2023新教材數(shù)學(xué)高考第一輪專題練習(xí)--專題七數(shù)列求和、數(shù)列的綜合專題檢測題組

1、2023新高考數(shù)學(xué)第一輪專題練習(xí)7.4數(shù)列求和、數(shù)列的綜合一、選擇題1.(2022屆河南期中聯(lián)考,8)設(shè)數(shù)列an和bn的前n項(xiàng)和分別為Sn,Tn,已知數(shù)列bn是等差數(shù)列,且bn=an+n2an,a3=3,b4+b5=11,則Sn+Tn=()A.n2-2nB.2n2-nC.2n2+nD.n2+2n答案D由a3=3,得b3=a3+32a3=3+323=4,設(shè)等差數(shù)列bn的公差為d,所以b1+2d=4,2b1+7d=11,解得b1=2,d=1,所以bn=2+(n-1)×1=n+1,則bn=an+n2an=n+1,所以an=n.所以Sn=n(1+n)2=n22+n2,Tn=n(2+n+1)2

2、=n22+3n2,則Sn+Tn=n2+2n.故選D.2.(2021江西上饒二模,9)函數(shù)f(x)=2sin x-x(x>0)的所有極大值點(diǎn)從小到大排成數(shù)列an,設(shè)Sn是數(shù)列an的前n項(xiàng)和,則cos S2 021=()A.1B.12C.-12D.0答案Bf '(x)=2cos x-1(x>0),且f '(x)是周期為2的周期函數(shù).當(dāng)x(0,2時(shí), f '(x)=0的根為x=3和x=53. f '(x)的圖象如圖,易知, f(x)在(0,2上的極大值點(diǎn)為3.即an是以3為首項(xiàng),2為公差的等差數(shù)列,故S2 021=2 021×3+2 021

3、15;2 0202×2=672+53+2 021×2 020,則cos S2 021 =cos53=12,故選B.3.(2022屆山西長治第二中學(xué)月考,10)已知數(shù)列an、bn的前n項(xiàng)和分別為An、Bn,記cn=anBn+bnAn-anbn(n1).則數(shù)列cn的前10項(xiàng)和為()A.A10+B10B.A10+B102C.A10B10D.A10B10答案C當(dāng)n=1時(shí),c1=a1b1;當(dāng)n2時(shí),cn=(An-An-1)Bn+(Bn-Bn-1)An-(An-An-1)(Bn-Bn-1)=AnBn-An-1Bn-1.故c1+c2+c10=A1B1+(A2B2-A1B1)+(A3B3-

4、A2B2)+(A10B10-A9B9)=A10B10.故選C.4.(2022屆安徽安慶月考,9)已知等差數(shù)列an的公差為2,前n項(xiàng)和為Sn,且S1,S2,S4成等比數(shù)列.令bn=1anan+1,則數(shù)列bn的前50項(xiàng)和T50=()A.5051B.4950C.100101D.50101答案DS1=a1,S2=2a1+2×12×2=2a1+2,S4=4a1+4×32×2=4a1+12.由題意得(2a1+2)2=a1(4a1+12),解得a1=1,所以an=2n-1,則bn=1(2n-1)(2n+1)=1212n-1-12n+1,則T50=121-13+13-1

5、5+15-17+199-1101=50101.5.(2021甘肅名校模擬,12)已知數(shù)列an的前n項(xiàng)和Sn=2n+1-2,若nN*,an4+S2n恒成立,則實(shí)數(shù)的最大值是()A.3B.4C.5D.6答案C數(shù)列an的前n項(xiàng)和Sn=2n+1-2,當(dāng)n2時(shí),an=Sn-Sn-1=(2n+1-2)-(2n-2)=2n;當(dāng)n=1時(shí),a1=S1=22-2=2滿足上式,所以an=2n(nN*).又nN*,an4+S2n恒成立,所以nN*,4+S2nan恒成立.令bn=4+S2nan=4+22n+1-22n=22n+1+22n=2n+1+22n,則bn+1-bn=2n+2+22n+1-2n+1+22n=2n+

6、1-12n>0對任意nN*都成立,所以bn=2n+1+22n單調(diào)遞增,因此(bn)min=b1=22+22=5,即4+S2nan的最小值為5,所以5,即實(shí)數(shù)的最大值是5.6.(多選)(2020百師聯(lián)盟練習(xí)題五,6)對于數(shù)列an ,令bn=an-1an,下列說法正確的是()A.若數(shù)列an是單調(diào)遞增數(shù)列,則數(shù)列bn也是單調(diào)遞增數(shù)列B.若數(shù)列an是單調(diào)遞減數(shù)列,則數(shù)列bn也是單調(diào)遞減數(shù)列C.若an=3n-1,則數(shù)列bn有最小值D.若an=1-12n,則數(shù)列bn有最大值答案CD如果a1=-1,a2=1,則b1=b2=0,A不正確;如果a1=1,a2=-1,則b1=b2=0,B不正確;函數(shù)f(x)

7、=x-1x在(0,+)上為增函數(shù),若an=3n-1,則an為遞增數(shù)列,當(dāng)n=1時(shí),an取最小值,a1=2>0,從而數(shù)列bn有最小值,C正確;若an=1-12n,當(dāng)n=1時(shí),an取最大值32,且an>0,從而數(shù)列bn有最大值,D正確.故選CD.7.(多選)(2021長沙一中月考(三),12)將n2個(gè)數(shù)排成n行n列的一個(gè)數(shù)陣,如:a11a12a13a1na21a22a23a2na31a32a33a3nan1an2an3ann該數(shù)陣第一列的n個(gè)數(shù)從上到下構(gòu)成以m為公差的等差數(shù)列,每一行的n個(gè)數(shù)從左到右構(gòu)成以m為公比的等比數(shù)列(其中m>0).已知a11=2,a13=a61+1,記這n

8、2個(gè)數(shù)的和為S.下列結(jié)論正確的有()A.m=3B.a67=17×37C.aij=(3i-1)×3j-1D.S=14n(3n+1)(3n-1)答案ACD由該數(shù)陣第一列的n個(gè)數(shù)從上到下構(gòu)成以m為公差的等差數(shù)列,每一行的n個(gè)數(shù)從左到右構(gòu)成以m為公比的等比數(shù)列,且a11=2,a13=a61+1,可得a13=a11m2=2m2,a61=2+5m,所以2m2=2+5m+1,解得m=3或m=-12(舍去),所以A中結(jié)論正確;a67=a61m6=(2+5×3)×36=17×36,所以B中結(jié)論不正確;因?yàn)閍ij=ai1mj-1=a11+(i-1)×m&

9、#215;mj-1=2+(i-1)×3×3j-1=(3i-1)×3j-1,所以C中結(jié)論正確;S=(a11+a12+a1n)+(a21+a22+a2n)+(an1+an2+ann)=a11(1-3n)1-3+a21(1-3n)1-3+an1(1-3n)1-3=12(3n-1)·(2+3n-1)n2=14n(3n+1)(3n-1),所以D中結(jié)論正確,故選ACD.二、填空題8.(2021皖江名校聯(lián)盟考試,16)已知Sn是各項(xiàng)均不為零的等差數(shù)列an的前n項(xiàng)和,且S2n-1=an2(nN*),若存在nN*,使不等式1a1a2a3+1a2a3a4+1a3a4a5+1

10、anan+1an+214n2+12n成立,則實(shí)數(shù)的最大值是. 答案445解析S2n-1=(2n-1)(a1+a2n-1)2=(2n-1)an=an2,又an0,an=2n-1,nN*.等差數(shù)列an的首項(xiàng)a1=1,公差d=2.1anan+1an+2=141anan+1-1an+1an+2,原式=141a1a2-1a2a3+1a2a3-1a3a4+1anan+1-1an+1an+2=141a1a2-1an+1an+2=n2+2n3(2n+1)(2n+3).即n2+2n3(2n+1)(2n+3)14n2+12n,存在nN*,使43(2n+1)(2n+3)成立,43(2n+1)(2n+3)m

11、ax,當(dāng)n=1時(shí),43×(2+1)×(2+3)=445,故實(shí)數(shù)的最大值是445.9.(2022屆湖南名校10月聯(lián)考,16)已知正項(xiàng)數(shù)列an的前n項(xiàng)和為Sn,且2Sn=an2+an.若bn=(-1)n2n+12Sn,則數(shù)列bn的前2 021項(xiàng)和為. 答案-2 0232 022解析當(dāng)n=1時(shí),2a1=a12+a1,因?yàn)閍n>0,所以a1=1.當(dāng)n2時(shí),2Sn-1=an-12+an-1,所以2(Sn-Sn-1)=an2-an-12+an-an-1,整理得(an+an-1)(an-an-1-1)=0.因?yàn)閍n>0,所以an-an-1=1,所以數(shù)列an是首項(xiàng)為1

12、,公差為1的等差數(shù)列,則an=n,Sn=n(n+1)2,所以bn=(-1)n2n+1n(n+1)=(-1)n1n+1n+1,所以b1+b2+b3+b2 020+b2 021=-1-12+12+13-13-14+12 020+12 021-12 021-12 022=-1-12 022=-2 0232 022.三、解答題10.(2022屆河南尖子生二診,19)等比數(shù)列an的首項(xiàng)a1=2,前n項(xiàng)和記作Sn,且Sn+1也是等比數(shù)列,公比為q.(1)求an的通項(xiàng)公式;(2)記bn=log2an,求數(shù)列n+1bnbn+1的前n項(xiàng)和Tn.解析(1)由已知得Sn+1=(S1+1)qn-1,又S1=a1=2,

13、Sn=3×qn-1-1.當(dāng)n2時(shí),an=Sn-Sn-1=3(q-1)qn-2.an為等比數(shù)列,an+1an=3(q-1)qn-13(q-1)qn-2=q(n2,nN*),a2a1=q,即3(q-1)2=q,q=3,數(shù)列an的公比也為q,且q=3,an=2×3n-1.(2)bn=log2(2×3n-1)=1+(n-1)log23,bn為等差數(shù)列,首項(xiàng)為1,公差d為log23.1bnbn+1=1d1bn-1bn+1=1bn-1bn+1·log32,Tn=(1+2+3+n)+1b1-1b2+1b2-1b3+1bn-1bn+1·log32=(1+n)n

14、2+1b1-1bn+1·log32=(1+n)n2+1-11+nlog23· log32=(1+n)n2+n1+nlog23.11.(2022屆河南三市聯(lián)考,17)設(shè)等差數(shù)列an的前n項(xiàng)和為Sn,點(diǎn)(n,Sn)在函數(shù)f(x)=x2+Bx+C-1(B,CR)的圖象上,且a1=C.(1)求數(shù)列an的通項(xiàng)公式;(2)若bn=an(a2n-1+1),求數(shù)列bn的前n項(xiàng)和Tn.解析(1)設(shè)數(shù)列an的公差為d,則Sn=na1+n(n-1)2d=d2n2+a1-d2n,又Sn=n2+Bn+C-1,兩式對照得d2=1,C-1=0,即d=2,C=1,又a1=C,所以a1=1,所以數(shù)列an的通

15、項(xiàng)公式為an=2n-1.(2)bn=(2n-1)(2·2n-1-1+1)=(2n-1)2n,則Tn=1×2+3×22+(2n-1)2n,2Tn=1×22+3×23+(2n-3)2n+(2n-1)2n+1,-得Tn=(2n-1)2n+1-2(22+2n)-2=(2n-1)2n+1-2·22(1-2n-1)1-2-2=(2n-3)2n+1+6.12.(2022屆安徽蚌埠質(zhì)檢,20)已知數(shù)列an的前n項(xiàng)和為Sn,滿足Sn=2an-2.(1)求數(shù)列an的通項(xiàng)公式;(2)記bn=log2an,數(shù)列an·bn的前n項(xiàng)和為Tn,求證:Sn

16、+Tnanbn為定值.解析(1)當(dāng)n=1時(shí),a1=S1=2a1-2,解得a1=2.當(dāng)n2時(shí),Sn-1=2an-1-2,從而Sn-Sn-1=an=2an-2an-1,化簡得an=2an-1(n2),所以數(shù)列an是首項(xiàng)為2,公比為2的等比數(shù)列,則an=2×2n-1,即an=2n.(2)證明:bn=log2an=log22n=n,an·bn=n·2n,所以Tn=1×21+2×22+3×23+n×2n,從而2Tn=1×22+2×23+3×24+(n-1)×2n+n×2n+1,兩式相減

17、,得-Tn=2+22+23+2n-n·2n+1,即-Tn=(1-n)×2n+1-2,所以Tn=(n-1)2n+1+2,而Sn=2an-2=2n+1-2,所以Sn+Tnanbn=n·2n+1-2n+1+2+2n+1-2n·2n=2,為定值.13.(2022屆長春月考,19)在公差不為零的等差數(shù)列an中,a2=5,a1,a3,a11成等比數(shù)列.(1)求數(shù)列an的通項(xiàng)公式;(2)已知bn=9(an+1)(an+4),設(shè)數(shù)列bn的前n項(xiàng)和為Sn,求證:12Sn<1.解析(1)設(shè)等差數(shù)列an的公差為d(d0).因?yàn)閍2=5,a1a11=a32,所以a1+d=

18、5,a1(a1+10d)=(a1+2d)2,解得a1=2,d=3,所以an=3n-1,nN*.(2)證明:bn=9(an+1)(an+4)=93n(3n+3)=1n(n+1)=1n-1n+1,所以Sn=1-12+12-13+1n-1n+1=1-1n+1.因?yàn)?n+1>0,所以Sn<1.又知數(shù)列Sn是遞增數(shù)列,于是SnS1=12.綜上,12Sn<1.14.(2022屆河南安陽模擬,19)已知數(shù)列an的前n項(xiàng)和為Sn,滿足a1=1,Sn=12n+tn(t為常數(shù)).(1)求an的通項(xiàng)公式;(2)若bn=(-1)nlg(anan+1),求數(shù)列bn的前n項(xiàng)和Tn.解析(1)令n=1,得

19、S1=a1=12+t=1,可得t=12,所以Sn=12n+12n,當(dāng)n2時(shí),Sn-1=12(n-1)+12(n-1),可得an=Sn-Sn-1=n,又因?yàn)閍1=1滿足該式,所以an=n(nN*).(2)因?yàn)閎n=(-1)nlg(anan+1)=(-1)n(lg an+lg an+1),所以Tn=-(lg a1+lg a2)+(lg a2+lg a3)-(lg a3+lg a4)+(-1)n(lg an+lg an+1)=(-1)nlg an+1-lg a1=(-1)nlg(n+1).15.(2022屆通州期中,19)設(shè)等差數(shù)列an的前n項(xiàng)和是Sn,bn是各項(xiàng)均為正數(shù)的等比數(shù)列,且a1=b1=1,a5=3b2.在a3+b3=14,a1b5=81,S4=4S2這三個(gè)條件中任選一個(gè),求解下列問題