2020版高考數(shù)學(xué)一輪復(fù)習(xí) 第六章 數(shù)列 課時(shí)規(guī)范練30 數(shù)列求和 文 北師大版.doc

課時(shí)規(guī)范練30數(shù)列求和基礎(chǔ)鞏固組1.數(shù)列1,3,5,7116,(2n-1)+12n,的前n項(xiàng)和Sn的值等于()A.n2+1-12nB.2n2-n+1-12nC.n2+1-12n-1D.n2-n+1-12n2.(2018河北衡水中學(xué)金卷十模,3)已知數(shù)列an是各項(xiàng)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-2B.2n+1-2C.2n-1D.2n+1-13.(2018山東濰坊二模,4)設(shè)數(shù)列an的前n項(xiàng)和為Sn,若Sn=-n2-n,則數(shù)列2(n+1)an的前40項(xiàng)的和為()A.3940B.-3940C.4041D.-40414.已知函數(shù)f(x)=xa的圖像過點(diǎn)(4,2),令an=1f(n+1)+f(n),nN+.記數(shù)列an的前n項(xiàng)和為Sn,則S2 018=.5.(2018浙江余姚中學(xué)4月模擬,17)已知等差數(shù)列an的前n項(xiàng)和為Sn,且S5=30,S10=110.(1)求Sn;(2)記Tn=1S1+1S2+1Sn,求Tn.6.(2018山西晉城月考)已知數(shù)列an滿足a1=3,an+1=2an+(-1)n(3n+1).(1)求證:數(shù)列an+(-1)nn是等比數(shù)列;(2)求數(shù)列an的前10項(xiàng)和S10.7.(2018河北唐山一模,17)已知數(shù)列an是以1為首項(xiàng)的等差數(shù)列,數(shù)列bn是以q(q1)為公比的等比數(shù)列.且a2=b1,a3-a1=b2-b1,a2b2=b3.(1)求an和bn的通項(xiàng)公式;(2)若Sn=a1bn+a2bn-1+an-1b2+anb1,求Sn.綜合提升組8.(2018廣東中山期末)等比數(shù)列an中,已知對任意自然數(shù)n,a1+a2+a3+an=2n-1,則a12+a22+a32+an2等于()A.2n-1B. (3n-1)C. (4n-1)D.以上都不對9.(2018湖北重點(diǎn)中學(xué)五模)設(shè)等差數(shù)列an的前n項(xiàng)和為Sn,a4=4,S5=15,若數(shù)列1anan+1的前m項(xiàng)和為1011,則m=()A.8B.9C.10D.1110.(2018山東濰坊三模,17)已知數(shù)列an的前n項(xiàng)和為Sn,且1,an,Sn成等差數(shù)列.(1)求數(shù)列an的通項(xiàng)公式;(2)若數(shù)列bn滿足anbn=1+2nan,求數(shù)列bn的前n項(xiàng)和Tn.11.(2018江西上饒三模,17)已知等比數(shù)列an的前n項(xiàng)和為Sn,且6Sn=3n+1+a(nN+).(1)求a的值及數(shù)列an的通項(xiàng)公式;(2)若bn=(3n+1)an,求數(shù)列an的前n項(xiàng)和Tn.創(chuàng)新應(yīng)用組12.(2017全國1,理12)幾位大學(xué)生響應(yīng)國家的創(chuàng)業(yè)號召,開發(fā)了一款應(yīng)用軟件.為激發(fā)大家學(xué)習(xí)數(shù)學(xué)的興趣,他們推出了“解數(shù)學(xué)題獲取軟件激活碼”的活動(dòng).這款軟件的激活碼為下面數(shù)學(xué)問題的答案:已知數(shù)列1,1,2,1,2,4,1,2,4,8,1,2,4,8,16,其中第一項(xiàng)是20,接下來的兩項(xiàng)是20,21,再接下來的三項(xiàng)是20,21,22,依此類推.求滿足如下條件的最小整數(shù)N:N>100且該數(shù)列的前N項(xiàng)和為2的整數(shù)冪.那么該款軟件的激活碼是()A.440B.330C.220D.11013.(2018云南玉溪月考)數(shù)列an滿足:a1=,a2=,且a1a2+a2a3+anan+1=na1an+1對任何的正整數(shù)n都成立,則1a1+1a2+1a97的值為()A.5 032B.5 044C.5 048D.5 050課時(shí)規(guī)范練30數(shù)列求和1.A該數(shù)列的通項(xiàng)公式為an=(2n-1)+12n,則Sn=1+3+5+(2n-1)+12+122+12n=n2+1-12n.2.C由題意log2a2=2-1=1,可得a2=2,log2a5=5-1=4,可得a5=16,a5a2=q3=8q=2,a1=1Sn=1-2n1-2=2n-1,故選C.3.DSn=-n2-n,a1=S1=-2.當(dāng)n2時(shí),an=Sn-Sn-1=-n2-n+(n-1)2+(n-1)=-2n,則數(shù)列an的通項(xiàng)公式為an=-2n,2(n+1)an=2-2n(n+1)=-1n-1n+1,數(shù)列2(n+1)an的前40項(xiàng)的和為S40=-1-12+12-13+140-141=-4041.4.2 019-1由f(4)=2,可得4a=2,解得a=,則f(x)=x12.an=1f(n+1)+f(n)=1n+1+n=n+1-n,S2 018=a1+a2+a3+a2 018=(2-1)+(3-2)+(4-3)+(2 019-2 018)=2 019-1.5.解 (1)設(shè)an的首項(xiàng)為a1,公差為d,由題意得S5=5a1+10d=30,S10=10a1+45d=110,解得a1=2,d=2,所以Sn=n2+n.(2)1Sn=1n(n+1)=1n-1n+1,所以Tn=1-12+12-13+1n-1n+1=1-1n+1=nn+1.6.(1)證明 an+1=2an+(-1)n(3n+1),an+1+(-1)n+1(n+1)an+(-1)nn=2an+(-1)n(3n+1)-(-1)n(n+1)an+(-1)nn=2an+(-1)nnan+(-1)nn=2.又a1-1=3-1=2,數(shù)列an+(-1)nn是首項(xiàng)為2,公比為2的等比數(shù)列.(2)解 由(1)得an+(-1)nn=22n-1=2n,an=2n-(-1)nn,S10=(2+22+210)+(1-2)+(3-4)+(9-10)=2(1-210)1-2-5=211-7=2 041.7.解 (1)設(shè)an的公差為d,bn的首項(xiàng)為b1,則an=1+(n-1)d,bn=b1qn-1.依題意可得1+d=b1,2d=b1(q-1),(1+d)b1q=b1q2,解得d=1,b1=2,q=2,所以an=n,bn=2n.(2)Sn=12n+22n-1+n21,所以2Sn=12n+1+22n+n22,-可得,Sn=2n+1+(2n+2n-1+22)-n21=2n+1-2n+4(2n-1-1)2-1=2n+2-2n-4.8.C當(dāng)n=1時(shí),a1=21-1=1,當(dāng)n2時(shí),a1+a2+a3+an=2n-1,a1+a2+a3+an-1=2n-1-1,兩式做差可得an=2n-2n-1=2n-1,且n=1時(shí),21-1=20=1=a1,an=2n-1,故an2=4n-1,a12+a22+a32+an2=1(1-4n)1-4=13(4n-1).9.CSn為等差數(shù)列an的前n項(xiàng)和,設(shè)公差為d,則a4=4,S5=15=5a3,解得d=1,則an=4+(n-4)1=n.由于1anan+1=1n(n+1)=1n-1n+1,則Sm=1-12+12-13+1m-1m+1=1-1m+1=1011,解得m=10.10.解 (1)由已知1,an,Sn成等差數(shù)列,得2an=1+Sn,當(dāng)n=1時(shí),2a1=1+S1=1+a1,a1=1.當(dāng)n2時(shí),2an-1=1+Sn-1,由-,得2an-2an-1=an,anan-1=2,數(shù)列an是以1為首項(xiàng),2為公比的等比數(shù)列,an=a1qn-1=12n-1=2n-1.(2)由anbn=1+2nan得bn=1an+2n,Tn=b1+b2+bn=1a1+2+1a2+4+1an+2n=1a1+1a2+1an+(2+4+2n)=1-12n1-12+(2+2n)n2=n2+n+2-12n-1.11.解 (1)6Sn=3n+1+a(nN+),當(dāng)n=1時(shí),6S1=6a1=9+a;當(dāng)n2時(shí),6an=6(Sn-Sn-1)=23n,即an=3n-1,an為等比數(shù)列,a1=1,則9+a=6,a=-3,an的通項(xiàng)公式為an=3n-1.(2)由(1)得bn=(3n+1)3n-1,Tn=b1+b2+bn=430+731+(3n+1)3n-1,3Tn=431+732+(3n-2)3n-1+(3n+1)3n,-2Tn=4+32+33+3n-(3n+1)3n,Tn=(6n-1)3n+14.12.A設(shè)數(shù)列的首項(xiàng)為第1組,接下來兩項(xiàng)為第2組,再接下來三項(xiàng)為第3組,以此類推,設(shè)第n組的項(xiàng)數(shù)為n,則前n組的項(xiàng)數(shù)和為n(1+n)2.第n組的和為1-2n1-2=2n-1,前n組總共的和為2(1-2n)1-2-n=2n+1-2-n.由題意,N>100,令n(1+n)2>100,得n14且nN+,即N出現(xiàn)在第13組之后.若要使最小整數(shù)N滿足:N>100且前N項(xiàng)和為2的整數(shù)冪,則SN-Sn(1+n)2應(yīng)與-2-n互為相反數(shù),即2k-1=2+n(kN+,n14),所以k=log2(n+3),解得n=29,k=5.所以N=29(1+29)2+5=440,故選A.13.Ba1a2+a2a3+anan+1=na1an+1,a1a2+a2a3+anan+1+an+1an+2=(n+1)a1an+2,-,得-an+1an+2=na1an+1-(n+1)a1an+2,n+1an+1-nan+2=4,同理得nan-n-1an+1=4,n+1an+1-nan+2=nan-n-1an+1,整理得2an+1=1an+1an+2,1an是等差數(shù)列.a1=14,a2=15,等差數(shù)列1an的首項(xiàng)為4,公差為1,1an=4+(n-1)1=n+3,1a1+1a2+1a97=97(4+100)2=5 044.