(福建專(zhuān)用)高考數(shù)學(xué)一輪復(fù)習(xí)課時(shí)規(guī)范練31數(shù)列求和理新人教A版

課時(shí)規(guī)范練31數(shù)列求和一、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.在數(shù)列{an}中,a1=-60,an+1=an+3，則|a1|+|a2|+…+|a30|=()A.-495B.765C.1080D.31053.已知數(shù)列{an}的前n項(xiàng)和Sn滿足Sn+Sm=Sn+m，其中m,n為正整數(shù)，且a1=1，則a10等于()A.1B.9C.10D.554.已知函數(shù)f(x)=xa的圖象過(guò)點(diǎn)(4,2)，令an=,n∈N*.記數(shù)列{an}的前n項(xiàng)和為Sn，則S2018等于()A.-1B.+1C.-1D.+15.已知數(shù)列{an}中,an=2n+1，則+…+=()A.1+B.1-2nC.1-D.1+2n?導(dǎo)學(xué)號(hào)21500545?6.設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,a1=2，若Sn+1=Sn，則數(shù)列的前2018項(xiàng)和為.7.已知等差數(shù)列{an}滿足:a5=11,a2+a6=18.(1)求數(shù)列{an}的通項(xiàng)公式;(2)若bn=an+2n，求數(shù)列{bn}的前n項(xiàng)和Sn.二、綜合提升組8.如果數(shù)列1,1+2,1+2+4,…,1+2+22+…+2n-1,…的前n項(xiàng)和Sn>1020，那么n的最小值是()A.7B.8C.9D.109.(2017山東煙臺(tái)模擬)已知數(shù)列{an}中,a1=1，且an+1=，若bn=anan+1，則數(shù)列{bn}的前n項(xiàng)和Sn為()A.B.C.D.?導(dǎo)學(xué)號(hào)21500546?10.(2017福建龍巖一模)已知Sn為數(shù)列{an}的前n項(xiàng)和，對(duì)n∈N*都有Sn=1-an，若bn=log2an，則+…+=.11.(2017廣西模擬)已知數(shù)列{an}的前n項(xiàng)和為Sn，且Sn=an-1(n∈N*).(1)求數(shù)列{an}的通項(xiàng)公式;(2)設(shè)bn=2log3+1，求+…+.三、創(chuàng)新應(yīng)用組12.(2017全國(guó)Ⅰ，理12)幾位大學(xué)生響應(yīng)國(guó)家的創(chuàng)業(yè)號(hào)召，開(kāi)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(wèn)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(lái)的兩項(xiàng)是20,21，再接下來(lái)的三項(xiàng)是20,21,22，依此類(lèi)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.110?導(dǎo)學(xué)號(hào)21500547?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.B由a1=-60,an+1=an+3可得an=3n-63，則a21=0,|a1|+|a2|+…+|a30|=-(a1+a2+…+a20)+(a21+…+a30)=S30-2S20=765，故選B.3.A∵Sn+Sm=Sn+m,a1=1,∴S1=1.可令m=1，得Sn+1=Sn+1,∴Sn+1-Sn=1，即當(dāng)n≥1時(shí),an+1=1,∴a10=1.4.C由f(4)=2，可得4a=2，解得a=，則f(x)=∴an=,S2018=a1+a2+a3+…+a2018=()+()+()+…+()=-1.5.Can+1-an=2n+1+1-(2n+1)=2n+1-2n=2n,所以+…++…+=1-=1-6∵Sn+1=Sn，又a1=2,∴當(dāng)n≥2時(shí),Sn=…S1=…2=n(n+1).當(dāng)n=1時(shí)也成立,∴Sn=n(n+1).∴當(dāng)n≥2時(shí),an=Sn-Sn-1=n(n+1)-n(n-1)=2n.當(dāng)n=1時(shí),a1=2也成立，所以an=2n.則數(shù)列的前2018項(xiàng)和=7.解(1)設(shè){an}的首項(xiàng)為a1，公差為d.由a5=11,a2+a6=18,得解得a1=3,d=2，所以an=2n+1.(2)由an=2n+1得bn=2n+1+2n,則Sn=[3+5+7+…+(2n+1)]+(21+22+23+…+2n)=n2+2n+=n2+2n+2n+1-2.8.Dan=1+2+22+…+2n-1=2n-1.∴Sn=(21-1)+(22-1)+…+(2n-1)=(21+22+…+2n)-n=2n+1-n-2,∴S9=1013<1020,S10=2036>1020,∴使Sn>1020的n的最小值是10.9.B由an+1=，得+2,∴數(shù)列是以1為首項(xiàng),2為公差的等差數(shù)列,=2n-1，又bn=anan+1,∴bn=,∴Sn=，故選B.10對(duì)n∈N*都有Sn=1-an，當(dāng)n=1時(shí),a1=1-a1，解得a1=當(dāng)n≥2時(shí),an=Sn-Sn-1=1-an-(1-an-1)，化為an=an-1.∴數(shù)列{an}是等比數(shù)列，公比為，首項(xiàng)為an=∴bn=log2an=-n則+…++…+=1-11.解(1)當(dāng)n=1時(shí),a1=a1-1,∴a1=2.當(dāng)n≥2時(shí),∵Sn=an-1,①Sn-1=an-1-1(n≥2),②∴①-②得an=，即an=3an-1,∴數(shù)列{an}是首項(xiàng)為2，公比為3的等比數(shù)列,∴an=2·3n-1.(2)由(1)得bn=2log3+1=2n-1,+…++…+=+…+12.A設(shè)數(shù)列的首項(xiàng)為第1組，接下來(lái)兩項(xiàng)為第2組，再接下來(lái)三項(xiàng)為第3組，以此類(lèi)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,N>100，令>100，得n