2025年高考數(shù)學(xué)一輪復(fù)習(xí)-第七章-第四節(jié)-求通項(xiàng)公式-專項(xiàng)訓(xùn)練【含解析】

PAGE第七章-第四節(jié)-求通項(xiàng)公式-專項(xiàng)訓(xùn)練【原卷版】(時間:45分鐘分值:85分)【基礎(chǔ)落實(shí)練】1.(5分)數(shù)列{an}中,an+1=2an+1,a1=1,則a100=()A.2100+1 B.2101 C.2100-1 D.21002.(5分)已知數(shù)列{an}滿足a1=3,an+1=5an-8,則a2026的值為()A.52025-2 B.52025+2C.52026+2 D.52026-23.(5分)在數(shù)列{an}中,a1=1,an+1=2anan+2(n∈N*A.第6項(xiàng) B.第7項(xiàng)C.第8項(xiàng) D.第9項(xiàng)4.(5分)在數(shù)列an中,若a1=2,an+1=3an+2n+1,則an=(A.n·2nB.52-C.2·3n-2n+1D.4·3n-1-2n+15.(5分)已知數(shù)列{an}滿足a1=1,an+1=an+log3(1-22n+1),則a41A.-1 B.-2C.-3 D.1-log3406.(5分)(多選題)已知數(shù)列{an}的前n項(xiàng)和為Sn,a1=1,Sn+1=Sn+2an+1,數(shù)列{2nan·an+1}的前n項(xiàng)和為Tn,A.數(shù)列{an+1}是等差數(shù)列B.數(shù)列{an+1}是等比數(shù)列C.數(shù)列{an}的通項(xiàng)公式為an=2n-1D.Tn<17.(5分)已知數(shù)列{an}中,a1=1,an+1=3an+4,則an=.

8.(5分)若數(shù)列an中,a1=0,an+1=an+(2n-1)(n∈N*),則通項(xiàng)公式an=9.(5分)已知數(shù)列{an}滿足a1=1,2n-1an=an-1,則通項(xiàng)公式an=.

10.(10分)已知在數(shù)列{an}中,a1=1,an+1=an4an+3(n(1)求證:{1an+2(2)求數(shù)列{an}的通項(xiàng)公式.11.(10分)(2023·合肥模擬)已知等差數(shù)列an的各項(xiàng)均為正數(shù),a1=1,a2+a5+a8=a3a5(1)求an的前n項(xiàng)和Sn(2)若數(shù)列bn滿足b1=1,an+2bn+1=anbn,求bn.【能力提升練】12.(5分)已知函數(shù)f(x)=x3x+1,數(shù)列{an}滿足a1=1,an+1=f(an)(n∈N*),則數(shù)列{an13.(5分)設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,若S2=4,an+1=2Sn+1,n∈N*,則a1=,S5=.

14.(10分)設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,已知a1=1,Sn+1=4an+2.(1)設(shè)bn=an+1-2an,證明:數(shù)列{bn}是等比數(shù)列;(2)求數(shù)列{an}的通項(xiàng)公式.第七章-第四節(jié)-求通項(xiàng)公式-專項(xiàng)訓(xùn)練【解析版】(時間:45分鐘分值:85分)【基礎(chǔ)落實(shí)練】1.(5分)數(shù)列{an}中,an+1=2an+1,a1=1,則a100=()A.2100+1 B.2101 C.2100-1 D.2100【解析】選C.數(shù)列{an}中,an+1=2an+1,故an+1+1=2(an+1),a1+1=2≠0,所以{an+1}是首項(xiàng)為2,公比為2的等比數(shù)列,所以an+1=2n,即an=2n-1,故a100=2100-1.2.(5分)已知數(shù)列{an}滿足a1=3,an+1=5an-8,則a2026的值為()A.52025-2 B.52025+2C.52026+2 D.52026-2【解析】選B.因?yàn)閍n+1=5an-8,所以an+1-2=5(an-2),又a1-2=1,所以{an-2}是公比為5,首項(xiàng)為1的等比數(shù)列,所以an-2=5n-1,an=5n-1+2,所以a2026=52025+2.3.(5分)在數(shù)列{an}中,a1=1,an+1=2anan+2(n∈N*A.第6項(xiàng) B.第7項(xiàng)C.第8項(xiàng) D.第9項(xiàng)【解析】選B.由an+1=2anan+2,可得1即數(shù)列1an是以1為首項(xiàng),12為公差的等差數(shù)列,故1an=1+(n-1)×12即an=2n+1,由2n+1=14.(5分)在數(shù)列an中,若a1=2,an+1=3an+2n+1,則an=(A.n·2nB.52-C.2·3n-2n+1D.4·3n-1-2n+1【解析】選C.令bn=an則bn+1bn=an又b1=a12+2=3,所以bn所以bn=an2n+2=3×(32)n-1,得an=2·3n5.(5分)已知數(shù)列{an}滿足a1=1,an+1=an+log3(1-22n+1),則a41A.-1 B.-2C.-3 D.1-log340【解析】選C.因?yàn)閍n+1=an+log3(1-22n+1)=an+log32n-12n+1=an所以an+1-an=log3(2n-1)-log3(2n+1),則a41-a40=log379-log381,a40-a39=log377-log379,…,a3-a2=log33-log35,a2-a1=log31-log33,將以上40個式子相加得a41-a1=log31-log381.又a1=1,所以a41=log31-log381+1=-3.6.(5分)(多選題)已知數(shù)列{an}的前n項(xiàng)和為Sn,a1=1,Sn+1=Sn+2an+1,數(shù)列{2nan·an+1}的前n項(xiàng)和為Tn,A.數(shù)列{an+1}是等差數(shù)列B.數(shù)列{an+1}是等比數(shù)列C.數(shù)列{an}的通項(xiàng)公式為an=2n-1D.Tn<1【解析】選BCD.因?yàn)镾n+1=Sn+2an+1,所以Sn+1-Sn=2an+1,即an+1=2an+1,an+1+1=2(an+1).因?yàn)閍1=1,a1+1=2,所以數(shù)列{an+1}是公比為2的等比數(shù)列,所以選項(xiàng)B正確,A不正確.又an+1=2·2n-1=2n,所以an=2n-1,故選項(xiàng)C正確.2nanan+1=所以Tn=(12-1-122-1)+(122-7.(5分)已知數(shù)列{an}中,a1=1,an+1=3an+4,則an=.

【解析】設(shè)an+1+t=3(an+t),即an+1=3an+2t,又an+1=3an+4,根據(jù)對應(yīng)項(xiàng)系數(shù)相等,解得t=2,故an+1+2=3(an+2).令bn=an+2,則b1=a1+2=3,且bn+1b所以{bn}是以3為首項(xiàng),3為公比的等比數(shù)列,所以bn=3×3n-1=3n,即an=3n-2.答案:3n-28.(5分)若數(shù)列an中,a1=0,an+1=an+(2n-1)(n∈N*),則通項(xiàng)公式an=【解析】an=a1+(a2-a1)+…+(an-an-1)=0+1+3+…+(2n-3)=(n-1)2,所以該數(shù)列的通項(xiàng)公式為an=(n-1)2.答案:(n-1)29.(5分)已知數(shù)列{an}滿足a1=1,2n-1an=an-1,則通項(xiàng)公式an=.

【解析】方法一:因?yàn)閍n=anan-1·an-1a=(12)n-1·(12)n-2·…·(12)2·(12)=(12)1+2+…+(n-1)=1所以an=(12)

方法二:由2n-1an=an-1得an=(12)n-1an-1所以an=(12)n-1an-1=(12)n-1·(12)n-2an-2=…=(12)n-1·(12)n-2·…·(12)1a1=(12)(n-1)+(答案:(12)10.(10分)已知在數(shù)列{an}中,a1=1,an+1=an4an+3(n(1)求證:{1an+2(2)求數(shù)列{an}的通項(xiàng)公式.【解析】(1)因?yàn)閍1=1,an+1=an4an+3(n∈N*),所以1an+1因?yàn)?a1+2=3,所以{1an(2)由(1)知1an+2=3×3n-1=3n,所以an=11.(10分)(2023·合肥模擬)已知等差數(shù)列an的各項(xiàng)均為正數(shù),a1=1,a2+a5+a8=a3a5(1)求an的前n項(xiàng)和Sn(2)若數(shù)列bn滿足b1=1,an+2bn+1=anbn,求bn【解析】(1)等差數(shù)列an因?yàn)閍2+a5+a8=a3a5,所以3a5=a3a5,又因?yàn)榈炔顢?shù)列an的各項(xiàng)均為正數(shù).所以a3又因?yàn)閍1=1,所以d=a3所以an=n,所以Sn=n(【解析】(2)由(1)得an=n,因?yàn)閎1=1,且an+2bn+1=anbn,所以bn≠0,所以bn+1bn=所以bnb1=bnbn-1×…×b3b2所以bn=2n(n+1)(n≥2)所以bn的通項(xiàng)公式為bn=2【能力提升練】12.(5分)已知函數(shù)f(x)=x3x+1,數(shù)列{an}滿足a1=1,an+1=f(an)(n∈N*),則數(shù)列{an【解析】由已知得,an+1=an所以1an+1=1an所以數(shù)列{1an}是首項(xiàng)為1a1=1,公差d為3的等差數(shù)列,所以1an故an=13n-2(n∈答案:an=13n-2(n13.(5分)設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,若S2=4,an+1=2Sn+1,n∈N*,則a1=,S5=.

【解析】方法一:由a1+a2=4由an+1=Sn+1-Sn=2Sn+1,得Sn+1=3Sn+1,所以Sn+1+12=3(Sn+1所以Sn+12是以32為首項(xiàng),3為公比的等比數(shù)列,所以Sn+12即Sn=3n-12,所以方法二:由a1+a又an+1=2Sn+1,an+2=2Sn+1+1,兩式相減得an+2-an+1=2an+1,即an+2a所以{an}是首項(xiàng)為1,公比為3的等比數(shù)列,所以Sn=3n-12,所以答案:112114.(10分)設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,已知a1=1,Sn+1=4an+2.(1)設(shè)bn=an+1-2an,證明:數(shù)列{bn}是等比數(shù)列;(2)求數(shù)列{an}的通項(xiàng)公式.【解析】(1)由a1=1及Sn+1=4an+2,得a1+a2=S2