高考數(shù)學(xué)(文數(shù))二輪復(fù)習(xí)查漏補(bǔ)缺練習(xí)：第32講《數(shù)列的綜合問(wèn)題》(教師版)

1、課時(shí)作業(yè)(三十二)第32講數(shù)列的綜合問(wèn)題時(shí)間 / 45分鐘分值 / 100分基礎(chǔ)熱身1.已知等差數(shù)列an的公差為2,若a1,a3,a4成等比數(shù)列,則a3=()A.-10B.-6 C.-8D.-42.已知數(shù)列an是公差為1的等差數(shù)列,Sn為an的前n項(xiàng)和,若S8=4S4,則a10=()A.172 B.192 C.10 D.123.已知數(shù)列an是各項(xiàng)均為正數(shù)的等比數(shù)列,點(diǎn)M(2,log2a2),N(5,log2a5)都在直線y=x-1上,則數(shù)列an的前n項(xiàng)和為()A.2n-2 B.2n+1-2 C.2n-1 D.2n+1-1 4.張丘建算經(jīng)卷上第22題為“今有女善織,日益功疾,初日織五尺,今一月日

2、織九匹三丈.”其意思為:現(xiàn)有一善于織布的女子,從第2天開(kāi)始,每天比前一天多織相同量的布,第1天織了5尺布,現(xiàn)在一月(按30天計(jì)算)共織390尺布.則該女子第30天織布()A.20尺 B.21尺 C.22尺 D.23尺5.在等比數(shù)列an中,若3a1,12a5,2a3成等差數(shù)列,則a9+a10a7+a8=. 能力提升6.在公比為q的正項(xiàng)等比數(shù)列an中,a4=4,則當(dāng)2a2+a6取得最小值時(shí),log2q=()A.14 B.-14 C.18 D.-18 7.已知等比數(shù)列an的前n項(xiàng)和是Sn,則下列說(shuō)法一定正確的是()A.若a3>0,則a2017<0 B.若a4>0,則a20

3、18<0C.若a3>0,則S2017>0 D.若a4>0,則S2018>08.設(shè)實(shí)數(shù)b,c,d成等差數(shù)列,且它們的和為9,如果實(shí)數(shù)a,b,c成公比不為-1的等比數(shù)列,則a+b+c的取值范圍為()A.94,+ B.-,94 C.94,3(3,+) D.(-,-3)-3,94 9.記數(shù)列an的前n項(xiàng)和為Sn,若對(duì)任意正整數(shù)n,都有2Sn=an+1成立,則a2018=() A.1 B.-1 C.2 D.-210.在數(shù)列an中,an=2n-1,若一個(gè)7行8列的數(shù)表中,第i行第j列的元素為cij=ai·aj+ai+aj(i=1,2,7,j=1,2,8),則該數(shù)表中

4、所有不相等的元素之和為()A.216-10 B.216+10 C.216-18 D.216+13 11.已知an是首項(xiàng)為1的等比數(shù)列,數(shù)列bn滿足b1=2,b2=5,且an(bn+1-bn)=an+1,則數(shù)列bn的前n項(xiàng)和為. 12.已知Sn是等比數(shù)列an的前n項(xiàng)和,若S3,S9,S6成等差數(shù)列,a2+a5=4,則a8=. 13.在等差數(shù)列an中,a3=7,a9=19,Sn為數(shù)列an的前n項(xiàng)和,則Sn+10an+1的最小值為. 14.(12分)設(shè)Sn為數(shù)列an的前n項(xiàng)和,已知a3=7,an=2an-1+a2-2(n2).(1)證明:an+1為等比數(shù)列;(2)求an

5、的通項(xiàng)公式,并判斷n,an,Sn是否成等差數(shù)列.15.(13分)已知數(shù)列an滿足an+1=2an+2n+1,且a1=2.(1)證明:數(shù)列an2n是等差數(shù)列;(2)設(shè)數(shù)列cn=ann-log2ann,求數(shù)列cn的前n項(xiàng)和Sn.難點(diǎn)突破16.(已知數(shù)列an的前n項(xiàng)和為Sn,a1=1,且滿足anan+1=2Sn,數(shù)列bn滿足b1=15,bn+1-bn=2n,則數(shù)列bnan中第項(xiàng)最小. 17.已知數(shù)列an的前n項(xiàng)和為Sn,且數(shù)列Snn是首項(xiàng)為3,公差為2的等差數(shù)列,若bn=a2n,數(shù)列bn的前n項(xiàng)和為T(mén)n,則使得Sn+Tn268成立的n的最小值為. 課時(shí)作業(yè)(三十二)1.答案為：D

6、；解析：根據(jù)題意知a1=a3-4,a4=a3+2,因?yàn)閍1,a3,a4成等比數(shù)列,所以a32=a4·a1,即a32=(a3+2)·(a3-4),所以a3=-4,故選D.2.答案為：B；解析：設(shè)數(shù)列an的公差為d,由S8=4S4得8a1+28d=4(4a1+6d),又d=1,所以a1=12,所以a10=a1+9d=192.3.答案為：C；解析：設(shè)數(shù)列an的公比為q,因?yàn)辄c(diǎn)M(2,log2a2),N(5,log2a5)都在直線y=x-1上,所以log2a2=2-1=1,即a2=2,log2a5=5-1=4,即a5=16,則a5a2=q3=8,則q=2,a1=1,故Sn=1-2n

7、1-2=2n-1,故選C.4.答案為：B；解析：由題意,該女子每天織的布的長(zhǎng)度成等差數(shù)列,且a1=5,設(shè)公差為d,則由前30項(xiàng)的和S30=30×5+30×292d=390,解得d=1629,所以a30=5+29×1629=21,故選B.5.3解析 若3a1,12a5,2a3成等差數(shù)列,則a5=3a1+2a3.又an為等比數(shù)列,設(shè)公比為q,則q4=3+2q2,可得(q2+1)(q2-3)=0,解得q2=3(負(fù)值舍去),所以a9+a10a7+a8=(a7+a8)q2a7+a8=q2=3.6.答案為：A；解析：2a2+a622a2a6=22a42=82,當(dāng)且僅當(dāng)q4=2

8、即q=214(負(fù)值舍去)時(shí)取等號(hào),所以log2q=log2214=14,故選A.7.答案為：C；解析：設(shè)數(shù)列an的公比為q,當(dāng)a3=a1q2>0時(shí),a1>0,若q1,則S2017=a1(1-q2017)1-q.當(dāng)q<0時(shí),1-q>0,1-q2017>0,a1(1-q2017)1-q>0,即S2017>0;當(dāng)0<q<1時(shí),1-q>0,1-q2017>0,a1(1-q2017)1-q>0,即S2017>0;當(dāng)q>1時(shí),1-q<0,1-q2017<0,a1(1-q2017)1-q>0,即S2017&

9、gt;0.若q=1,則S2017=2017a1>0.綜上可得,當(dāng)a3>0時(shí),S2017>0,故選C.8.答案為：C；解析：實(shí)數(shù)b,c,d成等差數(shù)列,且它們的和為9,b+d=2c,則3c=9,即c=3,又實(shí)數(shù)a,b,c成等比數(shù)列,則b2=3a,且a-b0,a+b+c=b23+b+3,由二次函數(shù)的性質(zhì)可知,當(dāng)b=-32時(shí),b23+b+3取得最小值94,a-b0,a+b+c3,故a+b+c的取值范圍為94,3(3,+),故選C.9.答案為：B；解析：因?yàn)?Sn=an+1,所以2Sn-1=an-1+1(n2),所以2an=2Sn-2Sn-1=an+1-(an-1+1)=an-an-1

10、,即an=-an-1(n2),又由2Sn=an+1得a1=1,所以an是等比數(shù)列,其首項(xiàng)為1,公比為-1,所以a2018=1×(-1)2017=-1.故選B.10.答案為：C；解析：該數(shù)表中第i行第j列的元素cij=ai·aj+ai+aj=(2i-1)(2j-1)+2i-1+2j-1=2i+j-1 (i=1,2,7,j=1,2,8),數(shù)表如下所示.ji12345678122-123-124-125-126-127-128-129-1223-124-125-126-127-128-129-1210-1324-1 25-1 26-1 27-1 28-129-1210-1211-

11、1425-126-127-128-1 29-1210-1211-1212-1526-127-1 28-1 29-1 210-1211-1212-1213-1627-1 28-1 29-1 210-1211-1212-1213-1214-17 28-1 29-1 210-1211-1212-1213-1214-1215-1由表可知,該數(shù)表中所有不相等的元素之和為22-1+23-1+215-1=4×(1-214)1-2-14=216-18.11.3n2+n2解析 設(shè)數(shù)列an的公比為q,由題得bn+1-bn=an+1an,當(dāng)n=1時(shí),3=a21,a2=3,q=31=3,bn+1-bn=3=

12、d,數(shù)列bn是等差數(shù)列.故數(shù)列bn的前n項(xiàng)和為n×2+n(n-1)2×3=3n2+n2.12.2解析 因?yàn)镾3,S9,S6成等差數(shù)列,所以公比q1,且2×1-q91-q=1-q31-q+1-q61-q,整理得2q6=1+q3,所以q3=-12或q3=1(舍去),所以a2+a5=a2+a2·q3=a21-12=4,解得a2=8,故a8=a2·q6=8×14=2.13.3解析 a3=7,a9=19,公差d=a9-a39-3=19-76=2,an=a3+(n-3)d=7+2(n-3)=2n+1,Sn=n(3+2n+1)2=n(n+2),因此

13、Sn+10an+1=n(n+2)+102n+2=12(n+1)+9n+112×2(n+1)·9n+1=3,當(dāng)且僅當(dāng)n=2時(shí)取等號(hào).14.解:(1)證明:a3=7,a3=3a2-2,a2=3,an=2an-1+1,a1=1,又an+1an-1+1=2an-1+2an-1+1=2(n2),an+1是首項(xiàng)為2,公比為2的等比數(shù)列.(2)由(1)知,an+1=2n,an=2n-1.Sn=2-2n+11-2-n=2n+1-n-2,n+Sn-2an=n+2n+1-n-2-2(2n-1)=0,n+Sn=2an,即n,an,Sn成等差數(shù)列.15.解:(1)證明:an+12n+1-an2n=

14、2an+2n+12n+1-an2n=2an2n+1+2n+12n+1-an2n=1,且a121=1,an2n是以1為首項(xiàng),1為公差的等差數(shù)列. (2)由(1)得an2n=1+(n-1)×1=n,故an=n·2n,cn=2n-n.Sn=c1+c2+c3+cn=(21-1)+(22-2)+(23-3)+(2n-n)=(21+22+23+2n)-(1+2+3+n)=2(1-2n)1-2-(1+n)·n2=2n+1-n(n+1)2-2.故數(shù)列cn的前n項(xiàng)和為Sn=2n+1-n(n+1)2-2(nN*).16.4解析 當(dāng)n2時(shí),2an=2Sn-2Sn-1=anan+1-an

15、-1an,an0,an+1-an-1=2;當(dāng)n=1時(shí),a1a2=2a1,解得a2=2.數(shù)列an的奇數(shù)項(xiàng)與偶數(shù)項(xiàng)分別成等差數(shù)列,公差都為2,且a1=1,a2=2,進(jìn)而得到數(shù)列an為等差數(shù)列,其首項(xiàng)為1,公差為1,an=1+n-1=n.數(shù)列bn滿足b1=15,bn+1-bn=2n,當(dāng)n2時(shí),bn=2(n-1)+(n-2)+2+1+15=2×(n-1)n2+15=n(n-1)+15,當(dāng)n=1時(shí),上式也成立,bn=n(n-1)+15,bnan=n-1+15n215-1(當(dāng)且僅當(dāng)n=15時(shí)取等號(hào)),又nN*,且b4a4=3+154,b3a3=7,則數(shù)列bnan中第4項(xiàng)最小.17.5解析 因?yàn)閿?shù)列Snn是首項(xiàng)為3,公差為2的等差數(shù)列,所以Snn=3+(n-1)×2,化簡(jiǎn)得Sn=2n2+n,則當(dāng)n2時(shí),Sn-1=2(n-1)2+(n-1),所以an=Sn-Sn-1=(2n2+n)-2(n-1)2+(n-1)=4n-1,當(dāng)n=1時(shí),S1=a1=3,滿足上式,所以an=4n-1.因?yàn)閎n=a2n,所以b1=a2,b2=a4,b3=a8,b4=a16,b5=a32,b6=a64,所以Tn=a2+a4+a8+a16+a2n-1+a2n=(23-1)+