（福建專版）2019高考數(shù)學(xué)一輪復(fù)習(xí) 課時規(guī)范練30 數(shù)列求和 文.docx

課時規(guī)范練30數(shù)列求和基礎(chǔ)鞏固組1.數(shù)列112,314,518,7116,(2n-1)+12n,的前n項(xiàng)和Sn的值等于()A.n2+1-12nB.2n2-n+1-12nC.n2+1-12n-1D.n2-n+1-12n2.在數(shù)列an中,a1=-60,an+1=an+3,則|a1|+|a2|+|a30|=()A.-495B.765C.1 080D.3 1053.已知數(shù)列an的前n項(xiàng)和Sn滿足Sn+Sm=Sn+m,其中m,n為正整數(shù),且a1=1,則a10等于()A.1B.9C.10D.554.已知函數(shù)f(x)=xa的圖象過點(diǎn)(4,2),令an=1f(n+1)+f(n),nN*.記數(shù)列an的前n項(xiàng)和為Sn,則S2 018等于()A.2 018-1B.2 018+1C.2 019-1D.2 019+15.已知數(shù)列an中,an=2n+1,則1a2-a1+1a3-a2+1an+1-an=()A.1+12nB.1-2nC.1-12nD.1+2n6.設(shè)數(shù)列an的前n項(xiàng)和為Sn,a1=2,若Sn+1=n+2nSn,則數(shù)列1anan+1的前2 018項(xiàng)和為.7.已知等差數(shù)列an滿足:a5=11,a2+a6=18.(1)求數(shù)列an的通項(xiàng)公式;(2)若bn=an+2n,求數(shù)列bn的前n項(xiàng)和Sn.導(dǎo)學(xué)號241909158.設(shè)等差數(shù)列an的公差為d,前n項(xiàng)和為Sn,等比數(shù)列bn的公比為q,已知b1=a1,b2=2,q=d,S10=100.(1)求數(shù)列an,bn的通項(xiàng)公式;(2)當(dāng)d>1時,記cn=anbn,求數(shù)列cn的前n項(xiàng)和Tn.導(dǎo)學(xué)號241909169.Sn為數(shù)列an的前n項(xiàng)和,已知an>0,an2+2an=4Sn+3.(1)求an的通項(xiàng)公式;(2)設(shè)bn=1anan+1,求數(shù)列bn的前n項(xiàng)和.導(dǎo)學(xué)號24190917綜合提升組10.如果數(shù)列1,1+2,1+2+4,1+2+22+2n-1,的前n項(xiàng)和Sn>1 020,那么n的最小值是()A.7B.8C.9D.1011.(2017山東煙臺模擬)已知數(shù)列an中,a1=1,且an+1=an2an+1,若bn=anan+1,則數(shù)列bn的前n項(xiàng)和Sn為()A.2n2n+1B.n2n+1C.2n2n-1D.2n-12n+1導(dǎo)學(xué)號2419091812.(2017福建龍巖一模,文15)已知Sn為數(shù)列an的前n項(xiàng)和,對nN*都有Sn=1-an,若bn=log2an,則1b1b2+1b2b3+1bnbn+1=.13.(2017廣西模擬)已知數(shù)列an的前n項(xiàng)和為Sn,且Sn=32an-1(nN*).(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)bn=2log3an2+1,求1b1b2+1b2b3+1bn-1bn.導(dǎo)學(xué)號24190919創(chuàng)新應(yīng)用組14.(2017全國)幾位大學(xué)生響應(yīng)國家的創(chuàng)業(yè)號召,開發(fā)了一款應(yīng)用軟件.為激發(fā)大家學(xué)習(xí)數(shù)學(xué)的興趣,他們推出了“解數(shù)學(xué)題獲取軟件激活碼”的活動.這款軟件的激活碼為下面數(shù)學(xué)問題的答案:已知數(shù)列1,1,2,1,2,4,1,2,4,8,1,2,4,8,16,其中第一項(xiàng)是20,接下來的兩項(xiàng)是20,21,再接下來的三項(xiàng)是20,21,22,依此類推.求滿足如下條件的最小整數(shù)N:N>100且該數(shù)列的前N項(xiàng)和為2的整數(shù)冪.那么該款軟件的激活碼是()A.440B.330C.220D.11015.觀察下列三角形數(shù)表:1第1行22第2行343第3行4774第4行51114115第5行假設(shè)第n行的第二個數(shù)為an(n2,nN*).(1)歸納出an+1與an的關(guān)系式,并求出an的通項(xiàng)公式;(2)設(shè)anbn=1(n2),求證:b2+b3+bn<2.答案：1.A該數(shù)列的通項(xiàng)公式為an=(2n-1)+12n,則Sn=1+3+5+(2n-1)+12+122+12n=n2+1-12n.2.B由a1=-60,an+1=an+3可得an=3n-63,則a21=0,|a1|+|a2|+|a30|=-(a1+a2+a20)+(a21+a30)=S30-2S20=765,故選B.3.ASn+Sm=Sn+m,a1=1,S1=1.可令m=1,得Sn+1=Sn+1,Sn+1-Sn=1,即當(dāng)n1時,an+1=1,a10=1.4.C由f(4)=2,可得4a=2,解得a=12,則f(x)=x12.an=1f(n+1)+f(n)=1n+1+n=n+1-n,S2 018=a1+a2+a3+a2 018=(2-1)+(3-2)+(4-3)+(2 019-2 018)=2 019-1.5.Can+1-an=2n+1+1-(2n+1)=2n+1-2n=2n,所以1a2-a1+1a3-a2+1an+1-an=12+122+123+12n=121-12n1-12=1-12n=1-12n.6.1 0094 038Sn+1=n+2nSn,Sn+1Sn=n+2n.又a1=2,當(dāng)n2時,Sn=SnSn-1Sn-1Sn-2Sn-2Sn-3S3S2S2S1S1=n+1n-1nn-2n-1n-342312=n(n+1).當(dāng)n=1時也成立,Sn=n(n+1).當(dāng)n2時,an=Sn-Sn-1=n(n+1)-n(n-1)=2n.當(dāng)n=1時,a1=2也成立,所以an=2n.1anan+1=12n2(n+1)=141n-1n+1.則數(shù)列1anan+1的前2 018項(xiàng)和=141-12+12-13+12 018-12 019=141-12 019=1 0094 038.7.解 (1)設(shè)an的首項(xiàng)為a1,公差為d.由a5=11,a2+a6=18,得a1+4d=11,2a1+6d=18,解得a1=3,d=2,所以an=2n+1.(2)由an=2n+1得bn=2n+1+2n,則Sn=3+5+7+(2n+1)+(21+22+23+2n)=n2+2n+2(1-2n)1-2=n2+2n+2n+1-2.8.解 (1)由題意,有10a1+45d=100,a1d=2,即2a1+9d=20,a1d=2,解得a1=1,d=2或a1=9,d=29.故an=2n-1,bn=2n-1或an=19(2n+79),bn=929n-1.(2)由d>1,知an=2n-1,bn=2n-1,故cn=2n-12n-1,于是Tn=1+32+522+723+924+2n-12n-1,12Tn=12+322+523+724+925+2n-12n.-可得12Tn=2+12+122+12n-2-2n-12n=3-2n+32n,故Tn=6-2n+32n-1.9.解 (1)由an2+2an=4Sn+3,可知an+12+2an+1=4Sn+1+3.兩式相減可得an+12-an2+2(an+1-an)=4an+1,即2(an+1+an)=an+12-an2=(an+1+an)(an+1-an).由于an>0,可得an+1-an=2.又a12+2a1=4a1+3,解得a1=-1(舍去),a1=3.所以an是首項(xiàng)為3,公差為2的等差數(shù)列,故an的通項(xiàng)公式為an=2n+1.(2)由an=2n+1可知bn=1anan+1=1(2n+1)(2n+3)=1212n+1-12n+3.設(shè)數(shù)列bn的前n項(xiàng)和為Tn,則Tn=b1+b2+bn=1213-15+15-17+12n+1-12n+3=n3(2n+3).10.Dan=1+2+22+2n-1=2n-1.Sn=(21-1)+(22-1)+(2n-1)=(21+22+2n)-n=2n+1-n-2,S9=1 013<1 020,S10=2 036>1 020,使Sn>1 020的n的最小值是10.11.B由an+1=an2an+1,得1an+1=1an+2,數(shù)列1an是以1為首項(xiàng),2為公差的等差數(shù)列,1an=2n-1,又bn=anan+1,bn=1(2n-1)(2n+1)=1212n-1-12n+1,Sn=1211-13+13-15+12n-1-12n+1=n2n+1,故選B.12.nn+1對nN*都有Sn=1-an,當(dāng)n=1時,a1=1-a1,解得a1=12.當(dāng)n2時,an=Sn-Sn-1=1-an-(1-an-1),化為an=12an-1.數(shù)列an是等比數(shù)列,公比為12,首項(xiàng)為12.an=12n.bn=log2an=-n.1bnbn+1=1-n(-n-1)=1n-1n+1.則1b1b2+1b2b3+1bnbn+1=1-12+12-13+1n-1n+1=1-1n+1=nn+1.13.解 (1)當(dāng)n=1時,a1=32a1-1,a1=2.當(dāng)n2時,Sn=32an-1,Sn-1=32an-1-1(n2),-得an=32an-1-32an-1-1,即an=3an-1,數(shù)列an是首項(xiàng)為2,公比為3的等比數(shù)列,an=23n-1.(2)由(1)得bn=2log3an2+1=2n-1,1b1b2+1b2b3+1bn-1bn=113+135+1(2n-3)(2n-1)=121-13+13-15+12n-3-12n-1=n-12n-1.14.A設(shè)數(shù)列的首項(xiàng)為第1組,接下來兩項(xiàng)為第2組,再接下來三項(xiàng)為第3組,以此類推,設(shè)第n組的項(xiàng)數(shù)為n,則前n組的項(xiàng)數(shù)和為n(1+n)2.第n組的和為1-2n1-2=2n-1,前n組總共的和為2(1-2n)1-2-n=2n+1-2-n.由題意,N>100,令n(1+n)2>100,得n14且nN*,即N出現(xiàn)在第13組之后.若要使最小整數(shù)N滿足:N>100且前N項(xiàng)和為2的整數(shù)冪,則SN-Sn(1+n)2應(yīng)與-2-n互為相反數(shù),即2k-1=2+n(kN*,n14),所以k=log2(n+3),解得n=29,k=5.所以N=29(1+29)2+5=440,故選A.15.解 (1)由題意知an+1=an+n(n2),a2=2,an=a2+(a3-a2)+(a4-a3)+(an-an-1)=2+2+3+(n-1)=2+(n-2)(n+1)2,an=12n2-12n+1(n2).(2)anbn=1,bn=1an=2n2-n+2<2n2-n=21n-1-1n,b2+b3+b4+bn<211-12+12+13+1n-1-1n=21-1n<2.