（新课标）2018届高考数学二轮复习 专题四 数列 专题能力训练12 数列的通项与求和 理.doc

专题能力训练12数列的通项与求和能力突破训练1.(2017甘肃兰州一诊)已知等差数列an的前n项和为sn,若a1=2,a4+a10=28,则s9=()a.45b.90c.120d.752.(2017东北三校联考)已知数列an是等差数列,满足a1+2a2=s5,下列结论错误的是()a.s9=0b.s5最小c.s3=s6d.a5=03.已知数列an的前n项和sn=n2-2n-1,则a3+a17=()a.15b.17c.34d.3984.已知函数f(x)满足f(x+1)=32+f(x)(xr),且f(1)=52,则数列f(n)(nn*)前20项的和为()a.305b.315c.325d.3355.已知数列an,构造一个新数列a1,a2-a1,a3-a2,an-an-1,此数列是首项为1,公比为13的等比数列,则数列an的通项公式为()a.an=32-3213n,nn*b.an=32+3213n,nn*c.an=1,n=1,32+3213n,n2,且nn*d.an=1,nn*6.(2017山西大同豪洋中学三模)已知数列an满足a1=1,an-an+1=nanan+1(nn*),则an=.7.(2017河北石家庄一模)已知数列an中,a1=a,an+1=3an+8n+6,若an为递增数列,则实数a的取值范围为.8.已知sn是等差数列an的前n项和,若a1=-2 017,s20142014-s20082008=6,则s2 017=.9.已知在数列an中,a1=1,an+1=an+2n+1,且nn*.(1)求数列an的通项公式;(2)令bn=2n+1anan+1,数列bn的前n项和为tn.如果对于任意的nn*,都有tnm,求实数m的取值范围.10.已知数列an的前n项和为sn,且a1=0,对任意nn*,都有nan+1=sn+n(n+1).(1)求数列an的通项公式;(2)若数列bn满足an+log2n=log2bn,求数列bn的前n项和tn.11.设数列an的前n项和为sn.已知2sn=3n+3.(1)求an的通项公式;(2)若数列bn满足anbn=log3an,求bn的前n项和tn.思维提升训练12.给出数列11,12,21,13,22,31,1k,2k-1,k1,在这个数列中,第50个值等于1的项的序号是()a.4 900b.4 901c.5 000d.5 00113.设sn是数列an的前n项和,且a1=-1,an+1=snsn+1,则sn=.14.已知等差数列an的公差为2,其前n项和sn=pn2+2n(nn*).(1)求p的值及an;(2)若bn=2(2n-1)an,记数列bn的前n项和为tn,求使tn910成立的最小正整数n的值.15.已知数列an满足an+2=qan(q为实数,且q1),nn*,a1=1,a2=2,且a2+a3,a3+a4,a4+a5成等差数列.(1)求q的值和an的通项公式;(2)设bn=log2a2na2n-1,nn*,求数列bn的前n项和.16.设数列a:a1,a2,an(n2).如果对小于n(2nn)的每个正整数k都有aka1,则g(a);(3)证明:若数列a满足an-an-11(n=2,3,n),则g(a)的元素个数不小于an-a1.参考答案专题能力训练12数列的通项与求和能力突破训练1.b解析因为an是等差数列,设公差为d,所以a4+a10=a1+3d+a1+9d=2a1+12d=4+12d=28,解得d=2.s9=9a1+982d=18+362=90.故选b.2.b解析由题设可得3a1+2d=5a1+10d2a1+8d=0,即a5=0,所以d中结论正确.由等差数列的性质可得a1+a9=2a5=0,则s9=9(a1+a9)2=9a5=0,所以a中结论正确.s3-s6=3a1+3d-6a1-15d=-3(a1+4d)=-3a5=0,所以c中结论正确.b中结论是错误的.故选b.在求等差数列的前n项和的最值时,一定要注意nn*.3.c解析sn=n2-2n-1,a1=s1=12-2-1=-2.当n2时,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)=52,f(2)=32+52,f(3)=32+32+52,f(n)=32+f(n-1),f(n)是以52为首项,32为公差的等差数列.s20=2052+20(20-1)232=335.5.a解析因为数列a1,a2-a1,a3-a2,an-an-1,是首项为1,公比为13的等比数列,所以an-an-1=13n-1,n2.所以当n2时,an=a1+(a2-a1)+(a3-a2)+(an-an-1)=1+13+132+13n-1=1-13n1-13=32-3213n.又当n=1时,an=32-3213n=1,则an=32-3213n,nn*.6.2n2-n+2解析因为an-an+1=nanan+1,所以an-an+1anan+1=1an+1-1an=n,1an=1an-1an-1+1an-1-1an-2+1a2-1a1+1a1=(n-1)+(n-2)+3+2+1+1a1=(n-1)(n-1+1)2+1=n2-n+22(n2).所以an=2n2-n+2(n2).又a1=1也满足上式,所以an=2n2-n+2.7.(-7,+)解析由an+1=3an+8n+6,得an+1+4(n+1)+5=3(an+4n+5),即an+1+4(n+1)+5an+4n+5=3,所以数列an+4n+5是首项为a+9,公比为3的等比数列.所以an+4n+5=(a+9)3n-1,即an=(a+9)3n-1-4n-5.所以an+1=(a+9)3n-4n-9.因为数列an为递增数列,所以an+1an,即(a+9)3n-4n-9(a+9)3n-1-4n-5,即(a+9)3n6恒成立.因为nn*,所以(a+9)36,解得a-7.8.-2 017解析sn是等差数列an的前n项和,snn是等差数列,设其公差为d.s20142014-s20082008=6,6d=6,d=1.a1=-2017,s11=-2017.snn=-2017+(n-1)1=-2018+n.s2017=(-2018+2017)2017=-2017.故答案为-2017.9.解(1)an+1=an+2n+1,an+1-an=2n+1,an-an-1=2n-1,an=a1+(a2-a1)+(a3-a2)+(an-an-1)=1+3+5+(2n-1)=n(1+2n-1)2=n2.(2)由(1)知,bn=2n+1anan+1=2n+1n2(n+1)2=1n2-1(n+1)2,tn=112-122+122-132+1n2-1(n+1)2=1-1(n+1)2,数列tn是递增数列,最小值为1-1(1+1)2=34,只需要34m,m的取值范围是-,34.10.解(1)(方法一)nan+1=sn+n(n+1),当n2时,(n-1)an=sn-1+n(n-1),两式相减,得nan+1-(n-1)an=sn-sn-1+n(n+1)-n(n-1),即nan+1-(n-1)an=an+2n,得an+1-an=2.当n=1时,1a2=s1+12,即a2-a1=2.数列an是以0为首项,2为公差的等差数列.an=2(n-1)=2n-2.(方法二)由nan+1=sn+n(n+1),得n(sn+1-sn)=sn+n(n+1),整理,得nsn+1=(n+1)sn+n(n+1),两边同除以n(n+1),得sn+1n+1-snn=1.数列snn是以s11=0为首项,1为公差的等差数列,snn=0+n-1=n-1.sn=n(n-1).当n2时,an=sn-sn-1=n(n-1)-(n-1)(n-2)=2n-2.又a1=0适合上式,数列an的通项公式为an=2n-2.(2)an+log2n=log2bn,bn=n2an=n22n-2=n4n-1.tn=b1+b2+b3+bn-1+bn=40+241+342+(n-1)4n-2+n4n-1,4tn=41+242+343+(n-1)4n-1+n4n,由-,得-3tn=40+41+42+4n-1-n4n=1-4n1-4-n4n=(1-3n)4n-13.tn=19(3n-1)4n+1.11.解(1)因为2sn=3n+3,所以2a1=3+3,故a1=3.当n1时,2sn-1=3n-1+3,此时2an=2sn-2sn-1=3n-3n-1=23n-1,即an=3n-1,所以an=3,n=1,3n-1,n1.(2)因为anbn=log3an,所以b1=13,当n1时,bn=31-nlog33n-1=(n-1)31-n.所以t1=b1=13;当n1时,tn=b1+b2+b3+bn=13+(13-1+23-2+(n-1)31-n),所以3tn=1+(130+23-1+(n-1)32-n),两式相减,得2tn=23+(30+3-1+3-2+32-n)-(n-1)31-n=23+1-31-n1-3-1-(n-1)31-n=136-6n+323n,所以tn=1312-6n+343n.经检验,当n=1时也适合.综上可得tn=1312-6n+343n.思维提升训练12.b解析根据条件找规律,第1个1是分子、分母的和为2,第2个1是分子、分母的和为4,第3个1是分子、分母的和为6,第50个1是分子、分母的和为100,而分子、分母的和为2的有1项,分子、分母的和为3的有2项,分子、分母的和为4的有3项,分子、分母的和为99的有98项,分子、分母的和为100的项依次是:199,298,397,5050,5149,991,第50个1是其中第50项,在数列中的序号为1+2+3+98+50=98(1+98)2+50=4901.13.-1n解析由an+1=sn+1-sn=snsn+1,得1sn-1sn+1=1,即1sn+1-1sn=-1,则1sn为等差数列,首项为1s1=-1,公差为d=-1,1sn=-n,sn=-1n.14.解(1)(方法一)an是等差数列,sn=na1+n(n-1)2d=na1+n(n-1)22=n2+(a1-1)n.又由已知sn=pn2+2n,p=1,a1-1=2,a1=3,an=a1+(n-1)d=2n+1,p=1,an=2n+1.(方法二)由已知a1=s1=p+2,s2=4p+4,即a1+a2=4p+4,a2=3p+2.又等差数列的公差为2,a2-a1=2,2p=2,p=1,a1=p+2=3,an=a1+(n-1)d=2n+1,p=1,an=2n+1.(方法三)当n2时,an=sn-sn-1=pn2+2n-p(n-1)2+2(n-1)=2pn-p+2,a2=3p+2,由已知a2-a1=2,2p=2,p=1,a1=p+2=3,an=a1+(n-1)d=2n+1,p=1,an=2n+1.(2)由(1)知bn=2(2n-1)(2n+1)=12n-1-12n+1,tn=b1+b2+b3+bn=11-13+13-15+15-17+12n-1-12n+1=1-12n+1=2n2n+1.tn910,2n2n+1910,20n18n+9,即n92.nn*,使tn910成立的最小正整数n的值为5.15.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).又因为q1,故a3=a2=2,由a3=a1q,得q=2.当n=2k-1(kn*)时,an=a2k-1=2k-1=2n-12;当n=2k(kn*)时,an=a2k=2k=2n2.所以,an的通项公式为an=2n-12,n为奇数,2n2,n为偶数.(2)由(1)得bn=log2a2na2n-1=n2n-1.设bn的前n项和为sn,则sn=1120+2121+3122+(n-1)12n-2+n12n-1,12sn=1121+2122+3123+(n-1)12n-1+n12n,上述两式相减,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,数列bn的前n项和为4-n+22n-1,nn*.16.解(1)g(a)的元素为2和5.(2)因为存在an使得ana1,所以in*|2in,aia1.记m=minin*|2in,aia1,则m2,且对任意正整数km,aka1a1.由(2)知g(a).设g(a)