**高大項

高中列.(2018山呂梁一模17)已知{}首為等比數(shù)列數(shù)列{}足=2,b且=ab+annn+1n+1(1)求數(shù)列{}通項公式;n(2)求數(shù)列{}前n項.n.(2018福龍巖質(zhì)檢17)已知正項等比數(shù)列{}前和為S且S=2a-1(n).nn(1)求數(shù)列{}通項公式;n(2)若=lga,求數(shù)列{}前n項和Tnnn.(2018北海淀期末15)已知等差數(shù)列{}前n項S,且=ann23(1)數(shù)列{}通公式;n(2)若=,數(shù)列{}前和nn.(2018河唐山一模17)已知數(shù)列{}單調(diào)遞增數(shù)列S為其前和,n(1)求{}通項公式;n(2)若=,T為數(shù)列{}前n項,證明:T<nnn

****.(2018湖衡陽二模17)等差數(shù)列{}=9,S為比數(shù)列{}前n項且b=2,若n7,3S,2S成等差1(1)求數(shù)列{},通項公;n(2)設(shè)c|·b求數(shù)列{c}前項和Tnnnn.知數(shù){}前和為S=(m+1)-ma對任意的nN都成立其中m為數(shù)且nnn(1)求證數(shù)列{}等比數(shù)列n(2)記數(shù)列{}公比為設(shè)q=f(m),若數(shù)列{}足=f(b)(n≥2,n∈).求證數(shù)列nn11nn-1數(shù)列;(3)在(的條件,設(shè)c=b·b,列{}前n項為,證:Tnn+1

是等差.(2018宿十三所中學(xué)期已知數(shù){}前n項為S并且滿足a=1,na+n(n+1).nn+1n(1)求數(shù)列{}通項公式;n(2)若=數(shù)列{}前n項為求T;nnnn(3)在(的條件,是否存在常數(shù)λ使得數(shù)列

為等比數(shù)列存在,試求出λ;不存在說明理

n-1n-1n-1n3-222nnn-1n-1n-1n3-222nn高考大題專項三

高考中的數(shù)列.b=a∴a=a-ab=3a12112111{a}nbn+12,3,b=3n-1,n=n

-

.∈nn

*

S-1,11aa1=2aa+aa=2.22122{},nq={}=2n

n-1(1),=lganT=(b)+(b)…+)n112n2+2)…+[(n-1)lg]=[lg2+2lg2+1+2+)=

-

lg.{}a,na1+(n-1)d,=2n+1.nn,a=2n+1.n(1)=2n+1,nbn{}1n{+b}nnnT+b)+(a+b)+(a+b)=(a+…+a)+(b+…+b)=n112n.(1)解n=1,2S=+1,(aa=1.11{},n2S=nn+12S-2Sn+1n2a=+1,n+1=(a-1),n+1a=a-1,-a=1,nn+1n{},,=n.nn

--

=n+2n+.(2)證明因b=nT=()+(n

+

]=..{},a-a=4d=9-1=8,n73a=an{}q,=4S,q=2,n=2n∵.n

34n-1n45nn+14nn+1n+34n-1n45nn+14nn+1n+n=1T=6,1n=2T2n≥3,2n-5>0,T=10+1×2+3×2+…++(2n-5)2n+…+n①②,-+,nTn

①②Tn

-.明,a=1.1nn=(m+1)-ma(n2),n-1n-1-=ma-ma(n2),nn-1(m+1)a=mann-1a≠0,m<-1,a≠≠1n-1

①②

≥{}1,n

.∵=1,b=f(b)=1n-1

≥

≥2),

=1(n≥2),-1,(2)=n,b=,c·bn++,=n.∵nan+1nn≥2,(n-1)a+n(n-1),nn-1-a=2(na+1×(1+1)=3,n+1n1{}1,=2n-1.nn

①②