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课时作业5数列的递推公式(选学)时间:45分钟满分:100分课堂训练1.在数列{an}中,a1=13,an=(-1)n·2an-1(n≥2),则a5=()A.-163B.163C.-83D.83【答案】B【解析】由an=(-1)n·2an-1知a2=23,a3=-2a2=-43,a4=2a3=-83,a5=-2a4=163.2.某数列第一项为1,并且对所有n≥2,n∈N,数列的前n项之积为n2,则这个数列的通项公式是()A.an=2n-1B.an=n2C.an=n2n-12D.an=n+12n2【答案】C【解析】∵a1·a2·a3·…·an=n2,a1·a2·a3·…·an-1=(n-1)2,∴两式相除,得an=n2n-12.3.已知数列{an}满足:a4n-3=1,a4n-1=0,a2n=an,n∈N+,则a2009=________,a2014=________.【答案】10【解析】考查数列的通项公式.∵2009=4×503-3,∴a2009=1,∵2014=2×1007,∴a2014=a1007,又1007=4×252-1,∴a1007=a4×252-1=0.4.已知数列{an},a1=0,an+1=1+an3-an,写出数列的前4项,并归纳出该数列的通项公式.【解析】a1=0,a2=1+a13-a1=13,a3=1+a23-a2=1+133-13=12,a4=1+a33-a3=1+123-12=35.直接观察可以发现,把a3=12写成a3=24,这样可知an=n-1n+1(n≥2,n∈N+).当n=1时,1-11+1=0=a1,所以an=n-1n+1(n∈N+).课后作业一、选择题(每小题5分,共40分)1.已知数列{an}满足:a1=-14,an=1-1an-1(n≥2),则a4=()A.45B.14C.-14D.15【答案】C【解析】∵a1=-14,an=1-1an-1(n≥2),∴a2=1-1a1=1-1-14=5,a3=1-1a2=1-15=45,a4=1-1a3=1-145=1-54=-14.2.数列{an}满足a1=13,an=-1an-1(n≥2,n∈N+),则a2013=()A.13B.-13C.3D.-3【答案】A【解析】由已知得,a2=-3,a3=13,a4=-3,所以an=13,n为奇数,-3,n为偶数,故a2013=13,选A.3.数列1,3,6,10,15,…的递推公式是()A.a1=1an+1=an+nn∈N+B.a1=1an=an-1+nn∈N+,n≥2C.a1=1an+1=an+n+1n∈N+,n≥2D.a1=1an=an-1+n-1n∈N+【答案】B【解析】代入验证得B.4.一个数列{an},其中a1=3,a2=6,an+2=an+1-an,那么这个数列的第5项是()A.6B.-3C.-12D.-6【答案】D【解析】an+2=an+1-an,an+3=an+2-an+1=an+1-an-an+1=-an,故a5=a2+3=-a2=-6.5.观察下图,并阅读图形下面的文字,像这样10条直线相交,交点的个数最多是()A.40个B.45个C.50个D.55个【答案】B【解析】交点个数依次组成数列为1,3,6,即2×12,2×32,3×42,由此猜想an=nn-12(n≥2,n∈N+),∴a10=10×92=45.6.在数列{an}中,a1=5,an+1=an+4n-1(n∈N+),则通项an等于()A.2n2-3nB.2n2-3n+6C.n2-3n+6D.2n2-3n+9【答案】B【解析】∵an+1-an=4n-1,∴a2-a1=4×1-1,a3-a2=4×2-1,a4-a3=4×3-1,…,an-an-1=4(n-1)-1,累加上述各式,得an-a1=4(1+2+…+n-1)-(n-1),∴an=2n2-3n+6.7.已知{an}中,a1=1,anan-1=an-1+(-1)n(n≥2),则a3a5的值为()A.-3B.-4C.34D.43【答案】C【解析】由递推公式逐个求解.8.已知数列{an}满足a1=0,an+1=an-33an+1(n∈N+),则a2013等于()A.0B.-3C.3D.32【答案】C【解析】a1=0,a2=-3,a3=3,a4=0,…,T=3,∴a2013=a3=3.二、填空题(每小题10分,共20分)9.设数列{an}满足a1=1,an=2+1an-1(n1),则a4=________.【答案】177【解析】由递推公式a2=2+1a1=3,a3=2+1a2=73,a4=2+1a3=177.10.已知数列{an}对任意p,q∈N+,有ap+aq=ap+q,若a1=19,则a36=________.【答案】4【解析】由已知得,a2=a1+1=2a1=29;a4=a2+2=2a2=49;a8=a4+4=2a4=89;a9=a1+8=a1+a8=19+89=1,a36=4a9=4.三、解答题(每小题20分,共40分.解答应写出必要的文字说明、证明过程或演算步骤)11.已知数列{an}的前n项和Sn分别是:(1)Sn=n2+n+1;(2)Sn=2n-1,求通项an.【解析】(1)当n=1时,a1=S1=3.当n≥2时,an=Sn-Sn-1=2n.∵a1不适合an,∴an=3,n=12n,n≥2.(2)当n=1时,a1=S1=1.当n≥2时,an=Sn-Sn-1=(2n-1)-(2n-1-1)=2n-1.∵a1适合an,∴an=2n-1(n≥1).12.求满足下列条件的数列{an}的通项公式.(1)已知{an}满足an+1=an+14n2-1,且a1=12,求an;(2)已知{an}满足an+1=3nan,且a1=3,求an.【解析】(1)由已知条件有an+1-an=14n2-1=12(12n-1-12n+1),∴an=a1+(a2-a1)+(a3-a2)+…+(an-an-1)=12+12[(1-13)+(13-15)+…+(12n-3-12n-1)]=12+12·(1-12n-1)=4n-34n-2.(2)由an+1=3nan,得an+1an=3n,∴an=a1·a2a1·a3a2·…·anan-1=3·3·32·…·3n-1=31+1+2+…+n-1=3n2-n+22.
本文标题:数列的递推公式练习
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