（新教材適用）高中數(shù)學(xué)復(fù)習(xí)課第1課時(shí)數(shù)列課后習(xí)題新人教A版選擇性

第1課時(shí)數(shù)列課后訓(xùn)練鞏固提升1.數(shù)列16,1A.an=1n B.an=nC.an=n3 D.an=解析:數(shù)列16,13,12答案:B2.若lga,lgb,lgc成等差數(shù)列,則()A.b=a+c2 B.C.2b=ac D.2lgb=lg(a+c)解析:若lga,lgb,lgc成等差數(shù)列,則2lgb=lga+lgc=lgac,即b2=ac,故選B.答案:B3.我國(guó)古代數(shù)學(xué)著作《九章算術(shù)》中有如下問題:“今有金棰,長(zhǎng)五尺,斬本一尺,重四斤,斬末一尺,重二斤,問次一尺各重幾何?”意思是:“現(xiàn)有一根金杖,長(zhǎng)5尺,一頭粗,一頭細(xì),在粗的一端截下1尺,質(zhì)量為4斤;在細(xì)的一端截下1尺,質(zhì)量為2斤.問依次每一尺的質(zhì)量各為多少斤.”設(shè)該金杖由粗到細(xì)是均勻變化的,其質(zhì)量為M斤,現(xiàn)將該金杖截成長(zhǎng)度相等的10段,記第i段的質(zhì)量為ai(i=1,2,…,10),且a1<a2<a3<…<a10,若80ai=7M,則i等于()(尺、斤現(xiàn)為非國(guó)際通用單位,1米=3尺,1斤=500克)A.4 B.5 C.6 D.7解析:由題意知,由細(xì)到粗每段的質(zhì)量成等差數(shù)列,設(shè)公差為d,則a可以求得a1=1516,d=1所以金杖總質(zhì)量M=10×1516+10因?yàn)?0ai=7M,所以80×1516+解得i=4.故選A.答案:A4.用數(shù)學(xué)歸納法證明不等式“1+12+13+…+12n>n+22(n≥2,n∈N*)”的過程中,由n=kA.1B.1C.12k+1+D.12k+1+解析:當(dāng)n=k時(shí),左邊=1+12+13+當(dāng)n=k+1時(shí),左邊=1+12+13+…+12k+12k+1故選D.答案:D5.已知等差數(shù)列{an}的前n項(xiàng)和Sn有最大值,且a7a6<1,則滿足Sn>0的最大正整數(shù)nA.6 B.7 C.11 D.12解析:因?yàn)榈炔顢?shù)列{an}的前n項(xiàng)和Sn有最大值,所以公差d<0,因?yàn)閍7a6<1,所以a6>0,a7又因?yàn)閍6+a7a6<0,所以a6+a7<0,所以S11=11(a1+a11)2=11a6>所以滿足Sn>0的最大正整數(shù)n的值為11.答案:C6.定義在(∞,0)∪(0,+∞)上的函數(shù)f(x),若對(duì)于任意給定的等比數(shù)列{an},{f(an)}仍是等比數(shù)列,則稱f(x)為“保等比數(shù)列函數(shù)”.現(xiàn)有定義在(∞,0)∪(0,+∞)上的如下函數(shù):①f(x)=x2;②f(x)=ex;③f(x)=|x|;④f(x)=則其中是“保等比數(shù)列函數(shù)”的是()A.①② B.③④ C.①③ D.②④解析:根據(jù)題意,由等比數(shù)列的性質(zhì),知an·an+2=an+1①f(x)=x2,f(an)f(an+2)=an2an+22=(an+12)2=f2(an+1②f(x)=ex,當(dāng)an=2n1時(shí),f(an)f(an+2)=ean·ean+2=ean+an+2≠(ean+1③f(x)=|x|,f(an)f(an+2)=|an||an+2|=(|an+1|④f(x)=ln|x|,當(dāng)an=2n1時(shí),f(an)f(an+2)=ln|an|·ln|an+2|≠(ln|an+1|)2=f2(an+1),故④不是“保等比數(shù)列函數(shù)”.故選C.答案:C7.(多選題)已知數(shù)列{an}的前n項(xiàng)和為Sn,Sn=2an3,若存在兩項(xiàng)am,an,使得aman=A.數(shù)列{an}為等比數(shù)列B.數(shù)列{an}的通項(xiàng)公式為an=2nC.mn為定值D.設(shè)數(shù)列{bn}的前n項(xiàng)和為Tn,bn=log2an,則數(shù)列Tn解析:數(shù)列{an}的前n項(xiàng)和為Sn,Sn=2an3,當(dāng)n=1時(shí),解得a1=3,當(dāng)n≥2時(shí),Sn1=2an13,所以an=SnSn1=2an2an1,整理得an=2an1,即anan-1=所以an=3·2n1,又a1=3符合上式,所以an=3·2n1,故選項(xiàng)A正確,選項(xiàng)B錯(cuò)誤.由于an=3·2n1,故存在兩項(xiàng)am,an,使得aman=128,即2mn=27,故mn=7因?yàn)閎n=log2an=log23+(n1),所以Tn=nlog23+n(n-1)2,所以Tnn=log故選項(xiàng)D正確.故選ACD.答案:ACD8.設(shè)Sn為等比數(shù)列{an}的前n項(xiàng)和,8a2a5=0,則公比q=,S4S2=解析:∵等比數(shù)列{an}中,8a2a5=0,∴a5a2=a1q∴S4S2=a1(1答案:259.已知等比數(shù)列{an}的前n項(xiàng)和為Sn,若S10=10,S30=30,則S20=.

解析:根據(jù)題意,在等比數(shù)列{an}中,設(shè)其公比為q,若S10=10,則有S10=a1+a2+a3+…+a10=10,S30=30,則有S30=a1+a2+a3+…+a30=(a1+a2+a3+…+a10)+(a11+a12+a13+…+a20)+(a21+a22+a23+…+a30)=(1+q10+q20)(a1+a2+a3+…+a10)=30,則有1+q10+q20=3,解得q10=1或2(舍),則S20=a1+a2+a3+…+a20=(a1+a2+a3+…+a10)+(a11+a12+a13+…+a20)=(1+q10)(a1+a2+a3+…+a10)=20.答案:2010.若數(shù)列{an}滿足an+2an+1+an+1an=q(q為常數(shù)),則稱數(shù)列{an}為等比和數(shù)列,q稱為公比和,已知數(shù)列{an}是以3為公比和的等比和數(shù)列,其中a1=解析:由題意可得a1=1,a2=2,a3a故a3=2,又a4a3+a3同理,a5=4,a6=8,a7=8,……,總結(jié)規(guī)律:當(dāng)n=2k1(k∈N*)時(shí),an=2k1,當(dāng)n=2k(k∈N*)時(shí),an=2k,故當(dāng)n=2022時(shí),k=1011,a2022=21011.答案:2101111.若各項(xiàng)均為正數(shù)的數(shù)列{an}的前n項(xiàng)和Sn滿足an+12=2Sn+n+2(n∈N*),且a3+a5(1)判斷數(shù)列{an}是否為等差數(shù)列,并說明理由;(2)求數(shù)列{an}的通項(xiàng)公式;(3)若bn=2nan,求數(shù)列{bn}的前n項(xiàng)和Tn.解:(1)數(shù)列{an}不是等差數(shù)列.理由如下:因?yàn)閍n+12=2Sn+n+2,所以當(dāng)n≥2時(shí),an2=2Sn1+(n1)+2,所以an+12-an2=2an+1,即an+21=因?yàn)閍n>0,所以an+1=an+1,即an+1an=1,所以當(dāng)n≥2時(shí),{an}是公差d=1的等差數(shù)列.因?yàn)閍3+a5=10,所以a4=5,所以a2=3.當(dāng)n=1時(shí),a22=2a1+1+2,所以a1=3,因?yàn)閍2a1=0所以數(shù)列{an}不是等差數(shù)列.(2)由(1)知,數(shù)列{an}從第2項(xiàng)開始是等差數(shù)列,當(dāng)n≥2時(shí),an=n+1,故數(shù)列{an}的通項(xiàng)公式an=3(3)由(2)得bn=6當(dāng)n≥2時(shí),Tn=6+3×22+4×23+…+n·2n1+(n+1)·2n,①2Tn=12+3×23+4×24+…+n·2n+(n+1)·2n+1,②①②得,Tn=6+23+24+…+2n(n+1)·2n+1,所以Tn=6(23+24+…+2n)+(n+1)·2n+1=(n+1)·2n+123(1-2n-2)當(dāng)n=1時(shí),T1=6,滿足上式,故Tn=n·2n+1+2.12.已知Sn為數(shù)列{an}的前n項(xiàng)和,2Sn+2n=3an(n∈N*).(1)求數(shù)列{an}的通項(xiàng)公式;(2)設(shè)bn=1+anan·an+1,數(shù)列{bn}的前n項(xiàng)和為(1)解:∵2Sn+2n=3an,∴2Sn+1+2(n+1)=3an+1,兩式相減得an+1=3an+2,∴an+1+1=3(an+1).∵2S1+2=3a1,∴a1=2.∴數(shù)列{an+1}是以3為首項(xiàng),3為公比的等比數(shù)列,∴an+1=3n,∴an=3n1.(2)證明:∵bn=3n∴Tn=1213-113.已知函數(shù)f(x)=x3-2x,在數(shù)列{an}中,若