課時(shí)跟蹤檢測(cè)(十一)構(gòu)造法求數(shù)列通項(xiàng)公式1.數(shù)列{an}滿(mǎn)足an=4an1+3(n≥2)且a1=0,則此數(shù)列第5項(xiàng)是()A.15 B.255C.16 D.632.在數(shù)列{an}中,a1=2,an+1=anan+1(n∈N*),則a20=(A.121 B.C.223 D.3.若正項(xiàng)數(shù)列{an}滿(mǎn)足a1=2,an+12=4an2+4an+1,則數(shù)列{a4.數(shù)列{an}中,a1=1,且an+1=an+3n+1,則通項(xiàng)公式an=.

5.已知數(shù)列{an}滿(mǎn)足:a1=1,且an=2an1+2n1(n≥2,n∈N),則an=.

6.已知數(shù)列{an}滿(mǎn)足a1=1,3an+1an=anan+1,則通項(xiàng)公式an=.

7.已知數(shù)列{an}的首項(xiàng)是a1=1,且an+1=nann+1,則數(shù)列{an8.記Sn為數(shù)列{an}的前n項(xiàng)和,若an=Sn21,則S7=9.在數(shù)列{an}中,a1=1,a2=3,且對(duì)任意的n∈N*,都有an+2=3an+12an,則數(shù)列{an}的通項(xiàng)公式為10.數(shù)列{an}的前n項(xiàng)和為Sn,已知a1=2,3Sn=an+12n+2+2(n∈N*(1)n∈N*時(shí),寫(xiě)出an+1與an之間的遞推關(guān)系;(2)求{an}的通項(xiàng)公式.11.已知數(shù)列{an}的前n項(xiàng)和為Sn,a1=3且Sn+Sn+1=2n2+6n+3,n∈N*.(1)求S9的值;(2)求數(shù)列{an}的通項(xiàng)公式.12.在數(shù)列{an}中,a1=4且an+1=3an+2an+4,求數(shù)列課時(shí)跟蹤檢測(cè)(十一)1.選B∵an=4an1+3(n≥2),∴an+1=4(an1+1)(n≥2),∴{an+1}是以1為首項(xiàng),4為公比的等比數(shù)列,則an+1=4n1.∴an=4n11,∴a5=441=255.故選B.2.選B因?yàn)閍n+1=anan+1,則1an+1=an+1an=1an+1,又1a1=12,故1an是首項(xiàng)為12,公差為1的等差數(shù)列.1an3.解析:在正項(xiàng)數(shù)列{an}中,an+12=4an2+4an+1=(2an+1)2,則有an+1=2an+1,于是得an+1+1=2(an+1),而a1+1=3,因此數(shù)列{an+1}是首項(xiàng)為3,公比為2的等比數(shù)列,則有an+1=3×2n1,即an=3×2n11,所以數(shù)列{an}的通項(xiàng)公式是an答案:an=3×2n114.解析:∵an+1=an+3n+1,∴an+132(n+1)2+n+12=an32n2+n2,∴數(shù)列an?32n2+n2為常數(shù)列,又a1=1,則a1答案:32n25.解析:由題設(shè),an2n=an?12n?1+12(n≥2),即an2nan?12n?1=12(n≥2),而a12=12,∴an2n是首項(xiàng)、公差均為1答案:n·2n16.解析:∵3an+1an=anan+1,∴1an+11an=3,且1a1=1,∴1an是以1為首項(xiàng),3為公差的等差數(shù)列,∴1an=1+(n1)答案:17.解析:由an+1=nann+1,即(n+1)an+1=nan,故數(shù)列{nan}為常數(shù)列,a1=1,即nan=1,a答案:an=18.解析:由已知an=Sn21,得SnSn1=Sn21,所以Sn2=2(Sn12)(n≥2),又a1=S121,即S1=2,S12=4,所以{Sn2}是以4為首項(xiàng),2為公比的等比數(shù)列,所以Sn2=4×2n1,即Sn=22n+1,所以答案:2549.解析:由an+2=3an+12an,得an+2an+1=2(an+1又a1=1,a2=3,所以a2a1=2≠0,所以{an+1an}是首項(xiàng)為2,公比為2的等比數(shù)列,所以an+1an=2n,所以an=a1+(a2a1)+…+(anan1)=1+2+22+…+2n1=2n1,n≥2,因?yàn)閍1=1符合上式,所以an=2n1.答案:an=2n110.解:(1)因?yàn)?Sn=an+12n+2+2①,所以當(dāng)n≥2時(shí),3Sn1=an2n+1+2②,①②得3an=an+1an2n+1(n≥2),即an+1=4an+2n+1(n≥2),在①中,令n=1得a2=3a1+232=12=4a1+22,也符合上式,所以an+1=4an+2n+1.(2)因?yàn)閍n+1=4an+2n+1,則an+1+2n+1=4(an+2n),且a1+2=4≠0,所以數(shù)列{an+2n}是以4為首項(xiàng),4為公比的等比數(shù)列,所以an+2n=4n,故an=4n2n.11.解:(1)因?yàn)镾n+Sn+1=2n2+6n+3,所以Sn+2+Sn+1=2(n+1)2+6(n+1)+3.兩式相減,得an+2+an+1=4(n+2),n∈N*.所以S9=a1+(a2+a3)+(a4+a5)+…+(a8+a9)=3+4[(1+2)+(3+2)+…+(7+2)]=3+4×(3+9)×42=99(2)由(1)知an+2+an+1=4(n+2)①,可得an+an+1=4(n+1)②,n≥2.因?yàn)閍1=3,S2+S1=11,所以a2=5,又S3+S2=23=2a1+2a2+a3,所以a3=7.又由①②得an+2an=4,n≥2.所以a2n=a2+4(n1)=4n+1,即an=2n+1,n為偶數(shù),則當(dāng)n≥3,且為奇數(shù)時(shí),an=4(n+1)an+1=4(n+1)[2(n+1)+1]=2n+1,又a1=3,a3=7符合上式,綜上得an=2n+1.12.解:由an+1=3an+2