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專題突破練13　求數(shù)列的通項(xiàng)及前n項(xiàng)和 1.(2018江西南昌三模,文17)已知數(shù)列{an}的各項(xiàng)均為正數(shù),且-2nan-(2n+1)=0,n∈N*. (1)求數(shù)列{an}的通項(xiàng)公式; (2)若bn=2nan,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=,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山東師大附中一模,文17)已知等差數(shù)列{an}是遞增數(shù)列,且滿足a4a7=15,a3+a8=8. (1)求數(shù)列{an}的通項(xiàng)公式; (2)令bn=(n≥2),b1=,求數(shù)列{bn}的前n項(xiàng)和Sn. 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{anbn}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=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)由-2nan-(2n+1)=0,得[an-(2n+1)](an+1)=0,∵數(shù)列{an}的各項(xiàng)均為正數(shù),∴an=2n+1. (2)由bn=2nan=2n(2n+1), ∴Tn=23+225+237+…+2n(2n+1), ① 2Tn=223+235+247+…+2n+1(2n+1), ② 由①-②得:-Tn=6+2(22+23+…+2n)-2n+1(2n+1) =6+2-2n+1(2n+1)=-2-2n+1(2n-1). 所以Tn=2+(2n-1)2n+1. 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+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) 解得 ∴d=, ∴an=1+(n-1)=n+. (2)bn= (n≥2),b1=滿足上式, ∴{bn}的通項(xiàng)公式為bn=.Sn=+…+. 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-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anbn=.Tn=+…+,① Tn=+…+,② ①-②,得Tn=+2+…+. ∴Tn=1+=3-=3-. 7.解 (1)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(shí),4a1=+2a1,即a1=2. 當(dāng)n≥2時(shí),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-<.