適用于新高考新教材備戰(zhàn)2025屆高考數(shù)學(xué)一輪總復(fù)習(xí)課時(shí)規(guī)范練40數(shù)列中的構(gòu)造問(wèn)題新人教A版

課時(shí)規(guī)范練40數(shù)列中的構(gòu)造問(wèn)題基礎(chǔ)鞏固練1.(2024·寧夏六盤山模擬)已知數(shù)列{an}中,a1=4,an+1=4an-6,則an等于()A.22n+1+2 B.22n+1-2C.22n-1+2 D.22n-1-22.(2024·江蘇鹽城高三期中)已知數(shù)列{an}滿足a1=2,an+1=an4,則a6的值為(A.220 B.224C.21024 D.240963.(2024·山東菏澤模擬)已知數(shù)列{an}中,a1=1且an+1=3anan+3(n∈N*),則aA.16 B.C.13 D.4.(多選題)(2024·廣東順德一中校考)已知數(shù)列{an}滿足a1=1,nan+1=2(n+1)an,設(shè)bn=ann,則下列結(jié)論正確的是(A.b3=4 B.{bn}是等差數(shù)列C.b4=16 D.an=n·2n-15.(2024·黑龍江哈爾濱模擬)設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,an+1+an=2n+3,且Sn=1450,若a2<4,則n的最大值為()A.50 B.51 C.52 D.536.(2024·廣西梧州模擬)在數(shù)列{an}中,已知a1=2,an+1=an3an+1,則{an7.(2024·江西景德鎮(zhèn)模擬)已知在數(shù)列{an}中,a1=1,當(dāng)n≥2時(shí),(n-1)an=2nan-1,則數(shù)列{an}的通項(xiàng)公式為.

8.已知數(shù)列{an}滿足a1=1,an+1=2an+3n,則數(shù)列{an}的通項(xiàng)公式為.

9.設(shè)數(shù)列{an}滿足a1=4,an=3an-1+2n-1(n≥2),則數(shù)列{an}的通項(xiàng)公式為.

10.已知數(shù)列{an}中,a1=1,a2=2,an+2=23an+1+13an,求{an}綜合提升練11.(2024·江西臨川模擬)已知數(shù)列{an}滿足a1=3,an+1=an+2an+1+1,則a10=(A.80 B.100 C.120 D.14312.已知數(shù)列{an}滿足an+1=2an+4·3n-1,a1=-1,則數(shù)列{an}的通項(xiàng)公式為.

13.已知a1=3,an+1=3an-4an-2創(chuàng)新應(yīng)用練14.用磚砌墻,第一層用去了全部磚塊的一半多一塊,第二層用去了剩下的一半多一塊,……以此類推,每一層都用去了上次剩下磚塊的一半多一塊,到第10層恰好把磚塊用完,則此次砌墻一共用了塊磚.

課時(shí)規(guī)范練40數(shù)列中的構(gòu)造問(wèn)題1.C解析因?yàn)閍n+1=4an-6,所以an+1-2=4(an-2),所以an+1-2an-2=4,又a1-2=2,所以數(shù)列{an-2}是一個(gè)以2為首項(xiàng),以4為公比的等比數(shù)列,所以an-2=2×4n-1,所以an2.C解析an+1=an4,a1=2,易知an>0,故lnan+1=4lnan,故{lnan}是首項(xiàng)為ln2,公比為4的等比數(shù)列,lnan=4n-1·ln2,lna6=45·ln2=ln21024,故a6=23.A解析由an+1=3anan+3得1an+1=an+33an=1an+13,又1a1=1,∴數(shù)列{1an}是以4.AD解析由條件可得an+1n+1=2ann,即bn+1=2bn,又b1=1,所以{bn}是首項(xiàng)為1,公比為2的等比數(shù)列,故B錯(cuò)誤;可得bn=ann=2n-1,所以an=n·2n-1,故D正確;則b3=4,b45.B解析∵an+1+an=2n+3,∴an+1-(n+2)=-(an-(n+1)),∴{an-(n+1)}是以-1為公比的等比數(shù)列,∴an-(n+1)=(a1-2)·(-1)n-1,an=(n+1)+(a1-2)·(-1)n-1,∴Sn=[2+3+…+(n+1)]+(a1-2)[1+(-1)+(-1)2+…+(-1)n-1]=n(n+3)2+(a1-2)1-(-1)n2.當(dāng)n為偶數(shù)時(shí),Sn=n(n+3)2=1450無(wú)解,當(dāng)n為奇數(shù)時(shí),Sn=n(n+3)2+a1-2=1450,∴a1=1452-n(n+3)2,又a1+a2=5,∴a2=5-a1<4,即a1>1,即n(n+3)<2902,y=n(n+3)6.an=26n-5解析由an+1=an3an+1,兩邊取倒數(shù)得1an+1=3an+1an=3+1an,即1an+1-1an=3,7.an=n·2n-1解析當(dāng)n≥2時(shí),(n-1)an=2nan-1,則ann=2·an-1n-1,而a11=1,因此數(shù)列{ann}是以1為首項(xiàng),2為公比的等比數(shù)列,則ann=1×2n-1=2n-8.an=3n-2n解析an+1=2an+3n兩邊同除以3n+1,得an+13n+1=23·an3n+13,令bn=an3n,則bn+1=23bn+13,設(shè)bn+1+λ=23(bn+λ),解得λ=-1,則bn+1-1=23(bn-1).而b1-1=-23,所以數(shù)列{bn-1}是以-239.an=2·3n-n-1解析設(shè)an+pn+q=3[an-1+p(n-1)+q],化簡(jiǎn)后得an=3an-1+2pn+(2q-3p),所以2p=2,2q-3p=-1,解得p=1,q=1,即an+n+1=3(an-1+n-1+1).令bn=an+n+1,則bn=3bn-1.又b1=6,故bn=6·3n-1=2·3n,又10.解設(shè)an+2-san+1=t(an+1-san),即an+2=(s+t)an+1-stan,所以s解得s=1,t不妨令an+2-an+1=-13(an+1-an),又a2-a1=1,所以{an+1-an}是以1為首項(xiàng),-13為公比的等比數(shù)列,則an+1-an=(-13)n-1,累加得an-a1=(-13)0+(-13)1+…+(-13)n-2=1-(-an=74-34·(-13)又a1=1符合上式,故an=74-34·(-11.C解析易得an>0,因?yàn)閍n+1=an+2an+1+1,所以an+1+1=(an+1)2+2an+1+1,即an+1+1=(an+1+1)2,等式兩邊開(kāi)方可得an+1+1=an+1+1,即an+1+1-an+1=1,所以數(shù)列{an+1}是以a1+1=2為首項(xiàng),1為公差的等差數(shù)列,所以an+1=212.an=4×3n-1-5×2n-1解析(方法一)設(shè)an+1+λ·3n=2(an+λ·3n-1),整理得an+1=2an-λ·3n-1,可得λ=-4,即an+1-4×3n=2(an-4×3n-1),又a1-4×31-1=-5≠0,所以數(shù)列{an-4·3n-1}是首項(xiàng)為-5,公比為2的等比數(shù)列,所以an-4×3n-1=-5×2n-1,即an=4×3n-1-5×2n-1.(方法二)(兩邊同除以qn+1)兩邊同時(shí)除以3n+1,得an+13n+1=23×an3n+49,整理得an+13n+1-43=23×(an3n-43),又a13-43=-5(方法三)(兩邊同除以pn+1)兩邊同時(shí)除以2n+1,得an+12n+1=an2n+(32)n-1,即an+12n+1-an2n=(32)n-1,當(dāng)n≥2時(shí),an2n=(an2n-an-12n-1)+(an-12n-1-an-22n-2)+…+(a222-a12)+a12=(32)n-2+(32)13.解an+1-1=3an-4anan+1-4=3an-4an由①÷②得an+1-1a又因?yàn)閍1-1a1-4=-2,所以{an-1an-4}是首項(xiàng)為-2,公比為-2的等比數(shù)列14.2046解析設(shè)此次砌墻一共用了S塊磚,砌好第n層