1、第2课时数列的递推公式课时过关能力提升基础巩固1在数列an中,a1=-1,an+1=an-3,则a3等于().A.-7B.-4C.-1D.2解析:a2=a1-3=-1-3=-4,a3=a2-3=-4-3=-7.答案:A2已知数列an满足an+1=an+12,则数列an是().A.递增数列B.递减数列C.摆动数列D.常数列解析:由an+1=an+12,知an+1-an=120,所以an+1an,即从第2项起,每一项都大于它的前一项.答案:A3在数列an中,a1=13,an=2an-1(n2),则a5等于 ().A.43B.83C.163D.323解析:a2=2a1=23,a3=2a2=43,a4
2、=2a4=83,a5=2a4=163.答案:C4在数列an中,an-1=man+1(n1),且a2=3,a3=5,则实数m等于().A.0B.25C.2D.5解析:由题意得a2=ma3+1,则3=5m+1,故m=25.答案:B5在正项数列an中,a1=1,a2=2,2an2=an+12+an-12(n2),则a6等于().A.16B.8C.22D.4解析:2an2=an+12+an-12,an+12=2an2-an-12.a32=2a22-a12=24-1=7,a42=2a32-a22=14-4=10,a52=210-7=13,a62=213-10=16.又a60,a6=4.答案:D6在数列a
3、n中,a1=1,对所有的n2,都有a1a2a3an=n2,则a3+a5等于()A.259B.2516C.6116D.3115解析:a1a2a3=32,a1a2=22,a1a2a3a4a5=52,a1a2a3a4=42,则a3=3222=94,a5=5242=2516,故a3+a5=6116.答案:C7已知数列an满足a1=2,a2=5,a3=23,且an+1=an+,则,的值分别为.解析:an+1=an+,a2=a1+,a3=a2+,即2+=5,5+=23,解得=6,=-7.答案:6,-78已知在数列an中,an+1=2anan+2对任意正自然数n都成立,且a7=12,则a5=_.解析:由已知
4、a7=2a6a6+2=12,解得a6=23.又因为a6=2a5a5+2=23,解得a5=1.答案:19在数列an中,a1=1,an+1=2+1an,试写出a2,a3,a4,a5.解a2=2+1a1=2+11=3,a3=2+1a2=2+13=73,a4=2+1a3=2+37=177,a5=2+1a4=2+717=4117.10在数列an中,a1=1,an+1=nn+1an.(1)写出数列的前5项;(2)猜想数列的一个通项公式.解(1)a1=1,a2=11+11=12,a3=21+212=13,a4=31+313=14,a5=41+414=15.(2)猜想:an=1n.能力提升1数列12,14,1
5、8,116,的递推公式可以是().A.an=12n+1(nN*)B.an=12n(nN*)C.an+1=12an(nN*)D.an+1=2an(nN*)解析:数列从第二项起,后一项是前一项的12,故递推公式为an+1=12an(nN*).答案:C2由1,3,5,2n-1,构成数列an,数列bn满足b1=2,当n2时,bn=abn-1,则b6的值是().A.9B.17C.33D.65解析:an=2n-1,bn=abn-1=2bn-1-1.又b1=2,b2=3,b3=5,b4=9,b5=17,b6=33.答案:C3已知数列an对任意的p,qN*满足ap+q=ap+aq,且a2=-6,则a10等于(
6、).A.-165B.-33C.-30D.-21解析:(方法一)由已知得a4=a2+a2=-12,a8=a4+a4=-24,a10=a8+a2=-24-6=-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在数列an中,a1=2,an+1=an+ln1+1n,则an等于().A.2+ln nB.2+(n-1)ln nC.2+nln nD.1+n+ln n解析:(方法一)由a2=a1+ln 2=2+ln 2,排除C,D;由a3=a2+ln1+12=2+ln 3,排除B.(方法二)an+1-an=ln n+1
7、n,an=(an-an-1)+(an-1-an-2)+(a3-a2)+(a2-a1)+a1=ln nn-1+ln n-1n-2+ln 32+ln 2+2=lnnn-1n-1n-2322+2=2+ln n.答案:A5数列a1=1,a2,a3,an(nN*)的法则如下:若an为自然数,则an+1=an-2,否则an+1=an+3,则a6=.解析:a1=1是自然数,a2=a1-2=1-2=-1.a2=-1不是自然数,a3=a2+3=-1+3=2.a3=2是自然数,a4=a3-2=2-2=0.a4=0是自然数,a5=a4-2=0-2=-2.a5=-2不是自然数,a6=a5+3=-2+3=1.答案:16
8、设an是首项为1的正项数列,且(n+1)an+12-nan2+an+1an=0(n=1,2,3,),则它的通项公式是.解析:(n+1)an+12-nan2+anan+1=0,(n+1)an+1-nan(an+1+an)=0.an0,an+an+10.(n+1)an+1-nan=0.方法一:an+1an=nn+1,a2a1a3a2a4a3a5a4anan-1=12233445n-1n,ana1=1n.又a1=1,an=1na1=1n.方法二:(n+1)an+1-nan=0,nan=(n-1)an-1=1a1=1,nan=1,an=1n.答案:an=1n7在数列an中,已知a1=2,a2=3,an+2=3an+1-2an(n1),写出此数列的前6项.解a1=2,a2=3,a3=3a2-2a1=33-22=5,a4=3a3-2a2=35-23=9,a5=3a4-2a3=39-25=17,a6=3a5-2a4=317-29=33.8已知数列an满足an+1=2an,0an12,2an-1,12an1.若a1=67,试求a2015+a2016.解a1=6712,1,a2=2a1-1=5712,1,a3=2a2-1=370,12,a4=2a3=67.数列an是周期数列,且周期为3.a2 015+a2 016=a6713+2+a6713+3=a2+a3=57+37=87.
