新高考新教材適用2025版高考數(shù)學二輪復習專題二數(shù)列考點突破練5數(shù)列求和及其綜合應用

考點突破練5數(shù)列求和及其綜合應用1.(2024·全國甲·理17)設Sn為數(shù)列{an}的前n項和.已知2Snn+n=2an+(1)證明:{an}是等差數(shù)列;(2)若a4,a7,a9成等比數(shù)列,求Sn的最小值.2.(2024·河北石家莊一模)已知等差數(shù)列{an}的各項均為正數(shù),公差d<3,若分別從下表第一、二、三行中各取一個數(shù),依次作為a3,a4,a5,且a3,a4,a5中任何兩個數(shù)都不在同一列.項目第一列其次列第三列第一行356其次行748第三行11129(1)求數(shù)列{an}的通項公式;(2)設bn=8(an+1)·(an+1+3),數(shù)列{b3.(2024·河北唐山一模)已知數(shù)列{an}的各項均不為零,Sn為其前n項和,且anan+1=2Sn-1.(1)證明:an+2-an=2;(2)若a1=-1,數(shù)列{bn}為等比數(shù)列,b1=a1,b2=a3,求數(shù)列{anbn}的前2022項和T2022.4.(2024·河北石家莊二模)已知數(shù)列{an}的前n項和為Sn,a1=1,2an+1=Sn+2(n∈N*).(1)求數(shù)列{an}的通項公式;(2)數(shù)列{bn}滿意bn=an·(log32an-1)(n∈N*),求數(shù)列{bn}的前n項和T5.(2024·廣東梅州二模)已知Sn是數(shù)列{an}的前n項和,a1=1,.

①?n∈N*,an+an+1=4n;②數(shù)列Snn為等差數(shù)列,且Snn(1)求an;(2)設bn=an+an+1(an·6.(2024·重慶二模)已知Sn為數(shù)列{an}的前n項和,an>0,an2+2an=4Sn+3(n∈N*).數(shù)列{bn}滿意b1=2,b2=4,bn+12=bnbn+2(n(1)求數(shù)列{an}和{bn}的通項公式;(2)設cn=1Sn(n=2k-1,k

考點突破練5數(shù)列求和及其綜合應用1.(1)證明由2Snn+n=2an+1,變形為2Sn=2nan+n-n2,記為①式,又當n≥2時,有2Sn-1=2(n-1)an-1+n-1-(n-1)2,記為②式,①-②并整理可得(2n-2)an-(2n-2)an-1=2n-2,n≥2,n∈即an-an-1=1,n≥2,n∈N*,所以{an}是等差數(shù)列.(2)解由題意可知a72=a4a9,即(a1+6)2=(a1+3)(a1+8),解得a1=-12,所以an=-12+(n-1)×1=n-13,其中a1<a2<…<a12<0,a13=0,則Sn的最小值為S12=S13=-2.(1)解由題意可知,數(shù)列{an}為遞增數(shù)列,又公差d<3,所以a3=5,a4=7,a5=9,則可求出a1=1,d=2,所以an=2n-1.(2)證明bn=8(Tn=1-13+12-14+13-13.(1)證明因為anan+1=2Sn-1, ①所以an+1an+2=2Sn+1-1, ②②-①得an+1(an+2-an)=2an+1,又an+1≠0,所以an+2-an=2.(2)解由a1=-1,得a3=1,于是b2=a3=1.由b1=a1=-1,得{bn}的公比q=-1.所以bn=(-1)n,anbn=(-1)nan.由a1a2=2a1-1,得a2=3.由an+2-an=2,得a2022-a2021=a2020-a2019=…=a2-a1=4.因此T2022=-a1+a2-a3+a4…-a2021+a2022=(a2-a1)+(a4-a3)+…+(a2022-a2021)=1011×(a2-a1)=1011×4=4044.4.解(1)當n≥2時,由2an+1=Sn+2,得2an=Sn-1+2,兩式相減得2an+1-2an=an,所以an+1an=32.因為a1=1,a2=a1+22=32,所以a(2)bn=an·(log32an-1)=(n-2)·32n-1,則Tn=-1×320+0×32+1×322+…+(n-2)·32n-1,32Tn=-1×32+0×322+1×323+…+(n-3)·32n-1+(n-2)·32n,兩式相減得-12Tn=-1+32+322+5.解(1)選條件①:由?n∈N*,an+an+1=4n,得an+1+an+2=4(n+1),所以an+2-an=4(n+1)-4n=4,即數(shù)列{a2k-1},{a2k}(k∈N*)均為公差為4的等差數(shù)列,于是a2k-1=a1+4(k-1)=4k-3=2(2k-1)-1.又a1+a2=4,所以a2=3,a2k=a2+4(k-1)=4k-1=2·(2k)-1.所以an=2n-1.選條件②:由數(shù)列Snn為等差數(shù)列,且Snn的前3項和為6,得S11+S2所以Snn的公差為d=S22-SSnn=1+(n-1)=n,則Sn=n當n≥2時,an=Sn-Sn-1=n2-(n-1)2=2n-1.又a1=1滿意an=2n-1,所以對隨意的n∈N*,an=2n-1.(2)因為bn=an所以Tn=b1+b2+…+bn=12112-1326.解(1)an>0,an2+2an=4Sn+當n=1時,a12-2a1-3=0,解得a1=3或a1當n≥2時,an-12+2an-1=4Sn-①-②得(an+an-1)(an-an-1)=2(an+an-1),所以an-an-1=2.所以數(shù)列{an}是以3為首項,2為公差的等差數(shù)列.所以an=2n+1(n∈N*).因為數(shù)列{bn}滿意b1=2,b2=