2019年高考數(shù)學(xué)二輪復(fù)習(xí) 專(zhuān)題四 數(shù)列 專(zhuān)題能力訓(xùn)練12 數(shù)列的通項(xiàng)與求和 文.doc

專(zhuān)題能力訓(xùn)練12數(shù)列的通項(xiàng)與求和一、能力突破訓(xùn)練1.已知數(shù)列an是等差數(shù)列,a1=tan 225,a5=13a1,設(shè)Sn為數(shù)列(-1)nan的前n項(xiàng)和,則S2 016=()A. 2 016B.-2 016C.3 024D.-3 0242.已知數(shù)列an的前n項(xiàng)和為Sn,且Sn=n2+n,數(shù)列bn滿足bn=1anan+1(nN*),Tn是數(shù)列bn的前n項(xiàng)和,則T9等于()A.919B.1819C.2021D.9403.已知數(shù)列an的前n項(xiàng)和Sn=n2-2n-1,則a3+a17=()A.15B.17C.34D.3984.已知函數(shù)f(x)滿足f(x+1)= +f(x)(xR),且f(1)=,則數(shù)列f(n)(nN*)前20項(xiàng)的和為()A.305B.315C.325D.3355.已知數(shù)列an,構(gòu)造一個(gè)新數(shù)列a1,a2-a1,a3-a2,an-an-1,此數(shù)列是首項(xiàng)為1,公比為的等比數(shù)列,則數(shù)列an的通項(xiàng)公式為()A.an=32-3213n,nN*B.an=32+3213n,nN*C.an=1,n=1,32+3213n,n>2,且nN*D.an=1,nN*6.植樹(shù)節(jié),某班20名同學(xué)在一段直線公路一側(cè)植樹(shù),每人植一棵,相鄰兩棵樹(shù)相距10 m.開(kāi)始時(shí)需將樹(shù)苗集中放置在某一樹(shù)坑旁邊,使每位同學(xué)從各自樹(shù)坑出發(fā)前來(lái)領(lǐng)取樹(shù)苗往返所走的路程總和最小,這個(gè)最小值為 m.7.數(shù)列an滿足an+1=11-an,a11=2,則a1=.8.數(shù)列an滿足a1+122a2+12nan=2n+5,nN*,則an=.9.設(shè)數(shù)列an的前n項(xiàng)和為Sn.已知S2=4,an+1=2Sn+1,nN*.(1)求通項(xiàng)公式an;(2)求數(shù)列|an-n-2|的前n項(xiàng)和.10.已知等差數(shù)列an的前n項(xiàng)和為Sn,等比數(shù)列bn的前n項(xiàng)和為T(mén)n,a1=-1,b1=1,a2+b2=2.(1)若a3+b3=5,求bn的通項(xiàng)公式;(2)若T3=21,求S3.11.已知數(shù)列an和bn滿足a1=2,b1=1,an+1=2an(nN*),b1+b2+b3+bn=bn+1-1(nN*).(1)求an與bn;(2)記數(shù)列anbn的前n項(xiàng)和為T(mén)n,求Tn.二、思維提升訓(xùn)練12.給出數(shù)列11,12,21,13,22,31,1k,2k-1, ,在這個(gè)數(shù)列中,第50個(gè)值等于1的項(xiàng)的序號(hào)是()A.4 900B.4 901C.5 000D.5 00113.設(shè)Sn是數(shù)列an的前n項(xiàng)和,且a1=-1,an+1=SnSn+1,則Sn=.14.設(shè)數(shù)列an的前n項(xiàng)和為Sn,已知a1=1,a2=2,且an+2=3Sn-Sn+1+3,nN*.(1)證明:an+2=3an;(2)求Sn.15.已知an是等比數(shù)列,前n項(xiàng)和為Sn(nN*),且1a1-1a2=2a3,S6=63.(1)求an的通項(xiàng)公式;(2)若對(duì)任意的nN*,bn是log2an和log2an+1的等差中項(xiàng),求數(shù)列(-1)nbn2的前2n項(xiàng)和.16.已知數(shù)列an滿足an+2=qan(q為實(shí)數(shù),且q1),nN*,a1=1,a2=2,且a2+a3,a3+a4,a4+a5成等差數(shù)列.(1)求q的值和an的通項(xiàng)公式;(2)設(shè)bn=log2a2na2n-1,nN*,求數(shù)列bn的前n項(xiàng)和.專(zhuān)題能力訓(xùn)練12數(shù)列的通項(xiàng)與求和一、能力突破訓(xùn)練1.C解析 a1=tan 225=1,a5=13a1=13,則公差d=a5-a15-1=13-14=3,an=3n-2.又(-1)nan=(-1)n(3n-2),S2 016=(a2-a1)+(a4-a3)+(a6-a5)+(a2 014-a2 013)+(a2 016-a2 015)=1 008d=3 024.2.D解析 數(shù)列an的前n項(xiàng)和為Sn,且Sn=n2+n,當(dāng)n=1時(shí),a1=2;當(dāng)n2時(shí),an=Sn-Sn-1=2n,an=2n(nN*),bn=1anan+1=12n(2n+2)=141n-1n+1,T9=141-12+12-13+19-110=141-110=940.3.C解析 Sn=n2-2n-1,a1=S1=12-2-1=-2.當(dāng)n2時(shí),an=Sn-Sn-1=n2-2n-1-(n-1)2-2(n-1)-1=n2-(n-1)2+2(n-1)-2n-1+1=n2-n2+2n-1+2n-2-2n=2n-3.an=-2,n=1,2n-3,n2.a3+a17=(23-3)+(217-3)=3+31=34.4.D解析 f(1)=,f(2)=32+52,f(3)=32+32+52,f(n)=32+f(n-1),f(n)是以52為首項(xiàng),32為公差的等差數(shù)列.S20=2052+20(20-1)232=335.5.A解析 因?yàn)閿?shù)列a1,a2-a1,a3-a2,an-an-1,是首項(xiàng)為1,公比為的等比數(shù)列,所以an-an-1=13n-1,n2.所以當(dāng)n2時(shí),an=a1+(a2-a1)+(a3-a2)+(an-an-1)=1+13+132+13n-1=1-13n1-13=32-3213n.又當(dāng)n=1時(shí),an=32-3213n=1,則an=32-3213n,nN*.6.2 000解析 設(shè)放在第x個(gè)坑邊,則S=20(|x-1|+|x-2|+|20-x|).由式子的對(duì)稱性討論,當(dāng)x=10或11時(shí),S=2 000.當(dāng)x=9或12時(shí),S=20102=2 040;當(dāng)x=1或19時(shí),S=3 800.Smin=2 000 m.7.解析 由a11=2及an+1=11-an,得a10=.同理a9=-1,a8=2,a7=12,.所以數(shù)列an是周期為3的數(shù)列.所以a1=a10=12.8.14,n=1,2n+1,n2解析 在a1+122a2+12nan=2n+5中用(n-1)代換n得a1+122a2+12n-1an-1=2(n-1)+5(n2),兩式相減,得12nan=2,an=2n+1,又a1=7,即a1=14,故an=14,n=1,2n+1,n2.9.解 (1)由題意得a1+a2=4,a2=2a1+1,則a1=1,a2=3.又當(dāng)n2時(shí),由an+1-an=(2Sn+1)-(2Sn-1+1)=2an,得an+1=3an.所以,數(shù)列an的通項(xiàng)公式為an=3n-1,nN*.(2)設(shè)bn=|3n-1-n-2|,nN*,b1=2,b2=1.當(dāng)n3時(shí),由于3n-1>n+2,故bn=3n-1-n-2,n3.設(shè)數(shù)列bn的前n項(xiàng)和為T(mén)n,則T1=2,T2=3.當(dāng)n3時(shí),Tn=3+9(1-3n-2)1-3-(n+7)(n-2)2=3n-n2-5n+112,所以Tn=2,n=1,3n-n2-5n+112,n2,nN*.10.解 設(shè)an的公差為d,bn的公比為q,則an=-1+(n-1)d,bn=qn-1.由a2+b2=2得d+q=3.(1)由a3+b3=5,得2d+q2=6.聯(lián)立和解得d=3,q=0(舍去),d=1,q=2.因此bn的通項(xiàng)公式為bn=2n-1.(2)由b1=1,T3=21得q2+q-20=0,解得q=-5或q=4.當(dāng)q=-5時(shí),由得d=8,則S3=21.當(dāng)q=4時(shí),由得d=-1,則S3=-6.11.解 (1)由a1=2,an+1=2an,得an=2n(nN*).由題意知:當(dāng)n=1時(shí),b1=b2-1,故b2=2.當(dāng)n2時(shí),1nbn=bn+1-bn,整理得bn+1n+1=bnn,所以bn=n(nN*).(2)由(1)知anbn=n2n,因此Tn=2+222+323+n2n,2Tn=22+223+324+n2n+1,所以Tn-2Tn=2+22+23+2n-n2n+1.故Tn=(n-1)2n+1+2(nN*).二、思維提升訓(xùn)練12.B解析 根據(jù)條件找規(guī)律,第1個(gè)1是分子、分母的和為2,第2個(gè)1是分子、分母的和為4,第3個(gè)1是分子、分母的和為6,第50個(gè)1是分子、分母的和為100,而分子、分母的和為2的有1項(xiàng),分子、分母的和為3的有2項(xiàng),分子、分母的和為4的有3項(xiàng),分子、分母的和為99的有98項(xiàng),分子、分母的和為100的項(xiàng)依次是:199,298,397,5050,5149,991,第50個(gè)1是其中第50項(xiàng),在數(shù)列中的序號(hào)為1+2+3+98+50=98(1+98)2+50=4 901.13.-解析 由an+1=Sn+1-Sn=SnSn+1,得1Sn-1Sn+1=1,即1Sn+1-1Sn=-1,則1Sn為等差數(shù)列,首項(xiàng)為1S1=-1,公差為d=-1,1Sn=-n,Sn=-.14.(1)證明 由條件,對(duì)任意nN*,有an+2=3Sn-Sn+1+3,因而對(duì)任意nN*,n2,有an+1=3Sn-1-Sn+3.兩式相減,得an+2-an+1=3an-an+1,即an+2=3an,n2.又a1=1,a2=2,所以a3=3S1-S2+3=3a1-(a1+a2)+3=3a1,故對(duì)一切nN*,an+2=3an.(2)解 由(1)知,an0,所以an+2an=3,于是數(shù)列a2n-1是首項(xiàng)a1=1,公比為3的等比數(shù)列;數(shù)列a2n是首項(xiàng)a2=2,公比為3的等比數(shù)列.因此a2n-1=3n-1,a2n=23n-1.于是S2n=a1+a2+a2n=(a1+a3+a2n-1)+(a2+a4+a2n)=(1+3+3n-1)+2(1+3+3n-1)=3(1+3+3n-1)=3(3n-1)2,從而S2n-1=S2n-a2n=3(3n-1)2-23n-1=32(53n-2-1).綜上所述,Sn=32(53n-32-1),當(dāng)n是奇數(shù),32(3n2-1),當(dāng)n是偶數(shù).15.解 (1)設(shè)數(shù)列an的公比為q.由已知,有1a1-1a1q=2a1q2,解得q=2或q=-1.又由S6=a11-q61-q=63,知q-1,所以a11-261-2=63,得a1=1.所以an=2n-1.(2)由題意,得bn=12(log2an+log2an+1)=12(log22n-1+log22n)=n-12,即bn是首項(xiàng)為12,公差為1的等差數(shù)列.設(shè)數(shù)列(-1)nbn2的前n項(xiàng)和為T(mén)n,則T2n=(-b12+b22)+(-b32+b42)+(-b2n-12+b2n2)=b1+b2+b3+b4+b2n-1+b2n=2n(b1+b2n)2=2n2.16.解 (1)由已知,有(a3+a4)-(a2+a3)=(a4+a5)-(a3+a4),即a4-a2=a5-a3,所以a2(q-1)=a3(q-1).又因?yàn)閝1,故a3=a2=2,由a3=a1q,得q=2.當(dāng)n=2k-1(kN*)時(shí),an=a2k-1=2k-1=2n-12;當(dāng)n=2k(kN*)時(shí),an=a2k=2k=2n2.所以,an的通項(xiàng)公式為an=2n-12,n為奇數(shù),2n2,n為偶數(shù).(2)由(1)得bn=log2a2na2n-1=n2n-1.設(shè)bn的前n項(xiàng)和為Sn,則Sn=1120+2121+3122+(n-1)12n-2+n12n-1,12Sn=1121+2122+3123+(n-1)12n-1+n12n,上述兩式相減,得12Sn=1+12+122+12n-1-n2n=1-12n1-12-n2n=2-22n-n2n,整理得,Sn=4-n+22n-1.所以,數(shù)列bn的前n項(xiàng)和為4-n+22n-1,nN*.