2021高考數(shù)學(xué)(文-江蘇專用)二輪復(fù)習(xí)-專題六-第一講-等差數(shù)列-等比數(shù)列21-【要點(diǎn)導(dǎo)學(xué)】

等差數(shù)列、等比數(shù)列的基本量運(yùn)算例1(2022·重慶卷文)已知{an}是首項(xiàng)為1、公差為2的等差數(shù)列,Sn表示{an}的前n項(xiàng)和.(1)求an及Sn;(2)設(shè){bn}是首項(xiàng)為2的等比數(shù)列,公比q滿足q2-(a4+1)q+S4=0,求數(shù)列{bn}的通項(xiàng)公式及其前n項(xiàng)和Tn.【分析】(1)已知等差數(shù)列的首項(xiàng)和公差,可直接利用公式an=a1+(n-1)d,Sn=na1+d求解.(2)利用(1)的結(jié)果求出a4,S4,解方程q2-(a4+1)q+S4=0,得出等比數(shù)列{bn}的公比q的值,從而可直接由公式bn=b1·qn-1,Tn=求{bn}的通項(xiàng)公式及其前n項(xiàng)和Tn.【解答】(1)由于{an}是首項(xiàng)a1=1,公差d=2的等差數(shù)列,所以an=a1+(n-1)d=2n-1,Sn=1+3+…+(2n-1)===n2.(2)由(1)得a4=7,S4=16.由于q2-(a4+1)q+S4=0,即q2-8q+16=0,所以(q-4)2=0,從而q=4.又b1=2,q=4,所以bn=b1qn-1=2·4n-1=22n-1,從而{bn}的前n項(xiàng)和Tn==(4n-1).變式1(2022·南通期末)已知等差數(shù)列{an}、等比數(shù)列{bn}滿足a1+a2=a3,b1b2=b3,且a3,a2+b1,a1+b2成等差數(shù)列,a1,a2,b2成等比數(shù)列,求數(shù)列{an}和數(shù)列{bn}的通項(xiàng)公式.【解答】(1)設(shè)等差數(shù)列{an}的公差為d,等比數(shù)列{bn}的公比為q.依題意,得 解得a1=d=1,b1=q=2.故an=n,bn=2n.變式2已知等差數(shù)列{an}中的前三項(xiàng)和為12,且2a1,a2,a3+1依次成等比數(shù)列,求數(shù)列{an}的公差d及前n項(xiàng)和Sn.【分析】可以依據(jù)題意列出關(guān)于a1和d的方程組來解得.【解答】設(shè)等差數(shù)列{an}的公差為d,由數(shù)列的前三項(xiàng)和為12,得3a2=12,所以a2=4.由于2a1,a2,a3+1成等比數(shù)列,所以2a1(a3+1)=,即2(a2-d)(a2+d+1)=,即2(4-d)(5+d)=16,所以d2+d-12=0,解得d=-4或d=3.①當(dāng)d=-4時(shí),a1=8,所以Sn=8n+×(-4)=-2n2+10n.②當(dāng)d=3時(shí),a1=1,所以Sn=n+×3=.綜上,當(dāng)d=-4時(shí),Sn=-2n2+10n;當(dāng)d=3時(shí),Sn=.【點(diǎn)評】在等差數(shù)列、等比數(shù)列的運(yùn)算中,常見常用的有六個(gè)基本量,它們分別是a1,d,q,n,an,Sn.把握這六個(gè)基本量之間的各種關(guān)系,結(jié)合嫻熟的運(yùn)算,是正確解決等差數(shù)列與等比數(shù)列基本問題的前提.等差數(shù)列、等比數(shù)列的推斷與證明例2(2022·徐州三模改編)已知數(shù)列{an},{bn}滿足a1=3,anbn=2,bn+1=an,n∈N*,證明數(shù)列是等差數(shù)列,并求數(shù)列{bn}的通項(xiàng)公式.【分析】結(jié)合等差數(shù)列的概念,要證明數(shù)列是等差數(shù)列,就是要證明-是一個(gè)常數(shù).將anbn=2代入bn+1=an轉(zhuǎn)化成-=,即證明數(shù)列是等差數(shù)列.【解答】(1)由于anbn=2,所以an=,則bn+1=anbn-=2-=2-=,所以-=.又a1=3,所以b1=.故是以首項(xiàng)為、公差為的等差數(shù)列,即=+(n-1)×=,所以bn=.【點(diǎn)評】推斷或證明一個(gè)數(shù)列是等差數(shù)列或等比數(shù)列最直接和常用的方法就是定義法.結(jié)合等差數(shù)列、等比數(shù)列的概念,推斷或證明數(shù)列是等差、等比數(shù)列的常用方法有以下三種:一、定義法;二、等比(差)中項(xiàng)法;三、依據(jù)數(shù)列通項(xiàng)特征求證.變式(2022·江西卷)已知數(shù)列{an}的前n項(xiàng)和Sn=,n∈N*.(1)求數(shù)列{an}的通項(xiàng)公式;(2)求證:對任意n>1,都有m∈N*,使得a1,an,am成等比數(shù)列.【解答】(1)當(dāng)n=1時(shí),a1=S1=1.當(dāng)n≥2時(shí),an=Sn-Sn-1=-=3n-2.當(dāng)n=1時(shí),a1=3×1-2=1,所以an=3n-2.(2)若a1,an,am成等比數(shù)列,則=a1am,所以(3n-2)2=3m-2,即3m=(3n-2)2+2=9n2-12n+6,所以m=3n2-4n+2.則對任意n>1,都有3n2-4n+2∈N*,所以對任意n>1,都有m∈N*,使得a1,an,am成等比數(shù)列.等差數(shù)列與等比數(shù)列求和例3已知數(shù)列{an}的前n項(xiàng)和為Sn,且Sn=2an-2(n=1,2,3…),在數(shù)列{bn}中,b1=1,點(diǎn)P(bn,bn+1)在直線x-y+2=0上.(1)求數(shù)列{an},{bn}的通項(xiàng)an和bn;(2)設(shè)cn=an·bn,求數(shù)列{cn}的前n項(xiàng)和Tn.【分析】(1)先由第n項(xiàng)與前n項(xiàng)和的關(guān)系,求出數(shù)列{an}的遞推關(guān)系an=2an-1,再由等比數(shù)列的定義判定數(shù)列{an}是等比數(shù)列,用等比數(shù)列的通項(xiàng)公式,求出數(shù)列{an}的通項(xiàng)公式,由點(diǎn)P(bn,bn+1)在直線x-y+2=0上,得bn+1-bn=2.依據(jù)等差數(shù)列定義知數(shù)列{bn}是等差數(shù)列,所以再依據(jù)等差數(shù)列的通項(xiàng)公式,求出bn的通項(xiàng)公式.(2)由(1)知cn=an·bn是等差數(shù)列與等比數(shù)列對應(yīng)項(xiàng)乘積構(gòu)成的新數(shù)列,其求和用錯位相減法.【解答】(1)由于Sn=2an-2,Sn-1=2an-1-2,又Sn-Sn-1=an(n≥2,n∈N*),所以an=2an-2an-1.由于an≠0,所以=2(n≥2,n∈N*),即數(shù)列{an}是等比數(shù)列.由于a1=S1,所以a1=2a1-2,得a1=2,所以an=2n.由于點(diǎn)P(bn,bn+1)在直線x-y+2=0上,所以bn-bn+1+2=0,所以bn+1-bn=2,即數(shù)列{bn}是以b1=1為首項(xiàng),2為公差的等差數(shù)列,所以bn=2n-1.(2)由于cn=(2n-1)2n,所以Tn=a1b1+a2b2+…+anbn=1×2+3×22+5×23+…+(2n-1)2n①,所以2Tn=1×22+3×23+…+(2n-3)2n+(2n-1)2n+1②.①-②得-Tn=1×2+(2×22+2×23+…+2×2n)-(2n-1)2n+1,即-Tn=1×2+(23+24+…+2n+1)-(2n-1)2n+1,所以Tn=(2n-3)2n+1+6.變式(2022·淮安、宿遷摸底)已知數(shù)列{an}的各項(xiàng)均為正數(shù),其前n項(xiàng)和Sn=(an-1)(an+2),n∈N*.(1)求數(shù)列{an}的通項(xiàng)公式;(2)設(shè)bn=(-1)nanan+1,求數(shù)列{bn}的前2n項(xiàng)和T2n.【解答】(1)當(dāng)n=1時(shí),S1=(a1-1)·(a1+2),解得a1=-1或a1=2.由于a1>0,所以a1=2.當(dāng)n≥2時(shí),Sn=(an-1)(an+2),Sn-1=(an-1-1)(an-1+2).兩式相減得(an+an-1)(an-an-1-1)=0.又由于an>0,所以an+an-1>0,所以an-an-1=1.所以數(shù)列{an}是公差為1的等差數(shù)列,所以