2019-2020年高考数学二轮复习 专题四 数列 第二讲 数列求和及数列的综合应用素能提升练 理.doc

2019-2020年高考数学二轮复习 专题四 数列 第二讲 数列求和及数列的综合应用素能提升练 理1.若数列an是公差为2的等差数列,则数列是()A.公比为4的等比数列B.公比为2的等比数列C.公比为的等比数列D.公比为的等比数列解析:an=a1+2(n-1),=22=4.是等比数列,公比为4.答案:A2.已知在数列an中,a1=-60,an+1=an+3,则|a1|+|a2|+|a3|+|a30|等于()A.445B.765C.1080D.3105解析:an+1=an+3,an+1-an=3.an是以-60为首项,3为公差的等差数列.an=a1+3(n-1)=3n-63.令an0,得n21.前20项都为负值.|a1|+|a2|+|a3|+|a30|=-(a1+a2+a20)+a21+a30=-2S20+S30.Sn=n=n,|a1|+|a2|+|a3|+|a30|=765.答案:B3.设数列an的前n项和Sn=2n-1,则的值为()A.B.C.D.解析:Sn=2n-1,.答案:A4.设某商品一次性付款的金额为a元,以分期付款的形式等额地分成n次付清,若每期利率r保持不变,按复利计算,则每期期末所付款是()A.(1+r)n元B.元C.(1+r)n-1元D.元解析:设每期期末所付款是x元,则各次付款的本利和为x(1+r)n-1+x(1+r)n-2+x(1+r)n-3+x(1+r)+x=a(1+r)n,即x=a(1+r)n,故x=.答案:B5.(xx黑龙江大庆第二次质检,10)已知数列an满足an+1=an-an-1(n2),a1=1,a2=3,记Sn=a1+a2+an,则下列结论正确的是()A.axx=-1,Sxx=2B.axx=-3,Sxx=5C.axx=-3,Sxx=2D.axx=-1,Sxx=5解析:由已知数列an满足an+1=an-an-1(n2),知an+2=an+1-an,an+2=-an-1(n2),an+3=-an,an+6=an,又a1=1,a2=3,a3=2,a4=-1,a5=-3,a6=-2,所以当kN时,ak+1+ak+2+ak+3+ak+4+ak+5+ak+6=a1+a2+a3+a4+a5+a6=0,axx=a4=-1,Sxx=a1+a2+a3+a4=1+3+2+(-1)=5.答案:D6.数列an的通项公式an=ncos,其前n项和为Sn,则Sxx等于()A.1006B.xxC.503D.0解析:函数y=cos的周期T=4,可分四组求和:a1+a5+axx=0,a2+a6+axx=-2-6-xx=-5031006,a3+a7+axx=0,a4+a8+axx=4+8+xx=5031008.故Sxx=0-5031006+0+5031008=503(-1006+1008)=1006.答案:A7.(xx河南郑州第二次质检,12)已知正项数列an的前n项和为Sn,若2Sn=an+(nN*),则Sxx=()A.xx+B.xx-C.xxD.解析:由题意可知,当n2时,an=Sn-Sn-1,则2Sn=an+=Sn-Sn-1+,整理,得=1.=1,即数列是公差为1的等差数列,又由2S1=2a1=a1+,解得a1=1(an0),即S1=1,=1,因此=n.故Sxx=.答案:D8.(xx河北唐山高三统考,16)若数列an的前n项和为Sn,且a1=3,an=2Sn-1+3n(n2),则该数列的通项公式为an=.解析:an=2Sn-1+3n,an-1=2Sn-2+3n-1(n3),相减得an-an-1=2an-1+23n-1,即an=3an-1+23n-1.(n3).又a2=2S1+32=2a1+32=15,即,数列是以1为首项,为公差的等差数列,=1+(n-1).an=(2n+1)3n-1.答案:(2n+1)3n-19.(xx贵州六校第一次联考,16)已知f(x)=,各项均为正数的数列an满足a1=1,an+2=f(an),若a12=a14,则a13+axx=.解析:由f(x)=,a1=1,an+2=f(an)可得a3=,a5=,同理可推得a7=,a9=,a11=,a13=,由a12=a14,得,a10=a12,依次推出a2=a4=a6=axx,由a4=f(a2),得a2=+a2-1=0,a2=.故a13+axx=.答案:10.(xx浙江高考,理18)在公差为d的等差数列an中,已知a1=10,且a1,2a2+2,5a3成等比数列.(1)求d,an;(2)若d0,求|a1|+|a2|+|a3|+|an|.解:(1)由题意得5a3a1=(2a2+2)2,即d2-3d-4=0,故d=-1或d=4.当d=-1时,an=-n+11.当d=4时,an=4n+6.所以d=-1,an=-n+11,nN*或d=4,an=4n+6,nN*.(2)设数列an的前n项和为Sn.因为d0,所以an=2n.(2)bn=nan=n2n,Sn=121+222+323+(n-1)2n-1+n2n,2Sn=122+223+324+(n-1)2n+n2n+1,由-,得-Sn=121+22+23+2n-n2n+1,即-Sn=-n2n+1,整理可得Sn=(n-1)2n+1+2.12.已知向量p=(an,2n),向量q=(2n+1,-an+1),nN*,向量p与q垂直,且a1=1.(1)求数列an的通项公式;(2)若数列bn满足bn=log2an+1,求数列anbn的前n项和Sn.解:(1)向量p与q垂直,2n+1an-2nan+1=0,即2nan+1=2n+1an.=2.an是以1为首项,2为公比的等比数列.an=2n-1.(2)bn=log2an+1,bn=n.anbn=n2n-1.Sn=1+22+322+423+n2n-1.2Sn=12+222+323+424+n2n.-,得-Sn=1+2+22+23+24+2n-1-n2n=-n2n=(1-n)2n-1.Sn=1+(n-1)2n.13.设正项数列an的前n项和是Sn,若an和都是等差数列,且公差相等.(1)求an的通项公式;(2)若a1,a2,a5恰为等比数列bn的前三项,记数列cn=,数列cn的前n项和为Tn.求证:对任意nN*,都有Tn2.(1)解:设an的公差为d,则n,且a1-=0.d=,d=,a1=,an=.(2)证明:b1=a1=,b2=a2=,b3=a5=,bn=3n-1.cn=.当n2时,=,当n2时,Tn=+=2-2,且T1=2.故对任意nN*,都有Tn2.