（普通班）高三數(shù)學(xué)一輪復(fù)習(xí) 第六篇 數(shù)列 第4節(jié) 數(shù)列求和及綜合應(yīng)用基礎(chǔ)對(duì)點(diǎn)練 理-人教版高三全冊(cè)數(shù)學(xué)試題

第4節(jié)數(shù)列求和及綜合應(yīng)用【選題明細(xì)表】知識(shí)點(diǎn)、方法題號(hào)公式法、并項(xiàng)法、分組法求和1,2,6裂項(xiàng)相消法求和3,10,11,13錯(cuò)位相減法求和4,9數(shù)列的綜合應(yīng)用5,8,12,14數(shù)列的實(shí)際應(yīng)用7基礎(chǔ)對(duì)點(diǎn)練(時(shí)間:30分鐘)1.數(shù)列{1+2n-1}的前n項(xiàng)和為(C)(A)1+2n (B)2+2n(C)n+2n-1 (D)n+2+2n解析:由題意令an=1+2n-1,所以Sn=n+1-2n2.數(shù)列{an}的前n項(xiàng)和為Sn,已知Sn=1-2+3-4+…+(-1)n-1·n,則S17等于(A)(A)9 (B)8 (C)17 (D)16解析:S17=(1-2)+(3-4)+…+(15-16)+17=-1×8+17=9.故選A.3.(2015鞍山校級(jí)四模)數(shù)列{an}的前n項(xiàng)和為Sn,若an=1n(n(A)1 (B)QUOTE56 (C)QUOTE16 (D)130解析:因?yàn)閍n=QUOTE1n-1n+1,所以Sn=(1-QUOTE12)+(QUOTE12-QUOTE13)+…+(QUOTE1n-1n+1)=1-1n+1=nn+1.所以S5=QUOTE56.故選B.4.Sn=QUOTE12+QUOTE12+QUOTE38+…+QUOTEn2n等于(B)(A)2n-(C)2n-解析:由Sn=QUOTE12+QUOTE222+QUOTE323+…+QUOTEn2n,①得QUOTE12Sn=QUOTE122+QUOTE223+…+n-12n+n2n①-②得,QUOTE12Sn=QUOTE12+QUOTE122+QUOTE123+…+QUOTE12n-n2n+1=121-所以Sn=2n5.(2015鄭州二模)已知等比數(shù)列{an}的首項(xiàng)為QUOTE32,公比為-QUOTE12,其前n項(xiàng)和為Sn,則Sn的最大值為(D)(A)QUOTE34 (B)QUOTE23 (C)QUOTE43 (D)QUOTE32解析:因?yàn)榈缺葦?shù)列{an}的首項(xiàng)為QUOTE32,公比為-QUOTE12,所以Sn=32[1-(-12)

n]1當(dāng)n取偶數(shù)時(shí),Sn=1-(QUOTE12)n<1;當(dāng)n取奇數(shù)時(shí),Sn=1+(QUOTE12)n≤1+QUOTE12=QUOTE32.所以Sn的最大值為QUOTE32.故選D.6.(2016寧夏石嘴山高三聯(lián)考)在數(shù)列{an}中,a1=1,a2=2,且an+2-an=1+(-1)n(n∈N*),則a1+a2+…+a51=.

解析:因?yàn)閿?shù)列{an}中,a1=1,a2=2,且an+2-an=1+(-1)n(n∈N*),所以a3-a1=0,a5-a3=0,…a51-a49=0,所以a1=a3=a5=…=a51=1.由a4-a2=2,得a4=2+a2=4,同理可得a6=6,a8=8,…,a50=50.所以a1+a2+a3+…+a51=(a1+a3+a5+…+a51)+(a2+a4+…+a50)=26+(=676.答案:6767.現(xiàn)有一根n節(jié)的竹竿,自上而下每節(jié)的長(zhǎng)度依次構(gòu)成等差數(shù)列,最上面一節(jié)長(zhǎng)為10cm,最下面的三節(jié)長(zhǎng)度之和為114cm,第6節(jié)的長(zhǎng)度是首節(jié)與末節(jié)長(zhǎng)度的等比中項(xiàng),則n=.

解析:設(shè)自上而下每節(jié)竹竿的長(zhǎng)度構(gòu)成的等差數(shù)列為{an},由題意知,a1=10,an+an-1+an-2=114,a62=a1·a所以3an-1=114,即an-1=38.(a1+5d)2=a1·(an-1+d),所以(10+5d)2=10×(38+d),即5d2+18d-56=0,解得d=2或d=-285所以an-1=10+(n-2)×2=2n+6=38,所以n=16.答案:168.對(duì)于每一個(gè)正整數(shù)n,設(shè)曲線y=xn+1在點(diǎn)(1,1)處的切線與x軸的交點(diǎn)的橫坐標(biāo)為xn,令an=lgxn,則a1+a2+…+a99=.

解析:對(duì)y=xn+1求導(dǎo)得y′=(n+1)xn,則曲線在點(diǎn)(1,1)處的切線方程為y-1=(n+1)·(x-1),令y=0,得xn=nn+1,則an=lgxn=lg所以a1+a2+…+a99=lg(QUOTE12×QUOTE23×…×99100(=lg1=-2.答案:-29.(2015高考山東卷)設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,已知2Sn=3n+3.(1)求{an}的通項(xiàng)公式;(2)若數(shù)列{bn}滿(mǎn)足anbn=log3an,求{bn}的前n項(xiàng)和Tn.解:(1)因?yàn)?Sn=3n+3,所以2a1=3+3,故a1=3,當(dāng)n≥2時(shí),2Sn-1=3n-1+3,此時(shí)2an=2Sn-2Sn-1=3n-3n-1=2×3n-1,即an=3n-1,所以an=3(2)因?yàn)閍nbn=log3an,所以b1=QUOTE13,當(dāng)n≥2時(shí),bn=31-nlog33n-1=(n-1)·31-n.所以T1=b1=QUOTE13;當(dāng)n≥2時(shí),Tn=b1+b2+b3+…+bn=QUOTE13+[1×3-1+2×3-2+…+(n-1)×31-n],所以3Tn=1+[1×30+2×3-1+…+(n-1)×32-n],兩式相減,得2Tn=QUOTE23+(30+3-1+3-2+…+32-n)-(n-1)×31-n=QUOTE23+1-31-n1=136-6所以Tn=1312-6經(jīng)檢驗(yàn),n=1時(shí)也適合.綜上可得Tn=1312-610.(2015大連市高三一模)已知數(shù)列{an}中,a1=1,其前n項(xiàng)的和為Sn,且滿(mǎn)足an=2Sn2(1)求證:數(shù)列{QUOTE1Sn}是等差數(shù)列;(2)證明:S1+QUOTE12S2+QUOTE13S3+…+QUOTE1nSn<QUOTE32.證明:(1)當(dāng)n≥2時(shí),Sn-Sn-1=2SSn-1-Sn=2SnSn-1,QUOTE1Sn-1Sn從而{QUOTE1Sn}構(gòu)成以1為首項(xiàng),2為公差的等差數(shù)列.(2)當(dāng)n≥2時(shí),QUOTE1nSn=1n(2n-1)<1n(2n-2)=QUOTE12·1n(n-1)=QUOTE從而S1+QUOTE12S2+QUOTE13S3+…+QUOTE1nSn<1+QUOTE12{1-QUOTE12+QUOTE12-QUOTE13+…+1n-1-QUOTE1n}<QUOTE32-12n<QUOTE32.能力提升練(時(shí)間:15分鐘)11.(2016哈爾濱六中高三期中)數(shù)列{an}滿(mǎn)足a1=1,對(duì)任意的n∈N*都有an+1=a1+an+n,則QUOTE1a1+QUOTE1a2+…+1a2(A)20152解析:an+1=a1+an+n,a1=1,an=an-1+n,即an-an-1=n,an-1-an-2=n-1,…,a2-a1=2(n≥2),以上n-1個(gè)等式分別相加得an-a1=(n-1所以an=(n-1所以QUOTE1an=2n2+n=2(QUOTE1n-1n所以QUOTE1a1+QUOTE1a2+…+1a2016=2{1-QUOTE12+QUOTE12-QUOTE13+…+12016-12017}=12.(2016防城港模擬)已知{an}是等差數(shù)列,{bn}是正項(xiàng)等比數(shù)列,若a11=b10,則(D)(A)a13+a9=b14b6 (B)a13+a9=b14+b6(C)a13+a9≥b14+b6 (D)a13+a9≤b14+b6解析:設(shè){an}是公差為d的等差數(shù)列,{bn}是公比為q的正項(xiàng)等比數(shù)列,即有a13+a9=2a11=2b10,b14b6=b10則a13+a9-b14b6=(2-b10)b10,當(dāng)b10≥2時(shí),a13+a9≤b14b6;當(dāng)0<b10<2時(shí),a13+a9>b14b6.又b14+b6=b1q13+b1q5,由a13+a9-(b14+b6)=2b1q9-b1q13-b1q5=-b1q5(q8-2q4+1)=-b1q5(q4-1)2≤0,則有a13+a9≤b14+b6.綜上可得,A,B,C均錯(cuò),D正確.13.(2015廣西南寧二模)在各項(xiàng)均為正數(shù)的等比數(shù)列{an}中,a1=2,且2a1,a3,3a2成等差數(shù)列.(1)求等比數(shù)列{an}的通項(xiàng)公式;(2)若數(shù)列{bn}滿(mǎn)足bn=(n+2)log2an,求數(shù)列{QUOTE1bn}的前n項(xiàng)和Tn.解:(1)設(shè)數(shù)列{an}的公比為q,因?yàn)?a1,a3,3a2成等差數(shù)列,所以2a1+3a2=2a3,2a1+3a1q=2a1q2,2q2-3q-2=0,解得q=2或q=-QUOTE12.因?yàn)閝>0,所以q=2.因?yàn)閍1=2,所以數(shù)列{an}的通項(xiàng)公式an=a1qn-1=2n,n∈N*.(2)因?yàn)閎n=(n+2)log2an=n(n+2),所以QUOTE1bn=1n(n+2)=QUOTE12(QUOTE1n-1n+2Tn=QUOTE1b1+QUOTE1b2+…+1bn-1+QUOTE1bn=QUOTE12[(1-QUOTE13)+(QUOTE12-QUOTE14)+(QUOTE13-QUOTE15)+…+(1n-2-QUOTE1n)+(1n-1-1n+1)+(QUOTE1n-1n+2)]=QUOTE12(1+QUOTE12-1n+1-1n+2)=QUOTE34-2n+3214.(2015鄭州第二次質(zhì)量預(yù)測(cè))已知數(shù)列{an}的前n項(xiàng)和為Sn,且Sn=2an-2.(1)求數(shù)列{an}的通項(xiàng)公式;(2)設(shè)bn=log2a1+log2a2+…+log2an,求使(n-8)bn≥nk對(duì)任意n∈N解:(1)由Sn=2an-2可得a1=2,因?yàn)镾n=2an-2,所以當(dāng)n≥2時(shí),an=Sn-Sn-1=2an-2an-1,即an數(shù)列{an}是以a1=2為首項(xiàng),公比為2的等比數(shù)列,所以an=2n(n∈N*).(2)bn=log2a1+log2a2+…+log2an=1+2+3+…+n=由(n-8)bn≥nk對(duì)任意n∈N*恒成立,即實(shí)數(shù)(n-8)(n+1設(shè)cn=QUOTE12(n-8)(n+1),則當(dāng)n=3或n=4時(shí),cn取得最小值為-10,所以k≤-10,即k的取值范圍為(-∞,-10].精彩5分鐘1.(2015石家莊二模)已知函數(shù)f(x)=1-|1-x|,x∈(-∞,2),2f(x-2),(A)n2 (B)n2+n (C)2n-1 (D)2n+1-1解題關(guān)鍵:解決本題的關(guān)鍵是正確畫(huà)出函數(shù)f(x)的圖象,找出f(x)=2x-1解析:函數(shù)f(x)=1-|x=1時(shí),f(x)=1,x=3時(shí),f(x)=2,x=5時(shí),f(x)=4,所以方程f(x)=2x-12的根從小到大依次為1,3,5,…,數(shù)列{f(x所以數(shù)列{f(xn)}的前n項(xiàng)和為1-2n2.(2015甘肅一診)設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,已知QUOTE1S1+QUOTE1S2+…+QUOTE1Sn=nn+1,設(shè)bn=(QUOTE12)

an,數(shù)列{bn}的前n項(xiàng)和為T(mén)n,若對(duì)一切n∈N*均有Tn∈(QUOTE1m,m2-6m+163),則實(shí)數(shù)m的取值范圍是.

解題關(guān)鍵:先求出Sn,再由an與Sn的關(guān)系求出an,進(jìn)而求出bn及Tn,然后求出Tn的范圍,則由Tn∈(QUOTE1m,m2-6m+163)可知Tn的范圍應(yīng)是集合(QUOTE1m,m2-6m+163)的子集,比較端點(diǎn)值即可求解.解析:因?yàn)镼UOTE1S1+QUOTE1S2+…+QUOTE1Sn=nn+1,①所以當(dāng)n≥2時(shí),QUOTE1S1+QUOTE1S2+…+1Sn-1=n所以①-②得QUOTE1Sn=nn+1-n-1所以Sn=n(n+1)(n≥2).當(dāng)n=1時(shí),QUOTE1S1=11+1=QUOTE12,所以a1=2,符合Sn=n(n+1)(n≥2),所以Sn=n(n+1)