（浙江專用）2020版高考數(shù)學(xué)大一輪復(fù)習(xí) 高考解答題專項(xiàng)練3 數(shù)列.docx

高考解答題專項(xiàng)練數(shù)列1.已知正數(shù)數(shù)列an的前n項(xiàng)和為Sn,滿足an2=Sn+Sn-1(n2),a1=1.(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)bn=(1-an)2-a(1-an),若bn+1>bn對(duì)任意nN*恒成立,求實(shí)數(shù)a的取值范圍.解(1)an2=Sn+Sn-1(n2),an-12=Sn-1+Sn-2(n3).兩式相減可得an2-an-12=Sn-Sn-2=an+an-1,an-an-1=1.a1=1,正數(shù)數(shù)列an是以1為首項(xiàng),以1為公差的等差數(shù)列,an=n.(2)bn=(1-an)2-a(1-an),bn+1=(1-an+1)2-a(1-an+1).即bn=(1-n)2-a(1-n)=n2+(a-2)n+1-a,bn+1=1-(n+1)2-a1-(n+1)=n2+an.故bn+1-bn=2n+a-1.再由bn+1>bn對(duì)任意nN*恒成立可得2n+a-1>0恒成立,故a>1-2n恒成立.而1-2n的最大值為1-2=-1,故a>-1,即實(shí)數(shù)a的取值范圍為(-1,+).2.已知數(shù)列an滿足a1=1,Sn=2an+1,其中Sn為an的前n項(xiàng)和(nN*).(1)求S1,S2及數(shù)列Sn的通項(xiàng)公式;(2)若數(shù)列bn滿足bn=(-1)nSn,且bn的前n項(xiàng)和為Tn,求證:當(dāng)n2時(shí),13|Tn|79.(1)解數(shù)列an滿足Sn=2an+1,則Sn=2an+1=2(Sn+1-Sn),即3Sn=2Sn+1,Sn+1Sn=32.即數(shù)列Sn為以1為首項(xiàng),以32為公比的等比數(shù)列,Sn=32n-1(nN*).S1=1,S2=32.(2)證明在數(shù)列bn中,bn=(-1)nSn=(-1)(-1)n-132n-1=-23n-1,Tn為bn的前n項(xiàng)和,則|Tn|=(-1)1+-23+49+-233+-23n-1=1+-23+49+-233+-23n-1.而當(dāng)n2時(shí),1-231+-23+49+-233+-23n-11+-23+49=79,即13|Tn|79.3.已知數(shù)列an滿足:an2-an-an+1+1=0,a1=2.(1)求a2,a3;(2)證明數(shù)列an為遞增數(shù)列;(3)求證:1a1+1a2+1a3+1an<1.(1)解a1=2,an+1=an2-an+1,a2=22-2+1=3,同理可得a3=7.(2)證明an+1-an=an2-2an+1=(an-1)2>0,對(duì)nN*恒成立,an+1>an.(3)證明an+1-1=an2-an,故1an+1-1=1an2-an=1an-1-1an,故1an=1an-1-1an+1-1,故1a1+1a2+1a3+1an=1a1-1-1a2-1+1a2-1-1a3-1+1an-1-1an+1-1=1a1-1-1an+1-1=12-1-1an+1-1=1-1an+1-1<1.4.(2018浙江桐鄉(xiāng)第一次模擬考)已知數(shù)列an滿足a1=1,2(n+1)an+1=(n+2)an.(1)求a2,a3,并求數(shù)列an的通項(xiàng)公式;(2)設(shè)Sn是數(shù)列an的前n項(xiàng)和,求Sn.解(1)a2=34,a3=12.2(n+1)an+1=(n+2)an,an+1an=n+22(n+1).由累乘法得a2a1a3a2a4a3anan-1=322423524n+12n,ana1=n+12n-12=n+12n.an=n+12n.(2)由題意知Sn=221+322+423+n+12n,12Sn=222+323+n2n+n+12n+1,-,得12Sn=1+122+123+12n-n+12n+1=1+1221-12n-11-12-n+12n+1=32-n+32n+1,故Sn=3-n+32n.5.(2018浙江“七彩陽光”聯(lián)盟期初聯(lián)考)已知數(shù)列an滿足a1=2,(2n+1)anan+1=2n+1(2an-an+1)(nN*).(1)求a2,a3的值;(2)如果數(shù)列bn滿足anbn=2n,求數(shù)列bn的通項(xiàng)公式bn.解(1)由已知得an+1=2n+1ann+12an+2n(nN*),因?yàn)閍1=2,所以a2=21+1a11+12a1+21=85,a3=22+1a22+12a2+22=85.(2)因?yàn)閍nbn=2n,且由已知可得an+12n+1=ann+12an+2n,把bn=2nan代入即得bn+1-bn=n+12,所以b2-b1=1+12,b3-b2=2+12,bn-bn-1=(n-1)+12,累加得bn-b1=1+2+3+(n-1)+n-12=(n-1)n2+n-12=n2-12,又b1=2a1=22=1,因此bn=n2-12+1=n2+12.6.(2018江蘇高考)設(shè)an是首項(xiàng)為a1,公差為d的等差數(shù)列,bn是首項(xiàng)為b1,公比為q的等比數(shù)列.(1)設(shè)a1=0,b1=1,q=2,若|an-bn|b1對(duì)n=1,2,3,4均成立,求d的取值范圍;(2)若a1=b1>0,mN*,q(1,m2,證明:存在dR,使得|an-bn|b1對(duì)n=2,3,m+1均成立,并求d的取值范圍(用b1,m,q表示).解(1)由條件知,an=(n-1)d,bn=2n-1.因?yàn)閨an-bn|b1對(duì)n=1,2,3,4均成立,即|(n-1)d-2n-1|1對(duì)n=1,2,3,4均成立,即11,1d3,32d5,73d9,得73d52.因此,d的取值范圍為73,52.(2)由條件知,an=b1+(n-1)d,bn=b1qn-1.若存在d,使得|an-bn|b1(n=2,3,m+1)成立,即|b1+(n-1)d-b1qn-1|b1(n=2,3,m+1),即當(dāng)n=2,3,m+1時(shí),d滿足qn-1-2n-1b1dqn-1n-1b1.因?yàn)閝(1,m2,則1<qn-1qm2,從而qn-1-2n-1b10,qn-1n-1b1>0,對(duì)n=2,3,m+1均成立.因此,取d=0時(shí),|an-bn|b1對(duì)n=2,3,m+1均成立.下面討論數(shù)列qn-1-2n-1的最大值和數(shù)列qn-1n-1的最小值(n=2,3,m+1).當(dāng)2nm時(shí),qn-2n-qn-1-2n-1=nqn-qn-nqn-1+2n(n-1)=n(qn-qn-1)-qn+2n(n-1),當(dāng)1<q21m時(shí),有qnqm2,從而n(qn-qn-1)-qn+2>0.因此,當(dāng)2nm+1時(shí),數(shù)列qn-1-2n-1單調(diào)遞增,故數(shù)列qn-1-2n-1的最大值為qm-2m.設(shè)f(x)=2x(1-x),當(dāng)x>0時(shí),f(x)=(ln2-1-xln2)2x<0,所以f(x)單調(diào)遞減,從而f(x)<f(0)=1.當(dāng)2nm時(shí),qnnqn-1n-1=q(n-1)n21n1-1n=f1n<1,因此,當(dāng)2nm+1時(shí),數(shù)列qn-1n-1單調(diào)遞減,故數(shù)列qn-1n-1的最小值為qmm.因此,d的取值范圍為b1(qm-2)m,b1qmm.