·江蘇南通海安學(xué)情檢測(cè))數(shù)列{an}中,a1=0,a2=1,an+2=12(an+1+an)(n∈N*),則下列結(jié)論正確的是()A.0≤an≤1 B.{an+1an}是等比數(shù)列C.a8<a10<a9 D.a9<a10<a8答案ABD解析因?yàn)閿?shù)列{an}中,a1=0,a2=1,an+2=12(an+1+an)(n∈N*),所以2(an+2an+1)=(an+1an),即an+2an+1=12(an+1an),則{an+1an}是以1為首項(xiàng),12為公比的等比數(shù)列,所以an+1an=-12由累加法得ana1=-120+-121+…+所以an=23當(dāng)n為奇數(shù)時(shí),an=231?12n-1是遞增數(shù)列,所以0=a1當(dāng)n為偶數(shù)時(shí),an=231+12n-1是遞減數(shù)列,所以23<a所以0≤an≤1,故A正確.又a8=231+127,a10=231+129,a9=231?128,三、填空題7.已知數(shù)列{an}滿足a1=1,n2an+1=2(n+1)2an,則數(shù)列{an}的通項(xiàng)公式為.

答案an=n2·2n1解析因?yàn)閿?shù)列{an}滿足a1=1,n2an+1=2(n+1)2an,則an+1(n+1)因此數(shù)列ann2是以a112=1為首項(xiàng),2為公比的等比數(shù)列,即ann2=2n1,則所以數(shù)列{an}的通項(xiàng)公式是an=n2·2n1.8.若數(shù)列{an}中,a1=3且an+1=an2(n為正整數(shù)),則數(shù)列{an}的通項(xiàng)公式為答案an=3解析由題意可知an>0,將an+1=an2兩邊取對(duì)數(shù)得lgan+1=2lgan,即lgan+1lgan=2,所以數(shù)列{lgan}是以lga1=lg3為首項(xiàng),2為公比的等比數(shù)列,所以lgan=2n1·四、解答題9.已知數(shù)列{an}滿足a1=1,a2=4,an+2=4an+13an(n∈N*),(1)證明:數(shù)列{an+1an}是等比數(shù)列;(2)求數(shù)列{an}的通項(xiàng)公式.解析(1)因?yàn)閍n+2=4an+13an(n∈N*),所以an+2-an+1a又因?yàn)閍1=1,a2=4,則a2a1=3,故數(shù)列{an+1an}是以3為首項(xiàng),3為公比的等比數(shù)列.(2)由(1)知,數(shù)列{an+1an}是以3為首項(xiàng),3為公比的等比數(shù)列,則an+1an=3×3n1=3n,anan1=3n1,an1an2=3n2,…,a3a2=32,a2a1=31,則(anan1)+(an1an2)+…+(a3a2)+(a2a1)=3n1+3n2+…+32+31,即ana1=3×(1?3n-1)1?3,所以a10.已知數(shù)列{an}的前n項(xiàng)和Sn滿足Sn=2an+(1)n,n≥1.(1)求數(shù)列{an}的前3項(xiàng)a1,a2,a3;(2)求數(shù)列{an}的通項(xiàng)公式.解析(1)當(dāng)n=1時(shí),S1=a1=2a1+(1),則a1=1;當(dāng)n=2時(shí),S2=a1+a2=2a2+(1)2,則a2=0;當(dāng)n=3時(shí),S3=a1+a2+a3=2a3+(1)3,則a3=2.綜上可得,a1=1,a2=0,a3=2.(2)由已知得當(dāng)n≥2時(shí),an=SnSn1=2an+(1)n2an1(1)n1,即an=2an1+2(1)n1,上式可化為an+23·(1)n=2an1+23·(1)n1.所以數(shù)列an+23·(-1)n是以a1+23·(1)1則an+23·(1)n=13·2n即an=13·2n123·(1)n=23[2n2(1)故數(shù)列{an}的通項(xiàng)公式為an=23[2n2(1)n]11.(2024·山東新高考聯(lián)考)已知正項(xiàng)數(shù)列{an}的前n項(xiàng)和為Sn,對(duì)一切正整數(shù)n,點(diǎn)Pn(an,Sn)都在函數(shù)f(x)=x+12的圖象上.求數(shù)列{an}的通項(xiàng)公式解析由題意知Sn=a當(dāng)n=