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1、2022年高考數(shù)學(xué)二輪復(fù)習(xí) 專題突破練13 求數(shù)列的通項(xiàng)及前n項(xiàng)和 理 1.(2018河南鄭州一模,理17)已知等差數(shù)列{an}的前n項(xiàng)和為Sn,且a2+a5=25,S5=55. (1)求數(shù)列{an}的通項(xiàng)公式; (2)設(shè)anbn=,求數(shù)列{bn}的前n項(xiàng)和Tn. 2.已知{an}為公差不為零的等差數(shù)列,其中a1,a2,a5成等比數(shù)列,a3+a4=12. (1)求數(shù)列{an}的通項(xiàng)公式; (2)記bn=,設(shè){bn}的前n項(xiàng)和為Sn,求最小的正整數(shù)n,使得Sn>. 3.(2018山西太原三模,17)已知數(shù)列{an}滿足a1

2、=,an+1=. (1)證明數(shù)列是等差數(shù)列,并求{an}的通項(xiàng)公式; (2)若數(shù)列{bn}滿足bn=,求數(shù)列{bn}的前n項(xiàng)和Sn. 4.(2018江西上饒三模,理17)已知等比數(shù)列{an}的前n項(xiàng)和為Sn,且6Sn=3n+1+a(n∈N*). (1)求a的值及數(shù)列{an}的通項(xiàng)公式; (2)若bn=(3n+1)an,求數(shù)列{an}的前n項(xiàng)和Tn. 5.已知數(shù)列{an}滿足a1=1,a2=3,an+2=3an+1-2an(n∈N*). (1)證明:數(shù)列{an+1-an}是等比數(shù)列; (2)求數(shù)列{an}的通項(xiàng)公

3、式和前n項(xiàng)和Sn. 6.已知等差數(shù)列{an}滿足:an+1>an,a1=1,該數(shù)列的前三項(xiàng)分別加上1,1,3后成等比數(shù)列,an+2log2bn=-1. (1)求數(shù)列{an},{bn}的通項(xiàng)公式; (2)求數(shù)列{an·bn}的前n項(xiàng)和Tn. 7.(2018寧夏銀川一中一模,理17)設(shè)Sn為數(shù)列{an}的前n項(xiàng)和,已知an>0,+2an=4Sn+3. (1)求{an}的通項(xiàng)公式: (2)設(shè)bn=,求數(shù)列{bn}的前n項(xiàng)和. 8.設(shè)Sn是數(shù)列{an}的前n項(xiàng)和,an>0,且4Sn

4、=an(an+2). (1)求數(shù)列{an}的通項(xiàng)公式; (2)設(shè)bn=,Tn=b1+b2+…+bn,求證:Tn<. 參考答案 專題突破練13　求數(shù)列的通項(xiàng)及前n項(xiàng)和 1.解 (1) 求得an=3n+2. (2)bn= Tn=b1+b2+…+bn=+…+,∴Tn= 2.解 (1)設(shè)等差數(shù)列{an}的公差為d, ∵a1,a2,a5成等比數(shù)列,a3+a4=12, 即 ∵d≠0,∴解得 ∴an=2n-1,n∈N*. (2)∵bn=,∴Sn=1-+…+=1- 令1-,解得n>1 008, 故所求的n=1 009. 3.(1)證明 ∵an+

5、1=, =2,是等差數(shù)列,+(n-1)×2=2+2n-2=2n,即an= (2)解 ∵bn=,∴Sn=b1+b2+…+bn=1++…+, 則Sn=+…+, 兩式相減得Sn=1++…+=2, ∴Sn=4- 4.解 (1)∵6Sn=3n+1+a(n∈N*), ∴當(dāng)n=1時,6S1=6a1=9+a; 當(dāng)n≥2時,6an=6(Sn-Sn-1)=2×3n,即an=3n-1.∵{an}為等比數(shù)列, ∴a1=1,則9+a=6,a=-3, ∴{an}的通項(xiàng)公式為an=3n-1. (2)由(1)得bn=(3n+1)3n-1, Tn=b1+b2+…+bn=4×30+7×31+…+(3n+1

6、)3n-1,① 3Tn=4×31+7×32+…+(3n-2)3n-1+(3n+1)3n,② 由①-②,得-2Tn=4+32+33+…+3n-(3n+1)3n, -2Tn=4+-(3n+1)3n, -2Tn=, ∴Tn= 5.(1)證明 ∵an+2=3an+1-2an(n∈N*), ∴an+2-an+1=2(an+1-an)(n∈N*),=2. ∵a1=1,a2=3,∴數(shù)列{an+1-an}是以a2-a1=2為首項(xiàng),公比為2的等比數(shù)列. (2)解 由(1)得,an+1-an=2n(n∈N*), ∴an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1=2n

7、-1+2n-2+…+2+1=2n-1,(n∈N*). Sn=(2-1)+(22-1)+(23-1)+…+(2n-1)=(2+22+23+…+2n)-n=-n=2n+1-2-n. 6.解 (1)設(shè)等差數(shù)列{an}的公差為d,且d>0,由a1=1,a2=1+d,a3=1+2d,分別加上1,1,3后成等比數(shù)列,得(2+d)2=2(4+2d),解得d=2,∴an=1+(n-1)×2=2n-1.∵an+2log2bn=-1, ∴l(xiāng)og2bn=-n,即bn= (2)由(1)得an·bn=Tn=+…+,① Tn=+…+,② ①-②,得Tn=+2+…+ ∴Tn=1+=3-=3- 7.解 (1)

8、由+2an=4Sn+3,可知+2an+1=4Sn+1+3. 兩式相減,得+2(an+1-an)=4an+1,即2(an+1+an)==(an+1+an)(an+1-an). ∵an>0,∴an+1-an=2. +2a1=4a1+3, ∴a1=-1(舍)或a1=3. 則{an}是首項(xiàng)為3,公差d=2的等差數(shù)列,∴{an}的通項(xiàng)公式an=3+2(n-1)=2n+1. (2)∵an=2n+1,∴bn=, ∴數(shù)列{bn}的前n項(xiàng)和Tn=+…+ 8.(1)解 4Sn=an(an+2),① 當(dāng)n=1時,4a1=+2a1,即a1=2. 當(dāng)n≥2時,4Sn-1=an-1(an-1+2).② 由①-②得4an=+2an-2an-1,即2(an+an-1)=(an+an-1)·(an-an-1).∵an>0,∴an-an-1=2, ∴an=2+2(n-1)=2n. (2)證明 ∵bn=, ∴Tn=b1+b2+…+bn=1-+…+1-<