【全程復(fù)習(xí)方略】2022屆高考數(shù)學(xué)(文科人教A版)大一輪單元評(píng)估檢測(cè)(五)第五章-

溫馨提示：此套題為Word版，請(qǐng)按住Ctrl,滑動(dòng)鼠標(biāo)滾軸，調(diào)整合適的觀看比例，答案解析附后。關(guān)閉Word文檔返回原板塊。單元評(píng)估檢測(cè)(五)第五章(120分鐘150分)一、選擇題(本大題共10小題,每小題5分,共50分.在每小題給出的四個(gè)選項(xiàng)中,只有一項(xiàng)是符合題目要求的)1.(2021·臨沂模擬)等差數(shù)列{an},{bn}中,a1=25,b1=75,a100+b100=100,則數(shù)列{an+bn}的前100項(xiàng)的和為()A.0 B.100 C.1000 D.10000【解析】選D.由等差數(shù)列的定義知{an+bn}也是等差數(shù)列,所以要求的和為S=100(25+75+100)2.已知數(shù)列{an}的通項(xiàng)公式為an=411-2n,則滿足an+1<an的n的取值為A.3 B.4 C.5 D.6【解析】選C.由an+1<an,得an+1-an=49-2n-411-2n=8(9-2n)(11-2n)<0,解得9n∈N*,所以n=5.3.(2021·鄭州模擬)若數(shù)列{an}的通項(xiàng)公式是an=(-1)n(2n-1),則a1+a2+a3+…+a100=()A.-200 B.-100 C.200 D.100【解析】選D.由題意知,a1+a2+a3+…+a100=-1+3-5+7+…+(-1)100(2×100-1)=(-1+3)+(-5+7)+…+(-197+199)=2×50=100.4.(2021·衡水模擬)已知數(shù)列{an}的通項(xiàng)公式為an=pn+qn(p,q為常數(shù)),且a2=32,a4=32,則aA.54 B.94 C.34【解析】選B.由題意知2p+q所以a8=8p+q8=8×14+285.已知數(shù)列32,54,76,9a-b,aA.(19,3) B.(19,-3)C.192,32【解析】選C.由a-b=8,a+b=11解得a=192,b=3【方法技巧】數(shù)列通項(xiàng)公式的一般求法用觀看—?dú)w納—猜想—證明的方法,找數(shù)列的通項(xiàng)公式時(shí),首先要留意觀看各個(gè)式子的特征,并據(jù)此把式子分解成幾個(gè)部分,然后各個(gè)擊破,對(duì)于正負(fù)相間的項(xiàng),用-1的指數(shù)式來(lái)予以調(diào)整.6.已知數(shù)列{an}為等差數(shù)列,數(shù)列{bn}是各項(xiàng)均為正數(shù)的等比數(shù)列,且公比q>1,若a5=b5,a2021=b2021,則a1010與b1010的大小關(guān)系是()A.a1010=b1010 B.a1010>b1010C.a1010<b1010 D.無(wú)法推斷【解析】選B.由題意知a1010=a5+a20152>a7.數(shù)列{an}的通項(xiàng)an=nn2+90,則數(shù)列{aA.310 B.19 C.119 D.【解析】選C.由于an=1n+1n+90n≤1290,由于n∈N*8.假如一個(gè)數(shù)列{an}滿足an+1+an=h(h為常數(shù),n∈N*),則稱數(shù)列{an}為等和數(shù)列,h為公和,Sn是其前n項(xiàng)和.已知等和數(shù)列{an}中,a1=1,h=-3,則S2021等于()A.3020 B.3021 C.-3020 【解析】選C.由公和h=-3,a1=1,得a2=-4,并且數(shù)列{an}是以2為周期的數(shù)列,則S2021=1007(a1+a2)+a1=-3021+1=-3020.【加固訓(xùn)練】設(shè)Sn為數(shù)列{an}的前n項(xiàng)和,若S2nSn(n∈N*)是非零常數(shù),則稱該數(shù)列為“和等比數(shù)列”;若數(shù)列{cn}是首項(xiàng)為2,公差為d(d≠0)的等差數(shù)列,且數(shù)列{cn}是“和等比數(shù)列”【解析】由題意可知,數(shù)列{cn}的前n項(xiàng)和為Sn=n(c1+cn)2,前2n項(xiàng)和為S2n=2n(c1+c2n)2,所以答案:49.(2021·煙臺(tái)模擬)已知數(shù)列{an}的前n項(xiàng)之和是Sn,且4Sn=(an+1)2,則下列說(shuō)法正確的是()A.數(shù)列{an}為等差數(shù)列B.數(shù)列{an}為等比數(shù)列C.數(shù)列{an}為等差或等比數(shù)列D.數(shù)列{an}可能既不是等差數(shù)列也不是等比數(shù)列【解析】選C.當(dāng)n=1時(shí),4a1=(a1+1)2,解得a1=1;當(dāng)n≥2時(shí),an=Sn-Sn-1=(an+1即4an=an2-an-12+2a即(an+an-1)(an-an-1-2)=0,所以an+an-1=0或an-an-1=2,當(dāng)an=-an-1時(shí),{an}是等比數(shù)列,當(dāng)an-an-1=2時(shí),{an}是等差數(shù)列.10.(2021·合肥模擬)設(shè){an}是公比為q的等比數(shù)列,其前n項(xiàng)積為T(mén)n,并滿足條件a1>1,a99a100-1>0,a99-1a100-1<0,給出下列結(jié)論:③a99a101<1;④使Tn<1成立的最小自然數(shù)n等于199,則其中正確的是A.①②③ B.①③④C.②③ D.①②③④【解析】選B.由a1>1,a99a100>1得q>0,又a99-1a100-1<0知:a99-1>0,a100-1<0,且0<q<1,①式成立;又T198=a1·a2·…·a198=(a99a100)99>1,②式不成立;a99·a101=a1002<1,③成立;由于T198>1,T199=a1·a2·…·a199二、填空題(本大題共5小題,每小題5分,共25分.請(qǐng)把正確答案填在題中橫線上)11.(2021·天水模擬)已知數(shù)列{an}滿足:a1=1,an=1n=2,3,4,…,設(shè)bn=a

2n-1+1,n=1,2,3,…,則數(shù)列{bn【解析】由題意得,對(duì)于任意的正整數(shù)n,bn=a2所以bn+1=a2又a2n+1=(2a2n2所以bn+1=2bn,又b1=a1+1=2,所以{bn}是首項(xiàng)為2,公比為2的等比數(shù)列,所以bn=2n.答案:bn=2n【加固訓(xùn)練】若數(shù)列{an}滿足a1=1,an+1=an+2n,則an=.【解析】由已知an+1-an=2n,故有a2-a1=2,a3-a2=22,a4-a3=23,…,an-an-1=2n-1.以上n-1個(gè)式子兩邊分別相加,則有an-a1=2+22+23+…+2n-1=2(1-2n-1所以an=2n-2+a1=2n-1.答案:2n-112.(2021·菏澤模擬)已知數(shù)列{an}滿足a1=0,an+1=an-33an+1(n∈【解析】由題意知,a1=0,a2=-3,a3=3,a4=0,a5=-3,a6=3,從而數(shù)列{an}的各項(xiàng)呈周期性排列,周期為3,所以a20=a2=-3.答案:-3【加固訓(xùn)練】(2021·成都模擬)數(shù)列{an}是等差數(shù)列,a1=f(x+1),a2=0,a3=f(x-1),其中f(x)=x2-4x+2,則此數(shù)列的前n項(xiàng)和Sn=.【解析】由題意可得f(x+1)+f(x-1)=0,即(x+1)2-4(x+1)+2+(x-1)2-4(x-1)+2=0,解得:x=1或x=3,當(dāng)x=1時(shí),此時(shí)a1=-2,a2=0,a3=2,則d=2,所以Sn=n2-3n.當(dāng)x=3時(shí),a1=2,a2=0,a3=-2,則d=-2,所以Sn=-n2+3n.答案:n2-3n或-n2+3n13.如圖所示是一個(gè)樹(shù)形圖的生長(zhǎng)過(guò)程,依據(jù)圖中所示的生長(zhǎng)規(guī)律,第15行的實(shí)心圓點(diǎn)的個(gè)數(shù)是.【解析】觀看圖中所示的生長(zhǎng)規(guī)律,發(fā)覺(jué):1個(gè)空心圓點(diǎn)到下一行僅生長(zhǎng)出1個(gè)實(shí)心圓點(diǎn),而1個(gè)實(shí)心圓點(diǎn)到下一行生長(zhǎng)出1個(gè)實(shí)心圓點(diǎn)和1個(gè)空心圓點(diǎn).假如設(shè)第n行的實(shí)心圓點(diǎn)的個(gè)數(shù)是an,空心圓點(diǎn)的個(gè)數(shù)是bn,則a1=0,b1=1,an+1=an+bn,bn+1=an,n∈N*.所以a1=0,a2=1,an+1=an+an-1,從而{an}為:0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,…,故填377.答案:377【加固訓(xùn)練】(2021·揚(yáng)州模擬)如圖所示是畢達(dá)哥拉斯(Pythagoras)的生長(zhǎng)程序:正方形上連接著等腰直角三角形,等腰直角三角形邊上再連接正方形,…,如此連續(xù),若共得到1023個(gè)正方形,設(shè)初始正方形的邊長(zhǎng)為22,則最小正方形的邊長(zhǎng)為【解析】設(shè)1+2+4+…+2n-1=1023,即1-2n1-2=1023,2n=1024,n=10.正方形邊長(zhǎng)構(gòu)成數(shù)列22,222,223答案:114.(2021·蚌埠模擬)已知正項(xiàng)等比數(shù)列{an}滿足a2021=2a2021+a2022,若存在兩項(xiàng)am,an使得aman=4a1,則1m+【解析】設(shè)公比為q,則a2021q2=2a2021+a2021q,由于an>0,所以q2-q-2=0,q=2或q=-1(舍去),aman=a1·2m-1·a1·2n-1=4a1,2m+n-2=24,即m+n=6,所以1m+答案:315.已知數(shù)列{2n-1·an}的前n項(xiàng)和Sn=9+2n,則數(shù)列{an}的通項(xiàng)公式為an=.【解析】由于Sn=9+2n①,所以當(dāng)n≥2時(shí),Sn-1=9+2(n-1)②,①-②得2n-1an=2,所以an=22n-1=22-n.當(dāng)n=1時(shí),a1=S1=9+2=11,不符合上式,所以an答案:11【方法技巧】含Sn,an問(wèn)題的求解策略當(dāng)已知含有Sn+1,Sn之間的等式時(shí),或者含有Sn,an的混合關(guān)系的等式時(shí),可以接受降級(jí)角標(biāo)或者升級(jí)角標(biāo)的方法再得出一個(gè)等式,兩個(gè)等式相減就把問(wèn)題轉(zhuǎn)化為數(shù)列的項(xiàng)之間的遞推關(guān)系式.【加固訓(xùn)練】在數(shù)列{an}中,Sn為{an}的前n項(xiàng)和,n(an+1-an)=an(n∈N*),且a3=π,則tanS4=.【解析】方法一:由n(an+1-an)=an,得nan+1=(n+1)an,則3a4=4a3,又a3=π,故a4=4π3,又由2a3=3a2,得a2=2π3,由a2=2a1,得a1=π3,故S4=a1+a2+a3+a4=10π3方法二:由n(an+1-an)=an,得nan+1=(n+1)an,即an+1n+1=ann,所以ann=an-1n-1=所以an=π3故S4=a1+a2+a3+a4=π3(1+2+3+4)=10tanS4=tan10π3=答案:3三、解答題(本大題共6小題,共75分.解答時(shí)應(yīng)寫(xiě)出必要的文字說(shuō)明、證明過(guò)程或演算步驟)16.(12分)(2022·江西高考)已知數(shù)列an的前n項(xiàng)和Sn=3n2-n2(1)求數(shù)列an(2)證明:對(duì)任意的n>1,都有m∈N*,使得a1,an,am成等比數(shù)列.【解題提示】(1)利用an=Sn-Sn-1(n≥2)解決.(2)a1,an,am成等比數(shù)列,轉(zhuǎn)化為an2=a1·a【解析】(1)當(dāng)n=1時(shí)a1=S1=1;當(dāng)n≥2時(shí)an=Sn-Sn-1=3n2-n對(duì)n=1也滿足,所以an的通項(xiàng)公式為an(2)由(1)得a1=1,an=3n-2,am=3m-2,要使a1,an,am成等比數(shù)列,需要an2=a1·a所以(3n-2)2=3m-2,整理得m=3n2-4n+2∈N*,所以對(duì)任意n>1,都有m∈N*使得an2=a1·am即a1,an,am成等比數(shù)列.【加固訓(xùn)練】已知等比數(shù)列{an}的首項(xiàng)為1,公比q≠1,Sn為其前n項(xiàng)和,a1,a2,a3分別為某等差數(shù)列的第一、其次、第四項(xiàng).(1)求an和Sn.(2)設(shè)bn=log2an+1,數(shù)列1bnbn+2的前n項(xiàng)和為T(mén)n,求證:T【解析】(1)由于a1,a2,a3為某等差數(shù)列的第一、其次、第四項(xiàng),所以a3-a2=2(a2-a1),所以a1q2-a1q=2(a1q-a1),由于a1=1,所以q2-3q+2=0,由于q≠1,所以q=2,所以an=a1qn-1=2n-1.所以Sn=a1(1-qn)(2)由(1)知an+1=2n,所以bn=log2an+1=log22n=n.所以1bn·bn+2所以Tn=12111213-15+1214-1=1=34-12117.(12分)(2021·龍巖模擬)在等比數(shù)列{an}中,a1=2,公比q=2.(1)設(shè)Sn為{an}的前n項(xiàng)和,證明:Sn+2=2an.(2)在△ABC中,角A,B,C所對(duì)的邊分別是3,a1,a2,求△ABC的面積.【解析】(1)由于a1=2,公比q=2,所以an=a1qn-1=2n,Sn=a1(1-qn)所以Sn+2=2n+1=2an.(2)由(1)得a2=4,在△ABC中,由余弦定理得cosC=32+a12所以sinC=1-cos2所以△ABC的面積為S△ABC=12absinC=318.(12分)(2021·日照模擬)設(shè)數(shù)列{an}是等差數(shù)列,且首項(xiàng)a1=3,a8-a3=10,Sn為數(shù)列{an}的前n項(xiàng)和.(1)求數(shù)列{an}的通項(xiàng)公式及Sn.(2)若數(shù)列4an2-1的前n項(xiàng)和為T(mén)【解析】(1)設(shè)等差數(shù)列{an}的公差為d,由于首項(xiàng)a1=3,a8-a3=10,所以5d=10,解得d=2,故an=a1+(n-1)d=3+2(n-1)=2n+1,Sn=n(3+2n+1)2=n(2)由于4an2-1=4(2n+1)2所以Tn=1-12+12-13+…19.(12分)設(shè)數(shù)列{an}的前n項(xiàng)和Sn=aqn+b(a,b為非零實(shí)數(shù),q≠0且q≠1).(1)當(dāng)a,b滿足什么關(guān)系時(shí),{an}是等比數(shù)列?(2)若{an}為等比數(shù)列,證明:以(an,Sn)為坐標(biāo)的點(diǎn)都落在同一條直線上.【解析】(1)由于Sn=aqn+b,所以n≥2時(shí),an=Sn-Sn-1=(aqn+b)-(aqn-1+b)=a(q-1)qn-1,所以an+1an故若數(shù)列{an}是等比數(shù)列,只要a1=S1=aq+b符合an=a1(q-1)qn-1的形式即可.所以aq+b=a(q-1)·q0,所以a+b=0.所以當(dāng)a+b=0時(shí),數(shù)列{an}是等比數(shù)列.(2)當(dāng){an}是等比數(shù)列時(shí),Sn=aqn-a,an=a(q-1)qn-1,a1=a(q-1),所以Sn-=aq(qn-1-1)a(q-1)(即以(an,Sn)為坐標(biāo)的點(diǎn)都落在恒過(guò)點(diǎn)(a1,S1)且斜率k=qq-1【加固訓(xùn)練】設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,其中an≠0,a1為常數(shù),且-a1,Sn,an+1成等差數(shù)列.(1)求{an}的通項(xiàng)公式.(2)設(shè)bn=1-Sn,問(wèn):是否存在a1,使數(shù)列{bn}為等比數(shù)列?若存在,求出a1的值;若不存在,請(qǐng)說(shuō)明理由.【解析】(1)依題意,得2Sn=an+1-a1,當(dāng)n≥2時(shí),有2Sn=an+1-a又由于a2=2S1+a1=3a1,an≠0,所以數(shù)列{an}是首項(xiàng)為a1,公比為3的等比數(shù)列.因此an=a1·3n-1(n∈N*).(2)由于Sn=a1(1-3n)所以bn=1-Sn=1+a12-要使{bn}為等比數(shù)列,當(dāng)且僅當(dāng)1+12a1=0,即a1=-2,所以存在a1=-2,使數(shù)列{bn20.(13分)(2021·廈門(mén)模擬)已知遞增的等比數(shù)列{an}的前n項(xiàng)和Sn滿足:S4=S1+28,且a3+2是a2和a4的等差中項(xiàng).(1)求數(shù)列{an}的通項(xiàng)公式.(2)若bn=anlog

12an,Tn=b1+b2+…+bn,求使Tn+n·【解析】(1)設(shè)等比數(shù)列{an}的首項(xiàng)為a1,公比為q,由題知a解得q=2,a1由于{an}是遞增數(shù)列,故an=2n.(2)bn=anlog12an=2nlog

122n由于bn+1-bn=-(n+1)2n+1+n·2n=bn-2n+1,bn-bn-1=bn-1-2n,…,b3-b2=b2-23,b2-b1=b1-22,b1=-2