天津市2018高考数学文二轮复习题型练4大题专项数列的通项求和问题

题型练4大题专项(二)数列的通项、求和问题1.设数列{an}的前n项和为Sn,满足(1-q)Sn+qan=1,且q(q-1)≠0.(1)求{an}的通项公式;(2)若S3,S9,S6成等差数列,求证:a2,a8,a5成等差数列.2.已知等差数列{an}的首项a1=1,公差d=1,前n项和为Sn,bn=.(1)求数列{bn}的通项公式;(2)设数列{bn}前n项和为Tn,求Tn.3.(2017江苏,19)对于给定的正整数k,若数列{an}满足:an-k+an-k+1+…+an-1+an+1+…+an+k-1+an+k=2kan对任意正整数n(nk)总成立,则称数列{an}是“P(k)数列”.(1)证明:等差数列{an}是“P(3)数列”;(2)若数列{an}既是“P(2)数列”,又是“P(3)数列”,证明:{an}是等差数列.4.已知等差数列{an}的前n项和为Sn,公比为q的等比数列{bn}的首项是,且a1+2q=3,a2+4b2=6,S5=40.(1)求数列{an},{bn}的通项公式an,bn;(2)求数列的前n项和Tn.5.已知函数f(x)=,数列{an}满足:2an+1-2an+an+1an=0,且anan+1≠0.在数列{bn}中,b1=f(0),且bn=f(an-1).(1)求证:数列是等差数列;(2)求数列{|bn|}的前n项和Tn.6.记U={1,2,…,100}.对数列{an}(n∈N*)和U的子集T,若T=⌀,定义ST=0;若T={t1,t2,…,tk},定义ST=+…+.例如:T={1,3,66}时,ST=a1+a3+a66.现设{an}(n∈N*)是公比为3的等比数列,且当T={2,4}时,ST=30.(1)求数列{an}的通项公式;(2)对任意正整数k(1≤k≤100),若T⊆{1,2,…,k},求证:STak+1;(3)设C⊆U,D⊆U,SC≥SD,求证:SC+SC∩D≥2SD.##题型练4大题专项(二)数列的通项、求和问题1.解(1)当n=1时,由(1-q)S1+qa1=1,a1=1.当n≥2时,由(1-q)Sn+qan=1,得(1-q)Sn-1+qan-1=1,两式相减,得an=qan-1.又q(q-1)≠0,所以{an}是以1为首项,q为公比的等比数列,故an=qn-1.(2)由(1)可知Sn=,又S3+S6=2S9,所以,化简,得a3+a6=2a9,两边同除以q,得a2+a5=2a8.故a2,a8,a5成等差数列.2.解(1)∵在等差数列{an}中,a1=1,公差d=1,∴Sn=na1+d=,∴bn=.(2)bn==2,∴Tn=b1+b2+b3+…+bn=2+…+=2+…+=2.故Tn=.3.证明(1)因为{an}是等差数列,设其公差为d,则an=a1+(n-1)d,从而,当n≥4时,an-k+an+k=a1+(n-k-1)d+a1+(n+k-1)d=2a1+2(n-1)d=2an,k=1,2,3,所以an-3+an-2+an-1+an+1+an+2+an+3=6an,因此等差数列{an}是“P(3)数列”.(2)数列{an}既是“P(2)数列”,又是“P(3)数列”,因此,当n≥3时,an-2+an-1+an+1+an+2=4an,①当n≥4时,an-3+an-2+an-1+an+1+an+2+an+3=6an.②由①知,an-3+an-2=4an-1-(an+an+1),③an+2+an+3=4an+1-(an-1+an).④将③④代入②,得an-1+an+1=2an,其中n≥4,所以a3,a4,a5,…是等差数列,设其公差为d'.在①中,取n=4,则a2+a3+a5+a6=4a4,所以a2=a3-d',在①中,取n=3,则a1+a2+a4+a5=4a3,所以a1=a3-2d',所以数列{an}是等差数列.4.解(1)设{an}公差为d,由题意得解得故an=3n-1,bn=.(2)∵+22n+1,∴Tn=+…+(22n+3-8)=.5.(1)证明∵2an+1-2an+an+1an=0,∴,故数列是以为公差的等差数列.(2)解∵b1=f(0)=5,∴=5,7a1-2=5a1,∴a1=1,=1+(n-1)·,∴an=,bn==7-(n+1)=6-n.当n≤6时,Tn=(5+6-n)=;当n≥7时,Tn=15+(1+n-6)=.故Tn=6.(1)解由已知得an=a1·3n-1,n∈N*.于是当T={2,4}时,ST=a2+a4=3a1+27a1=30a1.又ST=30,故30a1=30,即a1=1.所以数列{an}的通项公式为an=3n-1,n∈N*.(2)证明因为T⊆{1,2,…,k},an=3n-10,n∈N*,所以ST≤a1+a2+…+ak=1+3+…+3k-1=(3k-1)3k.因此,STak+1.(3)证明下面分三种情况证明.①若D是C的子集,则SC+SC∩D=SC+SD≥SD+SD=2SD.②若C是D的子集,则SC+SC∩D=SC+SC=2SC≥2SD.③若D不是C的子集,且C不是D的子集.令E=C∩∁UD,F=D∩∁UC,则E≠⌀,F≠⌀,E∩F=⌀.于是SC=SE+SC∩D,SD=SF+SC∩D,进而由SC≥SD得SE≥SF.设k为E中的最大数,l为F中的最大数,则k≥1,l≥1,k≠l.由(2)知,SEak+1.于是3l-1=al≤SF≤SEak+1=3k,所以l-1k,即l≤k.又k≠l,故l≤k-1.从而SF≤a1+a2+…+al=1+3+…+3l-1=,故SE≥2SF+1,所以SC-SC∩D≥2(SD-SC∩D)+1,即SC+SC∩D≥2SD+1.综上①②③得,SC+SC∩D≥2SD.