新教材2023-2024學(xué)年高中數(shù)學(xué)第五章數(shù)列培優(yōu)課2等比數(shù)列習(xí)題課分層作業(yè)課件新人教B版選擇性必修第三冊(cè)

第五章培優(yōu)課?等比數(shù)列習(xí)題課A級(jí)必備知識(shí)基礎(chǔ)練123456789101112131415161718192021221.[探究點(diǎn)一]在等比數(shù)列{an}中,a1+a3=10,a4+a6=,則數(shù)列{an}的通項(xiàng)公式為(

)A.an=24-n

B.an=2n-4 C.an=2n-3 D.an=23-nA1234567891011121314151617182.[探究點(diǎn)一]在等比數(shù)列{an}中,a1+a2=6,a3+a4=12,則數(shù)列{an}的前8項(xiàng)和為(

)A解析

∵數(shù)列{an}是等比數(shù)列,∴a1+a2,a3+a4,a5+a6,a7+a8也成等比數(shù)列.∵a1+a2=6,a3+a4=12,∴a5+a6=24,a7+a8=48,∴前8項(xiàng)和為a1+a2+a3+a4+a5+a6+a7+a8=90.故選A.1234567891011121314151617183.[探究點(diǎn)四]已知數(shù)列{an}的前n項(xiàng)和Sn=2n-1+1,則數(shù)列{an}的前10項(xiàng)中所有奇數(shù)項(xiàng)之和與所有偶數(shù)項(xiàng)之和的比為(

)C解析

當(dāng)n≥2時(shí),an=Sn-Sn-1=2n-2,又a1=S1=2,即前10項(xiàng)分別為2,1,2,4,8,16,32,64,128,256,所以數(shù)列{an}的前10項(xiàng)中1234567891011121314151617184.[探究點(diǎn)一]已知Sn為等比數(shù)列{an}的前n項(xiàng)和,a3與S2分別為方程x2+3x-4=0的兩個(gè)根,則S5=(

)A.-11 B.8

C.15

D.-15A解析

設(shè){an}的公比為q.由x2+3x-4=0解得x=1或-4.∵a3與S2分別為方程x2+3x-4=0的兩個(gè)根,123456789101112131415161718故選A.1234567891011121314151617185.[探究點(diǎn)四](多選題)若Sn為數(shù)列{an}的前n項(xiàng)和,且Sn=2an+1,則下列說法正確的是(

)A.a5=-16B.S5=-63C.數(shù)列{an}是等比數(shù)列D.數(shù)列{Sn+1}是等比數(shù)列AC123456789101112131415161718解析

因?yàn)镾n為數(shù)列{an}的前n項(xiàng)和,且Sn=2an+1,所以S1=2a1+1,因此a1=-1.當(dāng)n≥2時(shí),an=Sn-Sn-1=2an-2an-1,即an=2an-1,所以數(shù)列{an}是以-1為首項(xiàng),2為公比的等比數(shù)列,故C正確;因此,a5=-1×24=-16,故A正確;又Sn=2an+1=-2n+1,所以S5=-25+1=-31,故B錯(cuò)誤;因?yàn)镾1+1=0,所以數(shù)列{Sn+1}不是等比數(shù)列,故D錯(cuò)誤.故選AC.1234567891011121314151617186.[探究點(diǎn)四](多選題)[2023黑龍江一模]已知Sn是等比數(shù)列{an}的前n項(xiàng)和,且Sn=2n+1+a,則下列說法正確的是(

)A.a=-2B.a=-1AD123456789101112131415161718解析

當(dāng)n=1時(shí),a1=S1=4+a.當(dāng)n≥2時(shí),an=Sn-Sn-1=2n+1+a-(2n+a)=2n.因?yàn)閧an}是等比數(shù)列,所以a1=21=4+a,所以a=-2,an=2n(n∈N+),故A正確,B錯(cuò)誤;1234567891011121314151617187.[探究點(diǎn)三·2023江蘇南京高二期中]已知數(shù)列{an}中,a1=3,an+1=3an-2,則an=

.

2·3n-1+1解析

因?yàn)閍n+1=3an-2,所以an+1-1=3(an-1).又a1-1=2,所以{an-1}是一個(gè)以2為首項(xiàng),以3為公比的等比數(shù)列,所以an-1=2·3n-1,an=2·3n-1+1.1234567891011121314151617188.[探究點(diǎn)四]已知數(shù)列{an}的前n項(xiàng)和為Sn,且對(duì)任意n∈N+,有2Sn=3an-2,則a1=

;Sn=

.

23n-1解析

令n=1,則2S1=3a1-2,得a1=2;當(dāng)n≥2時(shí),2Sn-1=3an-1-2,2an=2Sn-2Sn-1=3an-3an-1,即當(dāng)n≥2時(shí),an=3an-1.又a1=2,故數(shù)列{an}是以2為首項(xiàng),3為公比的等比數(shù)列,1234567891011121314151617189.[探究點(diǎn)二·2023甘肅臨夏高二階段練習(xí)]在等差數(shù)列{an}中,已知a3=4,a5+a8=15.(1)求數(shù)列{an}的通項(xiàng)公式;123456789101112131415161718所以Tn=2×22+3×23+4×24+…+(n+1)×2n+1,①則2Tn=2×23+3×24+…+n×2n+1+(n+1)×2n+2,②123456789101112131415161718B級(jí)關(guān)鍵能力提升練A.4n-1

B.4n-1

C.2n-1

D.2n-1D12345678910111213141516171812345678910111213141516171811.數(shù)列{an}中,a1=2,am+n=aman.若ak+1+ak+2+…+ak+10=215-25,則k=(

)A.2 B.3

C.4

D.5C解析

∵am+n=aman,令m=1,∴an+1=a1an=2an,12345678910111213141516171812.(多選題)[2023江西鷹潭貴溪第一中學(xué)高二階段練習(xí)]已知數(shù)列{an},下列結(jié)論正確的有(

)A.若a1=2,an+1=an+n+1,則a20=211B.若a1=1,an+1=3an+2,則a4=53AB123456789101112131415161718解析

選項(xiàng)A,由an+1=an+n+1,得an+1-an=n+1,則a20=(a20-a19)+(a19-a18)+…+(a2-a1)+a1=20+19+…+2+2=211,故A正確;選項(xiàng)B,由an+1=3an+2得an+1+1=3(an+1).又a1+1=2,所以數(shù)列{an+1}是以2為首項(xiàng),3為公比的等比數(shù)列,則an+1=2×3n-1,即an=2×3n-1-1,所以a4=2×33-1=53,故B正確;12345678910111213141516171812345678910111213141516171813.數(shù)列{an}的前n項(xiàng)和為Sn,已知a1=,且對(duì)任意正整數(shù)m,n,都有am+n=aman,若Sn<t

恒成立,則實(shí)數(shù)t的最小值為

.

12345678910111213141516171814.[2023山西太原高三期中]設(shè)等比數(shù)列{an}的前n項(xiàng)和為Sn,且an+1=Sn+2,則an=

.

2n解析

因?yàn)閍n+1=Sn+2,所以當(dāng)n≥2時(shí),an=Sn-1+2,所以an+1-an=(Sn+2)-(Sn-1+2)=an,即an+1=2an,所以等比數(shù)列{an}的公比為2,所以當(dāng)n=1時(shí),a2=S1+2=a1+2,即2a1=a1+2,解得a1=2,所以an=2·2n-1=2n.12345678910111213141516171815.[2023廣西南寧高二期末]已知數(shù)列{an}的前n項(xiàng)和Sn,且滿足Sn=2an-1.(1)求數(shù)列{an}的通項(xiàng)公式;(2)若bn=2n+1,求數(shù)列{an·bn}的前n項(xiàng)和Tn.解(1)當(dāng)n=1時(shí),a1=S1=2a1-1,解得a1=1.當(dāng)n≥2時(shí),an=Sn-Sn-1=(2an-1)-(2an-1-1)=2an-2an-1,則an=2an-1,所以數(shù)列{an}是首項(xiàng)為1,公比為2的等比數(shù)列,所以an=1×2n-1=2n-1.123456789101112131415161718(2)由(1)得an·bn=(2n+1)·2n-1,所以Tn=3×20+5×21+7×22+…+(2n-1)×2n-2+(2n+1)×2n-1,①2Tn=3×21+5×22+7×23+…+(2n-1)×2n-1+(2n+1)×2n,②①-②得-Tn=3×20+2×21+2×22+…+2×2n-2+2×2n-1-(2n+1)×2n=3-4+2n+1-(2n+1)×2n=-1+(1-2n)×2n,所以Tn=1+(2n-1)·2n.12345678910111213141516171816.已知數(shù)列{an},用a1,a2,a3,…,an,…構(gòu)造一個(gè)新數(shù)列a1,(a2-a1),(a3-a2),…,(an-an-1),…,此數(shù)列是首項(xiàng)為1,公比為

的等比數(shù)列.求:(1)數(shù)列{an}的通項(xiàng);(2)數(shù)列{an}的前n項(xiàng)和Sn.12345678910111213141516171812345678910111213141516171817.設(shè)數(shù)列{an}滿足:a1=1,an+1=3an.設(shè)Sn為數(shù)列{bn}的前n項(xiàng)和,已知b1≠0,2bn-b1=S1Sn.(1)求數(shù)列{an},{bn}的通項(xiàng)公式;(2)設(shè)cn=bn·log3an,求數(shù)列{cn}的前n項(xiàng)和Tn;123456789101112131415161718(1)解∵an+1=3an,∴數(shù)列{an}是公比為3,首項(xiàng)為1的等比數(shù)列,∴an=3n-1.∵2bn-b1=S1Sn,∴當(dāng)n=1時(shí),2b1-b1=S1S1.∵S1=b1,b1≠0,∴b1=1,∴Sn=2bn-1.當(dāng)n≥2時(shí),bn=Sn-Sn-1=2bn-2bn-1,∴bn=2bn-1,∴數(shù)列{bn}是公比為2,首項(xiàng)為1的等比數(shù)列,∴bn=2n-1.123456789101112131415161718(2)解

cn=bn·log3an=2n-1lo