1、2017春高中数学第2章数列2.1数列第2课时数列的递推公式(选学)课时作业新人教B版必修5基础巩固一、选择题1.已知数列{an}的首项a1=1,且an=3an-1+1(n≥2),则a4为( D )A.13 B.15 C.30 D.40[解析] ∵a1=1,∴a2=3a1+1=4,a3=3a2+1=13,a4=3a3+1=40.2.已知数列{an}中,a1=1,a2=3,an=an-1+(n≥3),则a5=( A )A.B.C.4D.5[解析] ∵a1=1,a2=3,∴a3=a2+=3+1=4,∴a4=a3+=4+=,∴a5=a4+=+=.3.已知数列
2、{an}中,an+1=an+2+an,a1=2,a2=5,则a6=( A )A.-3B.-4C.-5D.2[解析] 由an+1=an+2+an得a3=3,a4=-2,a5=-5,a6=-3.4.数列{an}满足关系anan+1=1-an+1(n∈N+),且a2014=2,则a2012等于( A )A.-3B.3C.-2D.2[解析] 由已知,得an==-1,所以a2013=-1=-1=-,a2012=-1=-1=-3.5.数列{an}中,a1=1,以后各项由公式a1·a2·a3·…·an=n2给出,则a3+a5等于( C )A.B.C.D.[解析] ∵a1·
3、a2·a3·…·an=n2,∴a1·a2·a3=9,a1·a2=4,∴a3=.同理a5=,∴a3+a5=+=.6.已知数列{an}满足a1=0,an+1=(n∈N+),则a20=( B )A.0B.-C.D.[解析] ∵an+1=,a1=0,∴a2=-,a3=,a4=0,a5=-,a6=,…,故此数列周期为3,∴a20=a3×6+2=a2=-,故选B.二、填空题7.已知数列{an}满足a1=1,an=nan-1(n≥2),则a5=120.[解析] 因为an=nan-1,且n≥2,所以当n=2时,a2=2a1=2;当n=3时,a3=3a2=6;当n=4时,a4
5、-nan](an+1+an)=0.∵an>0,∴an+an+1>0,∴(n+1)an+1-nan=0,∴=.∴····…·=××××…×=,∴=.又a1=1,∴an=a1=.能力提升一、选择题1.正项数列{an}中,an+1=,a1=2,则a4=( B )A.B.C.D.[解析] 由递推关系可得a2=,a3=,a4=.2.数列{an}满足a1=1,an+1=2an-1(n∈N*),则a1000=( A )A.1B.1999C.1000D.-1[解析] a1=1,a2=2×1-1=1,a3=2×1-1=1,a4=2×1-1=1,…,可知an=1(n∈N*).故
6、选A.3.在数列{an}中,a1=1,an+1=a-1(n≥1),则a1+a2+a3+a4+a5等于( A )A.-1B.1C.0D.2[解析] ∵an+1=a-1=(an+1)(an-1),∴a2=(a1+1)(a1-1)=0,a3=(a2+1)(a2-1)=-1,a4=(a3+1)(a3-1)=0,a5=(a4+1)(a4-1)=-1,∴a1+a2+a3+a4+a5=-1.4.观察下图,并阅读图形下面的文字,像这样10条直线相交,交点的个数最多的是( B )A.40个B.45个C.50个D.55个[解析] 交点个数依次组成数列为1,3,6,即,,,由此猜