2020版高考數(shù)學(xué)一輪復(fù)習(xí)課時(shí)規(guī)范練31數(shù)列求和理北師大版.docx

課時(shí)規(guī)范練31數(shù)列求和基礎(chǔ)鞏固組1.數(shù)列1,3,5,7,(2n-1)+,的前n項(xiàng)和Sn的值等于()A.n2+1-B.2n2-n+1-C.n2+1-D.n2-n+1-2.(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)列的前40項(xiàng)的和為()A.B.-C.D.-4.已知函數(shù)f(x)=xa的圖像過點(diǎn)(4,2),令an=,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=+,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山東濰坊一模,17)公差不為0的等差數(shù)列an的前n項(xiàng)和為Sn,已知S4=10,且a1,a3,a9成等比數(shù)列.(1)求an的通項(xiàng)公式;(2)求數(shù)列的前n項(xiàng)和Tn.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+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)列的前m項(xiàng)和為,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.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:N100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=對任何的正整數(shù)n都成立,則+的值為()A.5 032B.5 044C.5 048D.5 050參考答案課時(shí)規(guī)范練31數(shù)列求和1.A該數(shù)列的通項(xiàng)公式為an=(2n-1)+,則Sn=1+3+5+(2n-1)+=n2+1-.2.C由題意log2a2=2-1=1,可得a2=2,log2a5=5-1=4,可得a5=16,=q3=8Sn=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,=-,數(shù)列的前40項(xiàng)的和為S40=-1-+-+-=-.4.-1由f(4)=2,可得4a=2,解得a=,則f(x)=.an=-,S2 018=a1+a2+a3+a2 018=(-)+(-)+(-)+(-)=-1.5.解 (1)設(shè)an的首項(xiàng)為a1,公差為d,由題意得解得所以Sn=n2+n.(2)=-,所以Tn=1-+-+-=1-=.6.(1)證明 an+1=2an+(-1)n(3n+1),=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)=-5=211-7=2 041.7.解 (1)設(shè)an的公差為d,由題設(shè)可得,解得an=n.(2)令cn=,則Tn=c1+c2+cn=+,Tn=+,-得:Tn=+-=-=-,Tn=-.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=4n-1,+=(4n-1).9.CSn為等差數(shù)列an的前n項(xiàng)和,設(shè)公差為d,則解得d=1,則an=4+(n-4)1=n.由于=-,則Sm=1-+-+-=1-=,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,=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=+2n,Tn=b1+b2+bn=+2+4+2n=+(2+4+2n)=+=n2+n+2-.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=.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.第n組的和為=2n-1,前n組總共的和為-n=2n+1-2-n.由題意,N100,令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:N100且前N項(xiàng)和為2的整數(shù)冪,則SN-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=