1、單元質(zhì)檢卷六數(shù)列(b)(時(shí)間:60分鐘滿(mǎn)分:76分)一、選擇題:本題共6小題,每小題5分,共30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.已知等比數(shù)列an的前n項(xiàng)和為sn,若1a1+1a2+1a3=2,a2=2,則s3=()a.8b.7c.6d.42.等差數(shù)列an中,a1+a4+a7=39,a3+a6+a9=27,則數(shù)列an的前9項(xiàng)的和s9等于()a.66b.99c.144d.2973.(2020河北衡水中學(xué)三模,理5)有這樣一道題目:“戴氏善屠,日益功倍.初日屠五兩,今三十日屠訖,向共屠幾何?”其意思為:“有一個(gè)姓戴的人善于屠肉,每一天屠完的肉是前一天的2倍,第一天屠了5兩

2、肉,共屠了30天,問(wèn)一共屠了多少兩肉?”在這個(gè)問(wèn)題中,該屠夫前5天所屠肉的總兩數(shù)為()a.35b.75c.155d.3154.(2020福建福州模擬,理6)已知數(shù)列an為等差數(shù)列,若a1,a6為函數(shù)f(x)=x2-9x+14的兩個(gè)零點(diǎn),則a3a4=()a.-14b.9c.14d.205.已知在等比數(shù)列an中,an0,a22+a42=900-2a1a5,a5=9a3,則a2 019的個(gè)位數(shù)字是()a.6b.7c.8d.96.(2020河北武邑中學(xué)三模,5)若數(shù)列an的前n項(xiàng)和為sn,且a1=1,a2=2,(sn+1)(sn+2+1)=(sn+1+1)2,則sn=()a.n(n+1)2b.2n-1

3、c.2n-1d.2n-1+1二、填空題:本題共2小題,每小題5分,共10分.7.(2020北京海淀期中,13)已知等差數(shù)列an的前n項(xiàng)和為sn,若a1=9,公差d=-2.則sn的最大值為.8.(2020廣東廣州一模,文16)記sn為數(shù)列an的前n項(xiàng)和,若2sn-an=12n-1,則a3+a4=,數(shù)列an+2-an的前n項(xiàng)和tn=.三、解答題:本題共3小題,共36分.解答應(yīng)寫(xiě)出文字說(shuō)明、證明過(guò)程或演算步驟.9.(12分)(2020湖南永州二模,理17)已知sn是公差不為零的等差數(shù)列an的前n項(xiàng)和,s3=6,a3是a1與a9的等比中項(xiàng).(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)數(shù)列bn=(-1)n4an

4、4n2-1(nn*),數(shù)列bn的前2n項(xiàng)和為p2n,若p2n+112020,求正整數(shù)n的最小值.10.(12分)(2020山東淄博4月模擬,18)已知數(shù)列an滿(mǎn)足a1=1,an+1=4an+3n-1,bn=an+n.(1)證明:數(shù)列bn為等比數(shù)列;(2)求數(shù)列an的前n項(xiàng)和.11.(12分)(2020江西名校大聯(lián)考,理18)已知數(shù)列an的前n項(xiàng)和為sn,且sn=n2+n,數(shù)列bn滿(mǎn)足an=b12+1+b222+1+bn2n+1.(1)求數(shù)列an,bn的通項(xiàng)公式;(2)若cn=anbn4-n,求數(shù)列cn的前n項(xiàng)和tn.參考答案單元質(zhì)檢卷六數(shù)列(b)1.a因?yàn)榈缺葦?shù)列an的前n項(xiàng)和為sn,且1a1

5、+1a2+1a3=2,a2=2,則1a1+1a2+1a3=a1+a3a1a3+1a2=a1+a2+a3a22=s34=2,則s3=8.故選a.2.b由等差數(shù)列的性質(zhì)得a1+a7=2a4,a3+a9=2a6,a1+a4+a7=3a4=39,a3+a6+a9=3a6=27,a4=13,a6=9,a4+a6=22,數(shù)列an前9項(xiàng)的和s9=9(a1+a9)2=9(a4+a6)2=9222=99.3.c由題意可得該屠夫每天屠的肉成等比數(shù)列,記首項(xiàng)為a1,公比為q,前n項(xiàng)和為sn,所以a1=5,q=2,因此前5天所屠肉的總兩數(shù)為a1(1-q5)1-q=5(1-25)1-2=155.故選c.4.d因?yàn)榉匠蘹

6、2-9x+14=0的兩個(gè)實(shí)數(shù)根為2,7.所以a1=2,a6=7或a1=7,a6=2,當(dāng)a1=2,a6=7時(shí),d=a6-a16-1=1,則a3=4,a4=5,所以a3a4=20.當(dāng)a1=7,a6=2時(shí),d=a6-a16-1=-1,則a3=5,a4=4,所以a3a4=20.故選d.5.d設(shè)等比數(shù)列an的公比q,首項(xiàng)為a1,由a22+a42=900-2a1a5,得a22+a42+2a2a4=900,解得a2+a4=30,即a1q+a1q3=30,由a5=9a3,得q=3,所以a1=1,所以an=3n-1,所以a1=30=1,a2=31=3,a3=32=9,a4=33=27,a5=34=81,a6=3

7、5=243,所以an的個(gè)位數(shù)是以4為周期重復(fù)出現(xiàn)的.所以a2019的個(gè)位數(shù)字是a3的個(gè)位數(shù)字9,故選d.6.c(sn+1)(sn+2+1)=(sn+1+1)2,令bn=sn+1,bnbn+2=bn+12,bn為等比數(shù)列,設(shè)其公比為q,b1=s1+1=a1+1=2,b2=s2+1=a1+a2+1=4,q=b2b1=2,bn=b1qn-1=22n-1=2n,sn=bn-1=2n-1.故選c.7.25等差數(shù)列an的前n項(xiàng)和為sn,a1=9,公差d=-2,sn=9n+n(n-1)2(-2)=-n2+10n=-(n-5)2+25,n=5時(shí),sn取最大值且最大值為25.8.-1812-12n+12sn-a

8、n=12n-1,2sn+1-an+1=12n,兩式相減,得an+1+an=-12n,a3+a4=-123=-18;an+1+an=-12n,an+2+an+1=-12n+1,兩式相減,得an+2-an=-12n+1+12n=12n+1,an+2-an是以14為首項(xiàng),12為公比的等比數(shù)列,tn=14(1-12n)1-12=12-12n+1.9.解(1)設(shè)等差數(shù)列an的公差為d,由a3是a1與a9的等比中項(xiàng),可得a1a9=a32,即a1(a1+8d)=(a1+2d)2,解得a1=d.又因?yàn)閟3=3a1+3d=6,所以a1=d=1,所以數(shù)列an是以1為首項(xiàng),1為公差的等差數(shù)列,所以an=n.(2)由

9、(1)可得bn=(-1)n4n4n2-1=(-1)n12n-1+12n+1,所以p2n=-1-13+13+15-15-17+-14n-3-14n-1+14n-1+14n+1=-1+14n+1.因?yàn)閨p2n+1|=14n+120194,所以正整數(shù)n的最小值為505.10.(1)證明bn=an+n,bn+1=an+1+n+1.又an+1=4an+3n-1,bn+1bn=an+1+n+1an+n=(4an+3n-1)+n+1an+n=4(an+n)an+n=4.又b1=a1+1=1+1=2,數(shù)列bn是首項(xiàng)為2,公比為4的等比數(shù)列.(2)解由(1)知,bn=24n-1,an=bn-n=24n-1-n,sn=a1+a2+an=2(1+4+42+4n-1)-(1+2+3+n)=2(1-4n)1-4-n(n+1)2=23(4n-1)-12n2-12n.11.解(1)因?yàn)閟n=n2+n,所以當(dāng)n=1時(shí),a1=s1=2,當(dāng)n2時(shí),an=sn-sn-1=n2+n-(n-1)2-(n-1)=2n,又a1=2也滿(mǎn)足上式,所以an=2n(nn*).因?yàn)閎12+1+b222+1+bn2n+1=an=2n,所以b12+1+b222+1+bn-12n-1+1=2n-2(n2,nn*),兩式作差,得bn2n+1=2,所以bn=2n+1+2(n2,nn*),當(dāng)n=1時(shí),b13=2,所以b1=