6.5構(gòu)造法求通項(xiàng)

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(guò)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=.

2.(對(duì)接教材)若數(shù)列{an}的首項(xiàng)a1=1,且滿(mǎn)足an+1=2an+1,則an=.

3.(對(duì)接教材)已知數(shù)列{an}的首項(xiàng)a1=35,且滿(mǎn)足an+1=3an2a4.(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.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(1)求{an}的通項(xiàng)公式.(2)證明:1a1+1aan+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{aan+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(mǎn)足a1=1,an+1=2an+n1,求數(shù)列{an}的通項(xiàng)公式.an+1=pan+rqn形式1.類(lèi)型一:p=q典例3已知數(shù)列{an}的首項(xiàng)a1=12,滿(mǎn)足an+1=12an+12n(n∈N2.類(lèi)型:p≠q典例4已知數(shù)列{an}滿(mǎn)足an+1=2an+4×3n1,a1=1,求數(shù)列{an}的通項(xiàng)公式.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(mǎn)足a1=2,an+12an=3n1.證明:{an+13an}是等比數(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{a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(mǎn)足2anan+1=an1(3an+1an)(n≥2),則an=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=an3.形如an+1=pan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(mǎn)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(mǎn)足nan+1(n+1)an=1(n∈N*),a5=4,則a2025=.

特征根法求通項(xiàng)公式典例設(shè)數(shù)列{an}滿(mǎn)足a1=2,an+1=5an-1形如an+1=aan+bcan+d訓(xùn)練設(shè)數(shù)列{an}滿(mǎn)足a1=3,an+1=4an-2一、單選題1.在數(shù)列{an}中,a1=1,an+1=2an2+anA.2n+1 B.2nn+1C.n+12n2.已知數(shù)列{an}的前n項(xiàng)和為Sn,滿(mǎn)足Sn+2n=2an,則a2022=().A.220222 B.220232C.220242 D.2202123.已知數(shù)列{an}滿(mǎn)足a1=a2=2,an=3an1+4an2(n≥3),則a9+a10=().A.47 B.48C.49 D.4104.已知在數(shù)列{an}中,a1=56,an+1=13an+12A.32n23n B.23n32n二、多選題5.(2023·江蘇鎮(zhèn)江考前模擬)已知數(shù)列{an}滿(mǎn)足a1=1,an+1=anA.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+26.(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*A.0≤an≤1 B.{an+1an}是等比數(shù)列C.a8<a10<a9 D.a9<a10<a8三、填空題7.已知數(shù)列{an}滿(mǎn)足a1=1,n2an+1=2(n+1)2an,則數(shù)列{an}的通項(xiàng)公式為.

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)公式為四、解答題9.已知數(shù)列{an}滿(mǎn)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.10.已知數(shù)列{an}的前n項(xiàng)和Sn滿(mǎn)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.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{a12.(2023·河北石家莊部分重點(diǎn)高中月考)已知數(shù)列