新高考數(shù)學(xué)一輪復(fù)習(xí)專題六數(shù)列6-4數(shù)列求和練習(xí)含答案

6.4數(shù)列求和五年高考高考新風(fēng)向(2024全國(guó)甲理,18,12分,易)記Sn為數(shù)列{an}的前n項(xiàng)和,已知4Sn=3an+4.(1)求{an}的通項(xiàng)公式;(2)設(shè)bn=(-1)n-1nan,求數(shù)列{bn}的前n項(xiàng)和Tn.解析(1)∵4Sn=3an+4①,∴當(dāng)n=1時(shí),4S1=4a1=3a1+4,得a1=4,當(dāng)n≥2時(shí),4Sn-1=3an-1+4②,由①-②得,4an=3an-3an-1,∴an=-3an-1,∴數(shù)列{an}是首項(xiàng)為4,公比為-3的等比數(shù)列.∴an=4×(-3)n-1.(2)由(1)得bn=(-1)n-1nan=4n·3n-1,∴Tn=4×30+4×2×31+4×3×32+…+4(n-1)·3n-2+4n·3n-1③,3Tn=4×31+4×2×32+4×3×33+…+4(n-1)·3n-1+4n·3n④,③-④得-2Tn=4+4×31+4×32+…+4×3n-1-4n×3n,∴-2Tn=4+4·3(1?3n?1)1?3∴-2Tn=4+(2-4n)·3n-6=-2+(2-4n)3n,∴Tn=1+(2n-1)3n.考點(diǎn)數(shù)列求和1.(多選)(2021新高考Ⅱ,12,5分,難)設(shè)正整數(shù)n=a0·20+a1·21+…+ak-1·2k-1+ak·2k,其中ai∈{0,1},記ω(n)=a0+a1+…+ak,則(ACD)A.ω(2n)=ω(n)B.ω(2n+3)=ω(n)+1C.ω(8n+5)=ω(4n+3)D.ω(2n-1)=n2.(2021新高考Ⅰ,16,5分,難)某校學(xué)生在研究民間剪紙藝術(shù)時(shí),發(fā)現(xiàn)剪紙時(shí)經(jīng)常會(huì)沿紙的某條對(duì)稱軸把紙對(duì)折.規(guī)格為20dm×12dm的長(zhǎng)方形紙,對(duì)折1次共可以得到10dm×12dm,20dm×6dm兩種規(guī)格的圖形,它們的面積之和S1=240dm2,對(duì)折2次共可以得到5dm×12dm,10dm×6dm,20dm×3dm三種規(guī)格的圖形,它們的面積之和S2=180dm2,以此類推.則對(duì)折4次共可以得到不同規(guī)格圖形的種數(shù)為5;如果對(duì)折n次,那么k=1nSk=240×3?n+33.(2021新高考Ⅰ,17,10分,易)已知數(shù)列{an}滿足a1=1,an+1=a(1)記bn=a2n,寫出b1,b2,并求數(shù)列{bn}的通項(xiàng)公式;(2)求{an}的前20項(xiàng)和.解析(1)由題設(shè)可得a2k+2=a2k+1+1,a2k+1=a2k+2(k∈N*),故a2k+2=a2k+3,即bn+1=bn+3,即bn+1-bn=3,b1=a2=a1+1=2,b2=b1+3=5,所以{bn}是首項(xiàng)為2,公差為3的等差數(shù)列,故bn=2+(n-1)×3=3n-1.(2)當(dāng)n為奇數(shù)時(shí),an=an+1-1.設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,則S20=a1+a2+…+a20=(a1+a3+…+a19)+(a2+a4+…+a20)=[(a2-1)+(a4-1)+…+(a20-1)]+(a2+a4+…+a20)=2(a2+a4+…+a20)-10=2(b1+b2+…+b10)-10=2×10×2+9×102即{an}的前20項(xiàng)和為300.4.(2020新高考Ⅰ,18,12分,中)已知公比大于1的等比數(shù)列{an}滿足a2+a4=20,a3=8.(1)求{an}的通項(xiàng)公式;(2)記bm為{an}在區(qū)間(0,m](m∈N*)中的項(xiàng)的個(gè)數(shù),求數(shù)列{bm}的前100項(xiàng)和S100.解析(1)已知數(shù)列{an}是公比大于1的等比數(shù)列,設(shè)公比為q(q>1),依題意有a1q+a1q3=20,a1q2=8,解得a1=2,q=2或a1=32所以an=2n,所以數(shù)列{an}的通項(xiàng)公式為an=2n.(2)由于21=2,22=4,23=8,24=16,25=32,26=64,27=128,所以b1對(duì)應(yīng)的區(qū)間為(0,1],則b1=0;b2,b3對(duì)應(yīng)的區(qū)間分別為(0,2],(0,3],則b2=b3=1,即有2個(gè)1;b4,b5,b6,b7對(duì)應(yīng)的區(qū)間分別為(0,4],(0,5],(0,6],(0,7],則b4=b5=b6=b7=2,即有22個(gè)2;b8,b9,…,b15對(duì)應(yīng)的區(qū)間分別為(0,8],(0,9],…,(0,15],則b8=b9=…=b15=3,即有23個(gè)3;b16,b17,…,b31對(duì)應(yīng)的區(qū)間分別為(0,16],(0,17],…,(0,31],則b16=b17=…=b31=4,即有24個(gè)4;b32,b33,…,b63對(duì)應(yīng)的區(qū)間分別為(0,32],(0,33],…,(0,63],則b32=b33=…=b63=5,即有25個(gè)5;b64,b65,…,b100對(duì)應(yīng)的區(qū)間分別為(0,64],(0,65],…,(0,100],則b64=b65=…=b100=6,即有37個(gè)6.所以S100=1×2+2×22+3×23+4×24+5×25+6×37=480.

三年模擬練速度1.(2024浙江杭州二模,5)設(shè)數(shù)列{an},{bn}滿足a1=b1=1,an+bn+1=2n,an+1+bn=2n.設(shè)Sn為數(shù)列{an+bn}的前n項(xiàng)的和,則S7=(A)A.110B.120C.288D.3062.(2024浙江金麗衢十二校第二次聯(lián)考,15)已知等差數(shù)列{an}的前n項(xiàng)和為Sn,且2Sn=2an+n2-1.(1)求an;(2)求數(shù)列1anan+1的前解析(1)因?yàn)?Sn=2an+n2-1,①所以當(dāng)n≥2時(shí),2Sn-1=2an-1+(n-1)2-1,②①-②得2an=2an-2an-1+2n-1,整理得an-1=n-12,n≥2所以an=n+12,n∈N*(2)由(1)知an=n+12所以1anan+1=1n+12所以Tn=1a1a2+1a2a3+…+1anan+1=23-3.(2024江蘇南通、徐州大聯(lián)考,15)已知正項(xiàng)數(shù)列{an}的前n項(xiàng)和為Sn,且4Sn=(an+1)2.(1)求數(shù)列{an}的通項(xiàng)公式;(2)若bn=an·a2n,求數(shù)列{bn}的前n項(xiàng)和T解析(1)因?yàn)?Sn=(a所以當(dāng)n≥2時(shí),4Sn-1=(a所以4an=(an+1)2-(an?1+1)2=an整理,得2an+2an-1=(an-an-1)(an+an-1),因?yàn)閍n>0,所以an-an-1=2,所以數(shù)列{an}是公差為2的等差數(shù)列.當(dāng)n=1時(shí),4S1=4a1=(a解得a1=1,所以數(shù)列{an}的通項(xiàng)公式為an=2n-1.(2)由(1)得bn=an·a2n=(2n-1)(2n+1-1)=(2n-1)2n+1-(2n記An=i=1n(2i-1)2i+1,Bn=i=1n(則Bn=i=1n(2i-1)=n因?yàn)锳n=i=1n(2i-1)2i+1,2An=i=1n(2i-1)所以-An=22+2(23+24+…+2n+1)-(2n-1)2n+2=-(2n-3)2n+2-12,所以An=(2n-3)2n+2+12,所以Tn=An-Bn=(2n-3)2n+2+12-n2.練思維1.(2024廣東汕頭一模,15)已知數(shù)列{an}和{bn},其中bn=2an,n∈N*,數(shù)列{an+bn}的前n項(xiàng)和為S(1)若an=2n,求Sn;(2)若Sn=3n,求數(shù)列{an}和{bn}的通項(xiàng)公式.解析(1)當(dāng)an=2n時(shí),{an}是首項(xiàng)為2,公差為2的等差數(shù)列.(1分)又bn=22n=4n,所以{bn}是首項(xiàng)為4,公比為4的等比數(shù)列,(2分)從而Sn=(a1+a2+…+an)+(b1+b2+…+bn)(3分)=n(2+2n=4n+13+n2+n-43.((2)當(dāng)n≥2時(shí),an+bn=Sn-Sn-1=3,(7分)當(dāng)n=1時(shí),a1+b1=S1=3,滿足上式,(8分)故an+bn=3(n∈N*),即an+2an=3,(9分令f(x)=x+2x,則f(x)在R上單調(diào)遞增,且f(1)=3,(11分)從而an=1,(12分)則bn=3-an=2.(13分)2.(2024山東日照一模,16)已知各項(xiàng)均為正數(shù)的數(shù)列{an}的前n項(xiàng)和為Sn,且an,Sn,an2(1)求a1及{an}的通項(xiàng)公式;(2)記集合anan+4an≤2k,k∈N*的元素個(gè)數(shù)為bk,求數(shù)列{bk}的前50項(xiàng)和.解析(1)因?yàn)閍n,Sn,an2所以2Sn=an+an2,且an當(dāng)n=1時(shí),2S1=2a1=a1+a12,可得a1=1.(2分當(dāng)n≥2時(shí),2(Sn-Sn-1)=2an=an+an2-an-1-an?12,則an+an-1=an2-an?12=(an+an-1)(an由an+an-1>0,得an-an-1=1,(6分)所以{an}是首項(xiàng)為1,公差為1的等差數(shù)列,故an=n.(7分)(2)原式等價(jià)于12n+因?yàn)?2n+4n≥2,當(dāng)且僅當(dāng)n所以b1=0,b2=1,(9分)當(dāng)k≥3時(shí),因?yàn)?k?12+22k?1=k-12+22k?1<k,所以能使n2+2n≤k成立的n的最大值為2k所以bk=2k-1(k≥3),(13分)所以{bk}的前50項(xiàng)和為0+1+5+7+…+99=0+1+(5+99)×482=2497.(15分)3.(2024山東聊城二模,17)已知數(shù)列{an},{bn}滿足a2n-1=b2n-1+12m,a2n=mb2n,m為常數(shù),若{an}為等差數(shù)列,且b4-b2=2(b3-b1)=2(a1+b1)=8.(1)求m的值及{an}的通項(xiàng)公式;(2)求{bn}的前2n項(xiàng)和S2n.解析(1)由題意知b4-b2=8,b3-b1=4,a1+b1=4,(1分)因?yàn)閍2n-1=b2n-1+12m,a2n=mb2n,所以a1=b1+12設(shè)等差數(shù)列{an}的公差為d,則a3?a1=b3所以an=5+(n-1)×2=2n+3,所以m的值為12,{an}的通項(xiàng)公式為an=2n+3.(7分)(2)由(1)知,an=2n+3,b2n-1=a2n-1-6,b2n=2a2n,(8分)所以S2n=(b1+b3+b5+…+b2n-1)+(b2+b4+b6+…+b2n)=(a1+a3+a5+…+a2n-1-6n)+2(a2+a4+a6+…+a2n)=n(a1+a2n?1)2-6n+2×n(a2+a2所以{bn}的前2n項(xiàng)和S2n=6n2+7n.(15分)練風(fēng)向(新定義理解)(2024湖南長(zhǎng)沙雅禮中學(xué)一模,19)約數(shù),又稱因數(shù).它的定義如下:若整數(shù)a除以整數(shù)m(m≠0)除得的商正好是整數(shù)而沒(méi)有余數(shù),我們就稱a為m的倍數(shù),m為a的約數(shù).設(shè)正整數(shù)a共有k個(gè)正約數(shù),即為a1,a2,…,ak-1,ak(a1<a2<…<ak).(1)當(dāng)k=4時(shí),若正整數(shù)a的k個(gè)正約數(shù)構(gòu)成等比數(shù)列,請(qǐng)寫出一個(gè)a的值;(2)當(dāng)k≥4時(shí),若a2-a1,a3-a2,…,ak-ak-1構(gòu)成等比數(shù)列,求正整數(shù)a(結(jié)果用a2表示);(3)記A=a1a2+a2a3+…+ak-1ak,求證:A<a2.解析(1)當(dāng)k=4時(shí),正整數(shù)a的4個(gè)正約數(shù)構(gòu)成等比數(shù)列,比如1,2,4,8為8的所有正約數(shù),即a=8.(2)由題意可知a1=1,ak=a,ak-1=aa2,ak-2=因?yàn)閗≥4,a3?a2a2?a1=ak?ak?1ak?1?a因?yàn)閍3∈N*,所以a3?a因此可知a3是完全平