PAGEPAGE10考點(diǎn)規(guī)范練32數(shù)列求和基礎(chǔ)鞏固1.數(shù)列112,314,518,7116,…,(2n-1)+12n,…的前nA.n2+1-12n B.2n2-n+1C.n2+1-12n-1 D.n22.已知數(shù)列{an}滿足a1=1,且對(duì)任意的n∈N*都有an+1=a1+an+n,則1an的前100項(xiàng)和為(A.100101 B.99100 C.1011003.已知數(shù)列{an}滿足an+1-an=2,a1=-5,則|a1|+|a2|+…+|a6|=()A.9 B.15 C.18 D.304.已知等差數(shù)列{an}滿足a1=-2021,其前n項(xiàng)和為Sn,若5S12-6S10=120,則S2022=()A.-4040 B.-2022C.0 D.40405.已知數(shù)列{an}為等差數(shù)列,Sn是其前n項(xiàng)和,a2=5,S5=35.數(shù)列1an·an+1的前n項(xiàng)和為Tn,若對(duì)一切n∈N*都有2m+1>TA.-1 B.0 C.1 D.26.(2021四川涼山三模)我國(guó)古代很早就有對(duì)等差數(shù)列和等比數(shù)列的研究成果.北宋大科學(xué)家沈括在《夢(mèng)溪筆談》中首創(chuàng)的“隙積術(shù)”,就是關(guān)于高階等差數(shù)列求和的問題.現(xiàn)有一物品堆,從上向下數(shù),第一層有1個(gè)貨物,第二層比第一層多2個(gè),第三層比第二層多3個(gè),…,以此類推.記第n層貨物的個(gè)數(shù)為an,則數(shù)列1an的前2021項(xiàng)和為(A.40412021 B.20211011 C.202120227.已知數(shù)列{an}滿足:a3=15,an-an+1=2anan+1,則數(shù)列{anan+1}前10項(xiàng)的和為.8.在數(shù)列{an}中,a1=3,{an}的前n項(xiàng)和Sn滿足Sn+1=an+n2.(1)求數(shù)列{an}的通項(xiàng)公式;(2)設(shè)數(shù)列{bn}滿足bn=(-1)n+2an,求數(shù)列{bn}的前n項(xiàng)和T9.(2021廣西梧州模擬預(yù)測(cè)(文))已知數(shù)列{an}是公差為2的等差數(shù)列,它的前n項(xiàng)和為Sn,且a1,a3,a7成等比數(shù)列.(1)求數(shù)列{an}的通項(xiàng)公式;(2)設(shè)數(shù)列{bn}滿足bn=Sn-n23n+1,求數(shù)列{b10.已知數(shù)列{an}是各項(xiàng)均為正數(shù)的等比數(shù)列,且a1+a2=21a1+1a2,a3+a4(1)求數(shù)列{an}的通項(xiàng)公式;(2)設(shè)bn=an2+log2an,求數(shù)列{bn}的前n項(xiàng)和T11.(2021四川瀘州二模)已知數(shù)列{an}是等比數(shù)列,a2=4,且a3+2是a2和a4的等差中項(xiàng).(1)求數(shù)列{an}的通項(xiàng)公式;(2)設(shè)bn=an(an-1)(an+1-1),數(shù)列{bn}的前能力提升12.在數(shù)列{an}中,a1=1,且an+1=an2an+1.若bn=anan+1,則數(shù)列{bn}的前n項(xiàng)和A.2n2n+1C.2n2n13.數(shù)列1,x,1,x,x,1,x,x,x,1,x,x,x,x,1,x,…,其中在第n個(gè)1與第n+1個(gè)1之間插入n個(gè)x,若該數(shù)列的前2020項(xiàng)的和為7891,則x=.

14.已知各項(xiàng)均為正數(shù)的數(shù)列{an}的前n項(xiàng)和為Sn,滿足an+12=2Sn+n+4,a2-1,a3,a7恰為等比數(shù)列{b(1)求數(shù)列{an},{bn}的通項(xiàng)公式;(2)若cn=(-1)nlog2bn-1anan+1,求數(shù)列{cn}的前15.(2021山東菏澤一模)已知等比數(shù)列{an}的前n項(xiàng)和為Sn,且an+1=2Sn+2,數(shù)列{bn}滿足b1=2,(n+2)bn=nbn+1,其中n∈N*.(1)分別求數(shù)列{an}和{bn}的通項(xiàng)公式;(2)在an與an+1之間插入n個(gè)數(shù),使這n+2個(gè)數(shù)組成一個(gè)公差為cn的等差數(shù)列,求數(shù)列{bncn}的前n項(xiàng)和Tn.高考預(yù)測(cè)16.(2021云南曲靖二模)已知數(shù)列{an}的前n項(xiàng)和為Sn.(1)請(qǐng)從①2Sn=3an-3-4n,②a1=-3,an+1=-an-4這兩個(gè)條件中任選一個(gè),證明數(shù)列{an+2}是等比數(shù)列;(2)數(shù)列{bn}為等差數(shù)列,b3=5,b5=9,記cn=(an+2)bn,求數(shù)列{cn}的前n項(xiàng)和Tn.答案：1.A解析該數(shù)列的通項(xiàng)公式為an=(2n-1)+12則Sn=[1+3+5+…+(2n-1)]+12+122+…2.D解析∵an+1=a1+an+n,a1=1,∴an+1-an=1+n.∴an-an-1=n(n≥2).∴an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1=n+(n-1)+…+2+1=n(∴1an=2∴1an的前100項(xiàng)和為2×1-12+12-13+…+13.C解析∵an+1-an=2,a1=-5,∴數(shù)列{an}是首項(xiàng)為-5,公差為2的等差數(shù)列.∴an=-5+2(n-1)=2n-7.∴數(shù)列{an}的前n項(xiàng)和Sn=n(-5+2n-7)令an=2n-7≥0,解得n≥72∴當(dāng)n≤3時(shí),|an|=-an;當(dāng)n≥4時(shí),|an|=an.∴|a1|+|a2|+…+|a6|=-a1-a2-a3+a4+a5+a6=S6-2S3=62-6×6-2(32-6×3)=18.4.C解析設(shè)等差數(shù)列{an}的公差為d.由5S12-6S10=120,得5×12a1+12×112d-6×10a1+10×92d=120,化簡(jiǎn)得60d=120,解得d=又已知a1=-2021,則S2022=2022×(-2021)+2022×20212×25.B解析設(shè)等差數(shù)列{an}的首項(xiàng)為a1,公差為d.由題意得a1+故an=3+2(n-1)=2n+1.所以1an所以Tn=1213-15+15-1對(duì)一切n∈N*都有2m+1>Tn恒成立,只需滿足2m+1≥16故m能取到的最小整數(shù)為0.6.B解析由題意知,a2-a1=2,a3-a2則由累加法可知an-a1=2+3+…+n,所以an=1+2+…+n=n+n(當(dāng)n=1時(shí),1×(1+1)2=1=a1,則an=n(n+1)2,記1an的前n項(xiàng)的和為Sn,則Sn=1a1+1a2+…+1an=21-7.1021解析∵an-an+1=2anan+1,∴an-an+1∴數(shù)列1an∵1a3=5,∴1an=5+2(n-3)=2n-1.∴a∴anan+1=1(∴數(shù)列{anan+1}前10項(xiàng)的和為128.解(1)由Sn+1=an+n2,①得Sn+1+1=an+1+(n+1)2,②②-①,得an=2n+1.a1=3滿足上式,所以數(shù)列{an}的通項(xiàng)公式為an=2n+1.(2)由(1)得bn=(-1)n+22n+1,所以Tn=b1+b2+…+bn=[(-1)+(-1)2+…+(-1)n]+(23+25+…+22n+1)=(-=(-1)n-19.解(1)因?yàn)閿?shù)列{an}是公差為2的等差數(shù)列,且a1,a3,a7成等比數(shù)列,所以a32=a1·a7,即(a1+4)2=a1(a1+12),解得a1所以an=4+(n-1)×2=2n+2.(2)由(1)可得Sn=4n+n(n-1)2×2=n2+3n,bn=Sn-n23則13Tn=132+23①-②,得23Tn=1所以23Tn=1因此Tn=3410.解(1)設(shè)等比數(shù)列{an}的公比為q(q>0),則an=a1qn-1,且an>0.由已知得a1+a2=21a1+1a2=2(a1a3+a4=321a3+1a4=32(a3+∵a3a4=a1a2q4=2q4=32,且q>0,∴q=2.由a1a2=a12q=2,且a1>0,得a1=∴數(shù)列{an}的通項(xiàng)公式為an=2n-1.(2)由(1)知bn=an2+log2an=4n-1∴Tn=(1+4+42+…+4n-1)+(0+1+2+3+…+n-1)=1-11.解(1)設(shè)數(shù)列{an}的公比為q,因?yàn)閍2=4,所以a2=a1q=4.因?yàn)閍3+2是a2和a4的等差中項(xiàng),所以2(a3+2)=a2+a4,即2(a1q2+2)=a1q+a1q3,所以q2-2q=0.因?yàn)閝≠0,所以q=2,所以an=a2qn-2=4·2n-2=2n(n∈N*).(2)因?yàn)閍n=2n,所以bn=2nTn=12-1-122-1由Tn>6364,得1-1所以2n+1>65,即n>5,所以使Tn>6364成立的最小整數(shù)n=612.B解析由an+1=an2an∴數(shù)列1a∴1an=2n-1,又bn=anan+∴bn=1(∴Sn=12故選B.13.4解析當(dāng)n≥2時(shí),前n個(gè)1之間共有n+[1+2+3+…+(n-1)]=n(n+1)2項(xiàng),當(dāng)n=63時(shí),有63×故該數(shù)列前2020項(xiàng)的和為63×1+(2020-63)x=7891,解得x=4.14.解(1)因?yàn)閍n+12=2Sn+n+4,所以an2=2Sn-1+n-1兩式相減,得an+12-an所以an+12=an2+2an+1=因?yàn)閧an}是各項(xiàng)均為正數(shù)的數(shù)列,所以an+1=an+1,即an+1-an=1.又a32=(a2-1)a7,所以(a2+1)2=(a2-1)(a2解得a2=3.由a22=2S1+1+4,可得a1所以{an}是以2為首項(xiàng),1為公差的等差數(shù)列,所以an=n+1.由題意知b1=2,b2=4,b3=8,故bn=2n.(2)由(1)得cn=(-1)nlog22n-1(n+1)(n+2)=故Tn=c1+c2+…+cn=[-1+2-3+…+(-1)nn]-12設(shè)Fn=-1+2-3+…+(-1)nn,則當(dāng)n為偶數(shù)時(shí),Fn=(-1+2)+(-3+4)+…+[-(n-1)+n]=n2當(dāng)n為奇數(shù)時(shí),Fn=Fn-1+(-n)=n-12設(shè)Gn=12×3+1則Gn=12-13+所以Tn=n15.解(1)設(shè)等比數(shù)列{an}的公比為q,由已知an+1=2Sn+2,可得an=2Sn-1+2(n≥2),兩式相減可得an+1-an=2Sn-2Sn-1,即an+1-an=2an,整理得an+1=3an,可知q=3.已知an+1=2Sn+2,令n=1,得a2=2a1+2,即a1q=2a1+2,解得a1=2.故等比數(shù)列{an}的通項(xiàng)公式為an=2·3n-1(n∈N*).由b1=2,(n+2)bn=nbn+1(n∈N*),得bn那么b2b1以上n-1個(gè)式子相乘,可得bnb1=3bn=n(n+1)(n≥2),又b1=2滿足上式,所以{bn}的通項(xiàng)公式bn=n(n+1)(n∈N*).(2)若在an與an+1之間插入n個(gè)數(shù),使這n+2個(gè)數(shù)組成一個(gè)公差為cn的等差數(shù)列,則an+1-an=(n+1)cn,即為2·3n-2·3n-1=(n+1)cn,整理得cn=4·所以bncn=4·n·3n-1,Tn=b1c1+b2c2+b3c3+…+bn-1cn-1+bncn=4×1×30+4×2×31+4×3×32+…+4·(n-1)·3n-2+4·n·3n-1=4×(1×30+2×31+3×32+…+(n-1)·3n-2+n·3n-1),3Tn=4×[1×31+2×32+…+(n-1)·3n-1+n·3n],兩式相減得-2Tn=4×(30+31+32+…+3n-1-n·3n)=41-所以Tn=2n·3n+1-3n2=1+(2n-16.(1)證明方案一:選條件①當(dāng)n=1時(shí),2a1=2S1=3a1-3-4,解得a1=7,∴a1+2=7+2=9;當(dāng)n≥2時(shí),由2Sn=3an-3-4n,可得2Sn-1=3an-1-3-4(n-1),兩式相減,可得2an=3an-3an-1-4,即an=3an-1+4.∴an+2=3an-1+4+2=3(an-1+2),∴數(shù)列{an+2}是以9為首項(xiàng),3為公比的等比數(shù)列.方案二:選條件②當(dāng)n=1時(shí),a1+2=-3+2=-1;當(dāng)n≥2時(shí),an+1+2=-an-4+2=-(an+2).∴數(shù)列{an+2}是以-1為首項(xiàng),-1為公比的等比數(shù)列.(2)解由題意,設(shè)等差數(shù)列{bn}的公差為d,則d=b5-b35-3=9-52=2,b1∴bn=1+2(n-1)=2n-1,n∈N*.方案一:選條件①由(1),可得an+2=9·3n-1=3n+1,則cn=(an+2)bn=(2n-1)·3n+1,∴Tn=c1+c2+c3+…+cn=1×32+3×33+5×34+…+(2n-1)·3n+1,3Tn=1×33+3×34+…+(2n-3)·3n+1+(2n-1)·3n+2,兩式相減,可得-2Tn=1×32+2×33+2×34+…+2·3n+1-(