新版高中數(shù)學(xué)人教A版必修5習(xí)題第二章數(shù)列2.1.2

第2課時(shí)數(shù)列的遞推公式課時(shí)過關(guān)·能力提升基礎(chǔ)鞏固1在數(shù)列{an}中,a1=1,an+1=an3,則a3等于().A.7 B.4 C.1 D.2解析:a2=a13=13=4,a3=a23=43=7.答案:A2已知數(shù)列{an}滿足an+1=an+1A.遞增數(shù)列 B.遞減數(shù)列C.擺動(dòng)數(shù)列 D.常數(shù)列解析:由an+1=an+12,知an+所以an+1>an,即從第2項(xiàng)起,每一項(xiàng)都大于它的前一項(xiàng).答案:A3在數(shù)列{an}中,a1=13,an=2an-1(nA.C.解析:a2=2a1=答案:C4在數(shù)列{an}中,an1=man+1(n>1),且a2=3,a3=5,則實(shí)數(shù)m等于().A.0 B.C.2 D.5解析:由題意得a2=ma3+1,則3=5m+1,故m=答案:B5在正項(xiàng)數(shù)列{an}中,a1=1,a2=2,2an2=an+12+an-A.16 B.8 C.2解析:∵2∴aaa又a6>0,∴a6=4.答案:D6已知數(shù)列{an}對(duì)于任意的m,n∈N*都有am+an=am+n,若a1=1A.答案:D7已知數(shù)列{an}滿足a1=2,a2=5,a3=23,且an+1=αan+β,則α,β的值分別為.

解析:∵an+1=αan+β,∴答案:6,78已知在數(shù)列{an}中,an+1=2解析:由已知a7=2a6又因?yàn)閍6=2a5a5答案:19在數(shù)列{an}中,a1=1,an+1=2+1解a2=2+a3=2+a4=2+a5=2+10在數(shù)列{an}中,a1=1,an+1=n(1)寫出數(shù)列的前5項(xiàng);(2)猜想數(shù)列的一個(gè)通項(xiàng)公式.解(1)a1=1,a2=a4=(2)猜想:an=能力提升1數(shù)列1A.an=12n+1(B.an=12n(C.an+1=12an(D.an+1=2an(n∈N*)解析:數(shù)列從第二項(xiàng)起,后一項(xiàng)是前一項(xiàng)的12,故遞推公式為an+1=1答案:C2由1,3,5,…,2n1,…構(gòu)成數(shù)列{an},數(shù)列{bn}滿足b1=2,當(dāng)n≥2時(shí),bn=aA.9 B.17 C.33 D.65解析:∵an=2n1,∴bn=又b1=2,∴b2=3,b3=5,b4=9,b5=17,b6=33.答案:C3已知數(shù)列{an}對(duì)任意的p,q∈N*滿足ap+q=ap+aq,且a2=6,則a10等于().A.165 B.33 C.30 D.21解析:(方法一)由已知得a4=a2+a2=12,a8=a4+a4=24,a10=a8+a2=246=30.(方法二)∵a2=a1+a1=6,∴a1=3,a3=a1+a2=9,a5=a2+a3=15,a10=a5+a5=30.答案:C4在數(shù)列{an}中,a1=2,an+1=an+ln1+A.2+lnn B.2+(n1)lnnC.2+nlnn D.1+n+lnn解析:(方法一)由a2=a1+ln2=2+ln2,排除C,D;由a3=a2+ln1+12=2(方法二)∵an+1an=lnn∴an=(anan1)+(an1an2)+…+(a3a2)+(a2a1)+a1=lnnn-1+lnn=ln=2+lnn.答案:A★5數(shù)列a1=1,a2,a3,…,an(n∈N*)的法則如下:若an為自然數(shù),則an+1=an2,否則an+1=an+3,則a6=.

解析:∵a1=1是自然數(shù),∴a2=a12=12=1.∵a2=1不是自然數(shù),∴a3=a2+3=1+3=2.∵a3=2是自然數(shù),∴a4=a32=22=0.∵a4=0是自然數(shù),∴a5=a42=02=2.∵a5=2不是自然數(shù),∴a6=a5+3=2+3=1.答案:16已知數(shù)列{an}滿足:a4n3=1,a4n1=0,a2n=an,n∈N*,則a2016=,a2025=.

答案:017在數(shù)列{an}中,已知a1=2,a2=3,an+2=3an+12an(n≥1),寫出此數(shù)列的前6項(xiàng).解a1=2,a2=3,a3=3a22a1=3×32×2=5,a4=3a32a2=3×52×3=9,a5=3a42a3=3×92×5=17,a6=3a52a4=3×172×9=33.★8已知數(shù)列{an}滿足an+