2019屆高考數(shù)學(xué)二輪復(fù)習(xí) 第二篇 專題通關(guān)攻略 專題3 數(shù)列 專題能力提升練九 2.3.2 數(shù)列求和及綜合應(yīng)用.doc

專題能力提升練 九數(shù)列求和及綜合應(yīng)用(45分鐘80分)一、選擇題(每小題5分,共30分)1.等比數(shù)列an的前n項(xiàng)和為Sn=a3n-1+b,則ab=()A.-3B.-1C.1D.3【解析】選A.因?yàn)榈缺葦?shù)列an的前n項(xiàng)和為Sn=a3n-1+b,所以a1=S1=a+b,a2=S2-S1=3a+b-a-b=2a,a3=S3-S2=9a+b-3a-b=6a,因?yàn)榈缺葦?shù)列an中,a22=a1a3,所以(2a)2=(a+b)6a,解得ab=-3.2.等比數(shù)列an中,a3=9,前3項(xiàng)和為S3=033x2dx,則公比q的值是 ()A.1B.-12C.1或-12D.-1或-12【解析】選C.S3=x303=27,則當(dāng)q1時(shí),S3=a1(1-q3)1-q=27,a3=a1q2=9,可得q=1(舍)或-12.當(dāng)q=1時(shí),a3=a2=a1=9,S3=27,也符合題意.3.設(shè)數(shù)列an的前n項(xiàng)和為Sn,且a1=1,Sn+nan為常數(shù)列,則an=()A.13n-1B.2n(n+1)C.6(n+1)(n+2)D.5-2n3【解析】選B.由題意知,Sn+nan=2,當(dāng)n2時(shí),Sn-1+(n-1)an-1=2,兩式相減整理得(n+1)an=(n-1)an-1,從而a2a1a3a2a4a3anan-1=1324n-1n+1,有an=2n(n+1),當(dāng)n=1時(shí)上式成立,所以an=2n(n+1).4.已知x>1,y>1,且lg x,14,lg y成等比數(shù)列,則xy有()A.最小值10B.最小值10C.最大值10D.最大值10【解析】選B.因?yàn)閘g x,14,lg y成等比數(shù)列,所以142=(lg x)(lg y),即(lg x)(lg y)=116,又x>1,y>1,所以lg x>0,lg y>0,所以lg x+lg y2(lgx)(lgy)=12,當(dāng)且僅當(dāng)lg x=lg y時(shí),即x=y取等號(hào),所以lg x+lg y=lg(xy)12,則xy10,即xy有最小值是10.5. 已知數(shù)列an滿足a1=1,a2=2,an+2=1+cos2n2an+sin2n2,則該數(shù)列的前18項(xiàng)和為 ()A.2 101B.1 067 C.1 012D.2 012【解析】選B.當(dāng)n為奇數(shù)時(shí),an+2=an+1,這是一個(gè)首項(xiàng)為1,公差為1的等差數(shù)列;當(dāng)n為偶數(shù)時(shí),an+2=2an+1,這是一個(gè)以2為首項(xiàng),公比為2的等比數(shù)列,所以S18=a1+a2+a17+a18=(a1+a3+a17)+(a2+a4+a18)=9+9(9-1)21+2(1-29)1-2 =9+36+1 022=1 067.6.已知函數(shù)f(x)=exex+1,an為等比數(shù)列,an>0且a1 009=1,則f(ln a1)+f(ln a2)+f(ln a2 017)=()A.2 007B.11 009C.1D.2 0172【解析】選D.因?yàn)閒(x)=exex+1,所以f(-x)+f(x)=exex+1+e-xe-x+1=1,因?yàn)閿?shù)列an是等比數(shù)列,所以a1a2 017=a2a2 016=a1 008a1 010=a1 0092=1,所以設(shè)S2 017=f(ln a1)+f(ln a2)+f(ln a2 017),因?yàn)镾2 017=f(ln a2 017)+f(ln a2 016)+f(ln a1),+得2S2 017=2 017,所以S2 017=2 0172.二、填空題(每小題5分,共10分)7.數(shù)列an的前n項(xiàng)和為Sn,若點(diǎn)(n,Sn)(nN*)在函數(shù)y=log2(x+1)的反函數(shù)的圖象上,則an=_.【解析】由題意得n=log2(Sn+1)Sn=2n-1.當(dāng)n2時(shí),an=Sn-Sn-1=2n-2n-1=2n-1,當(dāng)n=1時(shí),a1=S1=21-1=1也適合上式,所以數(shù)列an的通項(xiàng)公式an=2n-1.答案:2n-18.(2018河南一診)已知Sn為數(shù)列an的前n項(xiàng)和,a1=1,當(dāng)n2時(shí),恒有kan= anSn-Sn2成立,若S99=150,則k=_.【解析】當(dāng)n2時(shí),恒有kan=anSn-Sn2成立,即為(k-Sn)(Sn-Sn-1)=-Sn2,化為1Sn-1Sn-1=1k,可得1Sn=1+n-1k,可得Sn=kk+n-1.由S99=150,可得150=kk+98,解得k=2.答案:2三、解答題(每小題10分,共40分)9.(2018佛山一模)已知數(shù)列an是等比數(shù)列,數(shù)列bn滿足b1=-3,b2=-6, an+1+bn=n(nN*).(1)求an的通項(xiàng)公式.(2)求數(shù)列bn的前n項(xiàng)和Sn.【解析】(1)因?yàn)閍n+1+bn=n,則a2+b1=1,得a2=4,a3+b2=2,得a3=8,因?yàn)閿?shù)列an是等比數(shù)列,所以a1q=4,a1q2=8,解得a1=2,q=2,所以an=a1qn-1=2n.(2)由(1)可得bn=n-an+1=n-2n+1,所以Sn=(1-22)+(2-23)+(n-2n+1)=(1+2+3+n)-(22+23+2n+1)=n(n+1)2-22(1-2n)1-2=n2+n2+4-2n+2.10.(2018化州二模)設(shè)數(shù)列an滿足:a1=1,點(diǎn)(an,an+1)(nN*)均在直線y=2x+1上.(1)證明數(shù)列an+1為等比數(shù)列,并求出數(shù)列an的通項(xiàng)公式.(2)若bn=log2(an+1),求數(shù)列(an+1)bn的前n項(xiàng)和Tn.【解析】(1)因?yàn)辄c(diǎn)(an,an+1)(nN*)均在直線y=2x+1上,所以an+1=2an+1,變形為:an+1+1=2(an+1),又a1+1=2.所以數(shù)列an+1為等比數(shù)列,首項(xiàng)與公比都為2.所以an+1=2n,解得an=2n-1.(2)bn=log2(an+1)=n,所以(an+1)bn=n2n.數(shù)列(an+1)bn的前n項(xiàng)和Tn=2+222+323+n2n,2Tn=22+223+(n-1)2n+n2n+1,相減可得:-Tn=2+22+2n-n2n+1=2(2n-1)2-1-n2n+1,所以Tn=(n-1)2n+1+2.11.(2018大慶一模)已知數(shù)列an的前n項(xiàng)和為Sn,點(diǎn)(n,Sn)在曲線y=12x2+52x上,數(shù)列bn滿足bn+bn+2=2bn+1,b4=11,bn的前5項(xiàng)和為45.(1)求an,bn的通項(xiàng)公式.(2)設(shè)cn=1(2an-3)(2bn-8),數(shù)列cn的前n項(xiàng)和為Tn,求使不等式Tn>k54恒成立的最大正整數(shù)k的值.【解析】(1)由已知得:Sn=12n2+52n,當(dāng)n=1時(shí),a1=S1=12+52=3,當(dāng)n2時(shí),an=Sn-Sn-1=12n2+52n-12(n-1)2-52(n-1)=n+2,當(dāng)n=1時(shí),a1也符合上式.所以an=n+2.因?yàn)閿?shù)列bn滿足bn+bn+2=2bn+1,所以bn為等差數(shù)列.設(shè)其公差為d.則b4=b1+3d=11,5b3=5(b1+2d)=45,解得b1=5,d=2,所以bn=2n+3.(2)由(1)得,cn=1(2an-3)(2bn-8)=1(2n+1)(4n-2)=12(2n+1)(2n-1)=1412n-1-12n+1,Tn=141-13+13-15+12n-1-12n+1=141-12n+1,因?yàn)門n+1-Tn=1412n+1-12n+3=12(2n+1)(2n+3)>0,所以Tn是遞增數(shù)列.所以TnT1=16,故Tn>k54恒成立只要T1=16>k54恒成立.所以k<9,最大正整數(shù)k的值為8.【加固訓(xùn)練】(2018茂名一模)設(shè)正項(xiàng)等比數(shù)列an,a4=81,且a2,a3的等差中項(xiàng)為32(a1+a2).(1)求數(shù)列an的通項(xiàng)公式.(2)若bn=log3a2n-1,數(shù)列bn的前n項(xiàng)和為Sn,數(shù)列cn滿足cn=14Sn-1,Tn為數(shù)列cn的前n項(xiàng)和,若Tn<n恒成立,求的取值范圍.【解析】(1)設(shè)等比數(shù)列an的公比為q(q>0),由題意,得a4=a1q3=81,a1q+a1q2=3(a1+a1q),解得a1=3,q=3,所以an=a1qn-1=3n.(2) 由(1)得bn=log332n-1=2n-1,Sn=n(b1+bn)2=n1+(2n-1)2=n2.所以cn=14n2-1=1212n-1-12n+1,所以Tn=121-13+13-15+12n-1-12n+1=n2n+1,若Tn=n2n+1<n恒成立,則>12n+1(nN*)恒成立,則>12n+1max,所以>13.12.已知函數(shù)f(x)=ln x+cos x-6-92x的導(dǎo)數(shù)為f(x),且數(shù)列an滿足an+1+an=nf6+3(nN*).(1)若數(shù)列an是等差數(shù)列,求a1的值.(2)若對(duì)任意nN*,都有an+2n20成立,求a1的取值范圍.【解析】f(x)=1x-sin x-6+92,則f6=4,故an+1+an=4n+3.(1)設(shè)等差數(shù)列an的公差為d,則an=a1+(n-1)d,an+1=a1+nd,由an+1+an=4n+3得(a1+nd)+a1+(n-1)d=4n+3,解得d=2,a1=52.(2)由an+1+an=4n+3得an+2+an+1=4n+7,兩式相減得an+2-an=4,故數(shù)列a2n-1是首項(xiàng)為a1,公差為4的等差數(shù)列;數(shù)列a2n是首項(xiàng)為a2,公差為4的等差數(shù)列,又a1+a2=7,a2=7-a1,所以an=2n-2+a1(n為奇數(shù)),2n+3-a1(n為偶數(shù)).當(dāng)n為奇數(shù)時(shí),an=2n-2+a1,則有a1-2n2-2n+2對(duì)任意的奇數(shù)n恒成立,令f(n)=-2n2-2n+2=-2n+122+52,n為奇數(shù),則f(n)max=f(1)=-2,所以a1-2.當(dāng)n為偶數(shù)時(shí),an=2n+3-a1,則有-a1-2n2-2n-3對(duì)任意的偶數(shù)n恒成立,令g(n)=-2n2-2n-3=-2n+122-52,n為偶數(shù),則g(n)max=g(2)=-15,故-a1-15,解得a115.綜上,a1的取值范圍是-2,15.(建議用時(shí):50分鐘)1.(2018遂寧一模)在數(shù)列an中,a2=8,a5=2,且2an+1-an+2=an(nN*),則|a1|+|a2|+|a10|的值是()A.210B.10C.50D.90【解析】選C.因?yàn)?an+1-an+2=an(nN*),即2an+1=an+2+an(nN*),所以數(shù)列an是等差數(shù)列,設(shè)公差為d,則a1+d=8,a1+4d=2,聯(lián)立解得a1=10,d=-2,所以an=10-2(n-1)=12-2n.令an0,解得n6.Sn=n(10+12-2n)2=11n-n2.所以|a1|+|a2|+|a10|=a1+a2+a6-a7-a10=2S6-S10=2(116-62)-(1110-102)=50.【加固訓(xùn)練】(2018內(nèi)江一模)已知Sn是等差數(shù)列an的前n項(xiàng)和,a1=1,a8=3a3,則a2S1S2+a3S2S3 +a4S3S4+an+1SnSn+1=_.【解析】由a1=1,a8=3a3,得a1+7d=3(a1+2d),即1+7d=3+6d,得d=2,an+1SnSn+1=Sn+1-SnSnSn+1=1Sn-1Sn+1,則a2S1S2+a3S2S3+a4S3S4+an+1SnSn+1=1S1-1S2+1S2-1S3+1Sn-1Sn+1=1S1-1Sn+1=1-1(n+1)+(n+1)n22=1-1(n+1)2.答案:1-1(n+1)22.已知數(shù)列an的前n項(xiàng)和為Sn,若a1為函數(shù)f(x)=3sin x+cos x(xR)的最大值,且滿足an-anSn+1=a12-anSn,則數(shù)列an的前2 018項(xiàng)之積A2 018=()A.1B.12C.-1D.2【解析】選A.函數(shù)f(x)=3sin x+cos x=2sinx+6,當(dāng)x=2k+3,kZ時(shí),f(x)取得最大值2,則a1=2.由an-anSn+1=a12-anSn=1-anSn,即為an=anSn+1-anSn+1,即有an+1=an-1an=1-1an,an+2=1-1an+1=11-an,an+3=1-1an+2=an,則數(shù)列an是周期為3的數(shù)列,且a1=2,a2=12,a3=-1,則一個(gè)周期的乘積為-1,由于2 018=3672+2,則數(shù)列an的前2 018項(xiàng)之積A2 018=1212=1.3.已知無窮數(shù)列an,a1=1,a2=2,對(duì)任意nN*,有an+2=an,數(shù)列bn滿足bn+1-bn=an(nN*),若數(shù)列b2nan中的任意一項(xiàng)都在該數(shù)列中重復(fù)出現(xiàn)無數(shù)次,則滿足要求的b1的值為_.【解析】a1=1,a2=2,對(duì)任意nN*,有an+2=an,所以a3=a1=1,a4=a2=2,a5=a3=a1=1,所以an=1,n為奇數(shù),2,n為偶數(shù),所以bn+1-bn=an=1,n為奇數(shù),2,n為偶數(shù),所以b2n+2-b2n+1=a2n+1=1,b2n+1-b2n=a2n=2,所以b2n+2-b2n=3,b2n+1-b2n-1=3,所以b3-b1=b5-b3=b2n+1-b2n-1=3,b4-b2=b6-b4=b8-b6=b2n-b2n-2=3,b2-b1=1,b2a1=b2,b4a2=b42,b6a3=b6,b8a4=b82,b4n-2a2n-1=b4n-2,b4na2n=b4n2,因?yàn)閿?shù)列b2nan中的任意一項(xiàng)都在該數(shù)列中重復(fù)出現(xiàn)無數(shù)次,所以b2=b6=b10=b4n-2,b4=b8=b12=b4n,解得b8=b4=3,b2=3,因?yàn)閎2-b1=1,所以b1=2.答案:24.(2018菏澤一模) 已知等差數(shù)列an的前n項(xiàng)和為Sn,且S6=-9,S8=4,若滿足不等式nSn的正整數(shù)n有且僅有3個(gè),則實(shí)數(shù)的取值范圍為_.【解析】不妨設(shè)Sn=An2+Bn,由S6=-9,S8=4,得36A+6B=-9,64A+8B=4,則A=1,B=-152,所以nSn=n3-152n2,令f(x)=x3-152x2,則f(x)=3x2-15x=3x(x-5),易得數(shù)列nSn在n5時(shí)單調(diào)遞減;在n>5時(shí)單調(diào)遞增.令nSn=bn,有b3=-812,b4=-56,b5=-1252,b6=-54,b7=-492.若滿足題意的正整數(shù)n只有3個(gè),則n只能為4,5,6,故實(shí)數(shù)的取值范圍為-54,-812.答案:-54,-8125.(2018日照一模)已知等差數(shù)列an的公差d>0,其前n項(xiàng)和為Sn,且a2+a4=8, a3,a5,a8成等比數(shù)列.(1)求數(shù)列an的通項(xiàng)公式.(2)令bn=1a2n-1a2n+1+n,求數(shù)列bn的前n項(xiàng)和Tn.【解析】(1)因?yàn)閍2+a4=8,a1+2d=4,因?yàn)閍3,a5,a8為等比數(shù)列,則a52=a3a8,即(a1+4d)2=(a1+2d)(a1+7d),化簡得:a1=2d,聯(lián)立和得:a1=2,d=1,所以an=n+1.(2)因?yàn)閎n=1a2n-1a2n+1+n=12n(2n+2)+n=141n-1n+1+n,所以Tn=141-12+1+1412-13+2+1413-14+3+141n-1n+1+n=141-12+12-13+13-14+1n-1n+1+(1+2+3+n)=141-1n+1+n(n+1)2=n4(n+1)+n(n+1)2.6.(2018安慶二模)已知公差不為0的等差數(shù)列an的首項(xiàng)a1=2,且a1+1,a2+1, a4+1成等比數(shù)列.(1)求數(shù)列an的通項(xiàng)公式.(2)設(shè)bn=1anan+1,nN*,Sn是數(shù)列bn的前n項(xiàng)和,求使Sn<319成立的最大的正整數(shù)n.【解析】(1)設(shè)數(shù)列an的公差為d,則an=2+(n-1)d,nN*.由a1+1,a2+1,a4+1成等比數(shù)列,得(a2+1)2=(a1+1)(a4+1), 即(3+d)2=3(3+3d),得d=0(舍去)或d=3.所以數(shù)列an的通項(xiàng)公式an=3n-1,nN*.(2)因?yàn)閎n=1anan+1=1(3n-1)(3n+2)=1313n-1-13n+2,所以Sn=1312-15+1315-18+1313n-1-13n+2=1312-13n+2=n2(3n+2),由Sn<319,即n2(3n+2)<319,得n<12.所以使Sn<319成立的最大的正整數(shù)n=11.【加固訓(xùn)練】1.在ABC中,D是BC的中點(diǎn),點(diǎn)列Pn(nN*)在線段AC上,且滿足=an+1 +an,若a1=1,則數(shù)列an的通項(xiàng)公式an=_.【解析】=12(+)=12+12=12-(>0),所以=-1+12,所以an+1=12,an=-1,所以an+1=-12an,又a1=1,所以an是以1為首項(xiàng),-12為公比的等比數(shù)列,所以an=-12n-1.答案:-12n-12.(2018成都七中二診)等差數(shù)列an各項(xiàng)都為正數(shù),且其前9項(xiàng)之和為45,設(shè)bn=1an+4a10-n,其中1n9,若bn中的最小項(xiàng)為b3,則an的公差不能為()A.1B.56C.23D.12【解析】選D.設(shè)等差數(shù)列an的首項(xiàng)為a1,公差為d,由前9項(xiàng)之和為45,可得S9=9a1+982d=45,所以a1+4d=5,a1=5-4d,an=5-4d+(n-1)d=nd+5-5d,bn=1(n-5)d+5+45d+5-nd,要使b3最小,則b3<b2,b3<b4,b2=15-3d+45+3d,b3=15-2d+45+2d,b4=15-d+45+d,可驗(yàn)證d=23,d=56,d=1時(shí),都有b3<b2,b3<b4成立,而當(dāng)d=12時(shí),b2=8291b3=1112b4=9499b3>b2,所以b3不是最小值,所以an的公差不能是12.3.(2018內(nèi)江一模)設(shè)nN*,函數(shù)f1(x)=xex,f2(x)=f1(x),f3(x)=f2(x),fn+1(x)=fn(x),曲線y=fn(x)的最低點(diǎn)為Pn,PnPn+1Pn+2的面積為Sn,則()A.Sn是常數(shù)列B.Sn不是單調(diào)數(shù)列C.Sn是遞增數(shù)列D.Sn是遞減數(shù)列【解析】選D.根據(jù)題意,函數(shù)f1(x)=xex,其導(dǎo)函數(shù)f1(x)=(x)ex+x(ex)=(x+1)ex,分析可得在(-,-1)上,f1(x)<0,f1(x)為減函數(shù),在(-1,+)上,f1(x)>0,f1(x)為增函數(shù),曲線y=f1(x)的最低點(diǎn)P1-1,-1e,對(duì)于函數(shù)f2(x)=f1(x)=(x+1)ex,其導(dǎo)數(shù)f2(x)=(x+1)ex+(x+1)(ex)=(x+2)ex,分析可得在(-,-2)上,f2(x)<0,f2(x)為減函數(shù),在(-2,+)上,f2(x)>0,f2(x)為增函數(shù),曲線y=f2(x)的最低點(diǎn)P2-2,-1e2,分析可得曲線y=fn(x)的最低點(diǎn)Pn,其坐標(biāo)為-n,-1en;則Pn+1-n-1,-1en+1,Pn+2-n-2,-1en+2;所以|PnPn+1|=(-n-1+n)2+-1en+1+1en2=1+1e2n1-1e2,直線PnPn+1的方程為y+1en-1en+1+1en=x+n-n-1+n,分子、分母同乘en+1,得en+1y+e-1+e=x+n-1-(en+1y+e)=(e-1)x+(e-1)n.即(e-1)x+en+1y+e+en-n=0,故點(diǎn)Pn+2到直線PnPn+1的距離d=e+1e-2(e-1)2+e2n+2,所以Sn=12|PnPn+1|d=e2-2e+12en+2,設(shè)g(n)=e2-2e+1en+2,易知函數(shù)g(n)為單調(diào)遞減函數(shù),故Sn是遞減數(shù)列.