2022年高考數(shù)學(xué)二輪復(fù)習(xí) 專題四 數(shù)列 專題突破練13 求數(shù)列的通項(xiàng)及前n項(xiàng)和 文

2022年高考數(shù)學(xué)二輪復(fù)習(xí) 專題四 數(shù)列 專題突破練13 求數(shù)列的通項(xiàng)及前n項(xiàng)和 文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,nN*.(1)求數(shù)列an的通項(xiàng)公式;(2)若bn=2n·an,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a4·a7=15,a3+a8=8.(1)求數(shù)列an的通項(xiàng)公式;(2)令bn=(n2),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(nN*).(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=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=2n·an=2n·(2n+1),Tn=2×3+22×5+23×7+2n×(2n+1),2Tn=22×3+23×5+24×7+2n+1×(2n+1),由-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d0,解得an=2n-1,nN*.(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= (n2),b1=滿足上式,bn的通項(xiàng)公式為bn=.Sn=+.5.(1)證明 an+2=3an+1-2an(nN*),an+2-an+1=2(an+1-an)(nN*),=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(nN*),an=(an-an-1)+(an-1-an-2)+(a2-a1)+a1=2n-1+2n-2+2+1=2n-1,(nN*).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,log2bn=-n,即bn=.(2)由(1)得an·bn=.Tn=+,Tn=+,-,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),當(dāng)n=1時(shí),4a1=+2a1,即a1=2.當(dāng)n2時(shí),4Sn-1=an-1(an-1+2).由-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-<.