2022版高考数学二轮复习专题四数列专题突破练13求数列的通项及前n项和文

专题突破练13　求数列的通项及前n项和1.(2022江西南昌三模,文17)已知数列{an}的各项均为正数,且-2nan-(2n+1)=0,n∈N*.(1)求数列{an}的通项公式;(2)若bn=2n·an,求数列{bn}的前n项和Tn.2.已知{an}为公差不为零的等差数列,其中a1,a2,a5成等比数列,a3+a4=12.(1)求数列{an}的通项公式;(2)记bn=,设{bn}的前n项和为Sn,求最小的正整数n,使得Sn>.7\n3.(2022山西太原三模,17)已知数列{an}满足a1=,an+1=.(1)证明数列是等差数列,并求{an}的通项公式;(2)若数列{bn}满足bn=,求数列{bn}的前n项和Sn.4.(2022山东师大附中一模,文17)已知等差数列{an}是递增数列,且满足a4·a7=15,a3+a8=8.(1)求数列{an}的通项公式;(2)令bn=(n≥2),b1=,求数列{bn}的前n项和Sn.5.已知数列{an}满足a1=1,a2=3,an+2=3an+1-2an(n∈N*).(1)证明:数列{an+1-an}是等比数列;(2)求数列{an}的通项公式和前n项和Sn.7\n6.已知等差数列{an}满足:an+1>an,a1=1,该数列的前三项分别加上1,1,3后成等比数列,an+2log2bn=-1.(1)求数列{an},{bn}的通项公式;(2)求数列{an·bn}的前n项和Tn.7.(2022宁夏银川一中一模,理17)设Sn为数列{an}的前n项和,已知an>0,+2an=4Sn+3.(1)求{an}的通项公式:(2)设bn=,求数列{bn}的前n项和.7\n8.设Sn是数列{an}的前n项和,an>0,且4Sn=an(an+2).(1)求数列{an}的通项公式;(2)设bn=,Tn=b1+b2+…+bn,求证:Tn<.参考答案专题突破练13　求数列的通项及前n项和1.解(1)由-2nan-(2n+1)=0,得[an-(2n+1)](an+1)=0,∵数列{an}的各项均为正数,∴an=2n+1.(2)由bn=2n·an=2n·(2n+1),∴Tn=2×3+22×5+23×7+…+2n×(2n+1),①2Tn=22×3+23×5+24×7+…+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)设等差数列{an}的公差为d,∵a1,a2,a5成等比数列,a3+a4=12,∴7\n即∵d≠0,∴解得∴an=2n-1,n∈N*.(2)∵bn=,∴Sn=1-+…+=1-.令1-,解得n>1008,故所求的n=1009.3.(1)证明∵an+1=,∴=2,∴是等差数列,∴+(n-1)×2=2+2n-2=2n,即an=.(2)解∵bn=,∴Sn=b1+b2+…+bn=1++…+,则Sn=+…+,两式相减得Sn=1++…+=2,∴Sn=4-.4.解(1)7\n解得∴d=,∴an=1+(n-1)=n+.(2)bn=(n≥2),b1=满足上式,∴{bn}的通项公式为bn=.Sn=+…+.5.(1)证明∵an+2=3an+1-2an(n∈N*),∴an+2-an+1=2(an+1-an)(n∈N*),∴=2.∵a1=1,a2=3,∴数列{an+1-an}是以a2-a1=2为首项,公比为2的等比数列.(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)设等差数列{an}的公差为d,且d>0,由a1=1,a2=1+d,a3=1+2d,分别加上1,1,3后成等比数列,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=+…+,①7\nTn=+…+,②①-②,得Tn=+2+…+.∴Tn=1+=3-=3-.7.解(1)由+2an=4Sn+3,可知+2an+1=4Sn+1+3.两式相减,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.则{an}是首项为3,公差d=2的等差数列,∴{an}的通项公式an=3+2(n-1)=2n+1.(2)∵an=2n+1,∴bn=,∴数列{bn}的前n项和Tn=+…+.8.(1)解4Sn=an(an+2),①当n=1时,4a1=+2a1,即a1=2.当n≥2时,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)证明∵bn=,∴Tn=b1+b2+…+bn=1-+…+1-<.7