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1、6.4數(shù)列的綜合應(yīng)用挖命題【考情探究】考點(diǎn)內(nèi)容解讀5年考情預(yù)測(cè)熱度考題示例考向關(guān)聯(lián)考點(diǎn)1.數(shù)列求和掌握非等差、等比數(shù)列求和的幾種常見方法2017天津,182015天津,182015天津文,18錯(cuò)位相減法求數(shù)列的和等比數(shù)列及前n項(xiàng)和2016天津文,18分組轉(zhuǎn)化法求數(shù)列的和2.數(shù)列的綜合應(yīng)用1.能在具體的問題情境中識(shí)別數(shù)列的等差關(guān)系或等比關(guān)系,抽象出數(shù)列的模型,并能用有關(guān)知識(shí)解決相應(yīng)的問題2.能處理數(shù)列與函數(shù),數(shù)列與不等式等綜合問題2018天津,18裂項(xiàng)相消法求數(shù)列的和等差、等比數(shù)列的通項(xiàng)公式2014天津,19等比數(shù)列求和不等式證明2013天津文,18等差數(shù)列、等比數(shù)列與函數(shù)數(shù)列的基本性質(zhì)分析解讀
2、綜合運(yùn)用數(shù)列,特別是等差數(shù)列、等比數(shù)列的有關(guān)知識(shí),解答數(shù)列綜合問題和實(shí)際問題,培養(yǎng)學(xué)生的理解能力、數(shù)學(xué)建模能力和運(yùn)算能力.數(shù)列的應(yīng)用主要從以下幾個(gè)方面考查:1.會(huì)用公式法、倒序相加法、錯(cuò)位相減法、裂項(xiàng)相消法、分組轉(zhuǎn)化法求解不同類型數(shù)列的和.2.能綜合利用等差、等比數(shù)列的基本知識(shí)解決相關(guān)綜合問題.3.數(shù)列遞推關(guān)系、非等差、等比數(shù)列的求和是高考熱點(diǎn),特別是錯(cuò)位相減法和裂項(xiàng)相消法求和.破考點(diǎn)【考點(diǎn)集訓(xùn)】考點(diǎn)一數(shù)列求和1.已知數(shù)列an,bn,其中an是首項(xiàng)為3,公差為整數(shù)的等差數(shù)列,且a3a1+3,a42an(n1,nN*),給出下列命題:若數(shù)列an滿足a2a1,則anan-1(n1,nN*)成立;存
3、在常數(shù)c,使得anc(nN*)成立;若p+qm+n(其中p,q,m,nN*),則ap+aqam+an;存在常數(shù)d,使得ana1+(n-1)d(nN*)都成立.上述命題正確的是.(寫出所有正確命題的序號(hào))答案5.已知數(shù)列an的前n項(xiàng)和Sn=n2+n2,等比數(shù)列bn的前n項(xiàng)和為Tn,若b1=a1+1,b2-a2=2.(1)求數(shù)列an,bn的通項(xiàng)公式;(2)求滿足Tn+an300的最小的n值.解析(1)a1=S1=1,n1時(shí),an=Sn-Sn-1=n2+n2-(n-1)2+(n-1)2=n,又n=1時(shí),a1=1,所以an=n成立,an=n(nN*),則由題意可知b1=2,b2=4,bn的公比q=42
4、=2,bn=2n(nN*).(2)Tn=2(1-2n)1-2=2(2n-1),Tn+an=2(2n-1)+n,Tn+an隨n的增大而增大,又T7+a7=2127+7=261300,所求最小的n值為8.煉技法【方法集訓(xùn)】方法1錯(cuò)位相減法求和1.已知數(shù)列an的前n項(xiàng)和為Sn,a1=5,nSn+1-(n+1)Sn=n2+n.(1)求證:數(shù)列Snn為等差數(shù)列;(2)令bn=2nan,求數(shù)列bn的前n項(xiàng)和Tn.解析(1)證明:由nSn+1-(n+1)Sn=n2+n得Sn+1n+1-Snn=1,又S11=5,所以數(shù)列Snn是首項(xiàng)為5,公差為1的等差數(shù)列.(2)由(1)可知Snn=5+(n-1)=n+4,所
5、以Sn=n2+4n.當(dāng)n2時(shí),an=Sn-Sn-1=n2+4n-(n-1)2-4(n-1)=2n+3.又a1=5也符合上式,所以an=2n+3(nN*),所以bn=(2n+3)2n,所以Tn=52+722+923+(2n+3)2n,2Tn=522+723+924+(2n+1)2n+(2n+3)2n+1,由-得Tn=(2n+3)2n+1-10-(23+24+2n+1)=(2n+3)2n+1-10-23(1-2n-1)1-2=(2n+3)2n+1-10-(2n+2-8)=(2n+1)2n+1-2.2.已知數(shù)列an是等比數(shù)列,a2=4,a3+2是a2和a4的等差中項(xiàng).(1)求數(shù)列an的通項(xiàng)公式;(2
6、)設(shè)bn=2log2an-1,求數(shù)列anbn的前n項(xiàng)和Tn.解析(1)設(shè)數(shù)列an的公比為q,因?yàn)閍2=4,所以a3=4q,a4=4q2.因?yàn)閍3+2是a2和a4的等差中項(xiàng),所以2(a3+2)=a2+a4,即2(4q+2)=4+4q2,化簡(jiǎn)得q2-2q=0.因?yàn)楣萹0,所以q=2.所以an=a2qn-2=42n-2=2n(nN*).(2)因?yàn)閍n=2n,所以bn=2log2an-1=2n-1,所以anbn=(2n-1)2n,則Tn=12+322+523+(2n-3)2n-1+(2n-1)2n,2Tn=122+323+524+(2n-3)2n+(2n-1)2n+1.由-得,-Tn=2+222+2
7、23+22n-(2n-1)2n+1=2+24(1-2n-1)1-2-(2n-1)2n+1=-6-(2n-3)2n+1,所以Tn=6+(2n-3)2n+1.方法2裂項(xiàng)相消法求和3.已知數(shù)列an(nN*)是公差不為0的等差數(shù)列,a1=1,且1a2,1a4,1a8成等比數(shù)列.(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)數(shù)列1anan+1的前n項(xiàng)和為Tn,求證:Tn1.解析(1)設(shè)an的公差為d.因?yàn)?a2,1a4,1a8成等比數(shù)列,所以1a42=1a21a8,即1a1+3d2=1a1+d1a1+7d,化簡(jiǎn)得(a1+3d)2=(a1+d)(a1+7d),又a1=1,且d0,解得d=1.所以an=a1+(n-1
8、)d=n.(2)證明:由(1)得1anan+1=1n(n+1)=1n-1n+1,所以Tn=1-12+12-13+1n-1n+1=1-1n+11.因此Tn0,所以q=2.所以,bn=2n.由b3=a4-2a1,可得3d-a1=8.由S11=11b4,可得a1+5d=16,聯(lián)立,解得a1=1,d=3,由此可得an=3n-2.所以,數(shù)列an的通項(xiàng)公式為an=3n-2,數(shù)列bn的通項(xiàng)公式為bn=2n.(2)設(shè)數(shù)列a2nb2n-1的前n項(xiàng)和為Tn,由a2n=6n-2,b2n-1=24n-1,有a2nb2n-1=(3n-1)4n,故Tn=24+542+843+(3n-1)4n,4Tn=242+543+84
9、4+(3n-4)4n+(3n-1)4n+1,上述兩式相減,得-3Tn=24+342+343+34n-(3n-1)4n+1=12(1-4n)1-4-4-(3n-1)4n+1=-(3n-2)4n+1-8.得Tn=3n-234n+1+83.所以,數(shù)列a2nb2n-1的前n項(xiàng)和為3n-234n+1+83.方法總結(jié)(1)等差數(shù)列與等比數(shù)列中有五個(gè)量a1,n,d(或q),an,Sn,一般可以“知三求二”,通過(guò)列方程(組)求關(guān)鍵量a1和d(或q),問題可迎刃而解.(2)數(shù)列an是公差為d的等差數(shù)列,bn是公比q1的等比數(shù)列,求數(shù)列anbn的前n項(xiàng)和適用錯(cuò)位相減法.2.(2016天津文,18,13分)已知an
10、是等比數(shù)列,前n項(xiàng)和為Sn(nN*),且1a1-1a2=2a3,S6=63.(1)求an的通項(xiàng)公式;(2)若對(duì)任意的nN*,bn是log2an和log2an+1的等差中項(xiàng),求數(shù)列(-1)nbn2的前2n項(xiàng)和.解析(1)設(shè)數(shù)列an的公比為q.由已知,有1a1-1a1q=2a1q2,解得q=2,或q=-1.又由S6=a11-q61-q=63,知q-1,所以a11-261-2=63,得a1=1.所以an=2n-1.(2)由題意,得bn=12(log2an+log2an+1)=12(log22n-1+log22n)=n-12,即bn是首項(xiàng)為12,公差為1的等差數(shù)列.設(shè)數(shù)列(-1)nbn2的前n項(xiàng)和為T
11、n,則T2n=(-b12+b22)+(-b32+b42)+(-b2n-12+b2n2)=b1+b2+b3+b4+b2n-1+b2n=2n(b1+b2n)2=2n2.評(píng)析本題主要考查等差數(shù)列、等比數(shù)列及其前n項(xiàng)和公式等基礎(chǔ)知識(shí),考查數(shù)列求和的基本方法和運(yùn)算求解能力.3.(2015天津,18,13分)已知數(shù)列an滿足an+2=qan(q為實(shí)數(shù),且q1),nN*,a1=1,a2=2,且a2+a3,a3+a4,a4+a5成等差數(shù)列.(1)求q的值和an的通項(xiàng)公式;(2)設(shè)bn=log2a2na2n-1,nN*,求數(shù)列bn的前n項(xiàng)和.解析(1)由已知,得(a3+a4)-(a2+a3)=(a4+a5)-(
12、a3+a4),即a4-a2=a5-a3,所以a2(q-1)=a3(q-1).又因?yàn)閝1,故a3=a2=2,由a3=a1q,得q=2.當(dāng)n=2k-1(kN*)時(shí),an=a2k-1=2k-1=2n-12;當(dāng)n=2k(kN*)時(shí),an=a2k=2k=2n2.所以,an的通項(xiàng)公式為an=2n-12,n為奇數(shù),2n2,n為偶數(shù).(2)由(1)得bn=log2a2na2n-1=n2n-1.設(shè)bn的前n項(xiàng)和為Sn,則Sn=1120+2121+3122+(n-1)12n-2+n12n-1,12Sn=1121+2122+3123+(n-1)12n-1+n12n,上述兩式相減,得12Sn=1+12+122+12n
13、-1-n2n=1-12n1-12-n2n=2-22n-n2n,整理得,Sn=4-n+22n-1.所以,數(shù)列bn的前n項(xiàng)和為4-n+22n-1,nN*.評(píng)析本題主要考查等比數(shù)列及其前n項(xiàng)和公式、等差中項(xiàng)等基礎(chǔ)知識(shí).考查數(shù)列求和的基本方法、分類討論思想和運(yùn)算求解能力.4.(2015天津文,18,13分)已知an是各項(xiàng)均為正數(shù)的等比數(shù)列,bn是等差數(shù)列,且a1=b1=1,b2+b3=2a3,a5-3b2=7.(1)求an和bn的通項(xiàng)公式;(2)設(shè)cn=anbn,nN*,求數(shù)列cn的前n項(xiàng)和.解析(1)設(shè)數(shù)列an的公比為q,數(shù)列bn的公差為d,由題意知q0.由已知,有2q2-3d=2,q4-3d=10
14、,消去d,整理得q4-2q2-8=0.又因?yàn)閝0,解得q=2,所以d=2.所以數(shù)列an的通項(xiàng)公式為an=2n-1,nN*;數(shù)列bn的通項(xiàng)公式為bn=2n-1,nN*.(2)由(1)有cn=(2n-1)2n-1,設(shè)cn的前n項(xiàng)和為Sn,則Sn=120+321+522+(2n-3)2n-2+(2n-1)2n-1,2Sn=121+322+523+(2n-3)2n-1+(2n-1)2n,上述兩式相減,得-Sn=1+22+23+2n-(2n-1)2n=2n+1-3-(2n-1)2n=-(2n-3)2n-3,所以,Sn=(2n-3)2n+3,nN*.評(píng)析本題主要考查等差數(shù)列、等比數(shù)列及其前n項(xiàng)和公式等基礎(chǔ)
15、知識(shí).考查數(shù)列求和的基本方法和運(yùn)算求解能力.考點(diǎn)二數(shù)列的綜合應(yīng)用1.(2018天津,18,13分)設(shè)an是等比數(shù)列,公比大于0,其前n項(xiàng)和為Sn(nN*),bn是等差數(shù)列.已知a1=1,a3=a2+2,a4=b3+b5,a5=b4+2b6.(1)求an和bn的通項(xiàng)公式;(2)設(shè)數(shù)列Sn的前n項(xiàng)和為Tn(nN*).(i)求Tn;(ii)證明k=1n(Tk+bk+2)bk(k+1)(k+2)=2n+2n+2-2(nN*).解析(1)設(shè)等比數(shù)列an的公比為q.由a1=1,a3=a2+2,可得q2-q-2=0.由q0,可得q=2,故an=2n-1.設(shè)等差數(shù)列bn的公差為d.由a4=b3+b5,可得b1
16、+3d=4.由a5=b4+2b6,可得3b1+13d=16,從而b1=1,d=1,故bn=n.所以,數(shù)列an的通項(xiàng)公式為an=2n-1,數(shù)列bn的通項(xiàng)公式為bn=n.(2)(i)由(1),有Sn=1-2n1-2=2n-1,故Tn=k=1n(2k-1)=k=1n2k-n=2(1-2n)1-2-n=2n+1-n-2.(ii)證明:因?yàn)?Tk+bk+2)bk(k+1)(k+2)=(2k+1-k-2+k+2)k(k+1)(k+2)=k2k+1(k+1)(k+2)=2k+2k+2-2k+1k+1,所以k=1n(Tk+bk+2)bk(k+1)(k+2)=233-222+244-233+2n+2n+2-2n
17、+1n+1=2n+2n+2-2.方法總結(jié)解決數(shù)列求和問題的兩種思路(1)利用轉(zhuǎn)化的思想將一般數(shù)列設(shè)法轉(zhuǎn)化為等差或等比數(shù)列,這一思想方法往往通過(guò)通項(xiàng)分解或錯(cuò)位相減來(lái)完成.(2)不能轉(zhuǎn)化為等差或等比數(shù)列的,往往通過(guò)裂項(xiàng)相消法、倒序相加法等來(lái)求和.2.(2014天津,19,14分)已知q和n均為給定的大于1的自然數(shù).設(shè)集合M=0,1,2,q-1,集合A=x|x=x1+x2q+xnqn-1,xiM,i=1,2,n.(1)當(dāng)q=2,n=3時(shí),用列舉法表示集合A;(2)設(shè)s,tA,s=a1+a2q+anqn-1,t=b1+b2q+bnqn-1,其中ai,biM,i=1,2,n.證明:若anbn,則st.解
18、析(1)當(dāng)q=2,n=3時(shí),M=0,1,A=x|x=x1+x22+x322,xiM,i=1,2,3.可得,A=0,1,2,3,4,5,6,7.(2)證明:由s,tA,s=a1+a2q+anqn-1,t=b1+b2q+bnqn-1,ai,biM,i=1,2,n及anbn,可得s-t=(a1-b1)+(a2-b2)q+(an-1-bn-1)qn-2+(an-bn)qn-1(q-1)+(q-1)q+(q-1)qn-2-qn-1=(q-1)(1-qn-1)1-q-qn-1=-10.所以s0,d0時(shí),滿足am0,am+10的項(xiàng)數(shù)m,可使得Sn取得最大值,最大值為Sm;當(dāng)a10時(shí),滿足am0,am+10的
19、項(xiàng)數(shù)m,可使得Sn取得最小值,最小值為Sm.2.(2016課標(biāo),17,12分)Sn為等差數(shù)列an的前n項(xiàng)和,且a1=1,S7=28.記bn=lgan,其中x表示不超過(guò)x的最大整數(shù),如0.9=0,lg99=1.(1)求b1,b11,b101;(2)求數(shù)列bn的前1000項(xiàng)和.解析(1)設(shè)an的公差為d,據(jù)已知有7+21d=28,解得d=1.所以an的通項(xiàng)公式為an=n.b1=lg1=0,b11=lg11=1,b101=lg101=2.(2)因?yàn)閎n=0,1n10,1,10n100,2,100n0,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)
20、和.解析(1)由an2+2an=4Sn+3,可知an+12+2an+1=4Sn+1+3.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an0,可得an+1-an=2.又a12+2a1=4a1+3,解得a1=-1(舍去)或a1=3.所以an是首項(xiàng)為3,公差為2的等差數(shù)列,故通項(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
21、+3=n3(2n+3).4.(2014四川,19,12分)設(shè)等差數(shù)列an的公差為d,點(diǎn)(an,bn)在函數(shù)f(x)=2x的圖象上(nN*).(1)若a1=-2,點(diǎn)(a8,4b7)在函數(shù)f(x)的圖象上,求數(shù)列an的前n項(xiàng)和Sn;(2)若a1=1,函數(shù)f(x)的圖象在點(diǎn)(a2,b2)處的切線在x軸上的截距為2-1ln2,求數(shù)列anbn的前n項(xiàng)和Tn.解析(1)由已知得,b7=2a7,b8=2a8=4b7,有2a8=42a7=2a7+2.解得d=a8-a7=2.所以,Sn=na1+n(n-1)2d=-2n+n(n-1)=n2-3n.(2)函數(shù)f(x)=2x在(a2,b2)處的切線方程為y-2a2=
22、(2a2ln2)(x-a2),它在x軸上的截距為a2-1ln2.由題意得,a2-1ln2=2-1ln2,解得a2=2.所以d=a2-a1=1.從而an=n,bn=2n.所以Tn=12+222+323+n-12n-1+n2n,2Tn=11+22+322+n2n-1.因此,2Tn-Tn=1+12+122+12n-1-n2n=2-12n-1-n2n=2n+1-n-22n.所以,Tn=2n+1-n-22n.評(píng)析本題考查等差數(shù)列與等比數(shù)列的概念、等差數(shù)列與等比數(shù)列通項(xiàng)公式與前n項(xiàng)和、導(dǎo)數(shù)的幾何意義等基礎(chǔ)知識(shí),考查運(yùn)算求解能力.考點(diǎn)二數(shù)列的綜合應(yīng)用1.(2018浙江,20,15分)已知等比數(shù)列an的公比q
23、1,且a3+a4+a5=28,a4+2是a3,a5的等差中項(xiàng).數(shù)列bn滿足b1=1,數(shù)列(bn+1-bn)an的前n項(xiàng)和為2n2+n.(1)求q的值;(2)求數(shù)列bn的通項(xiàng)公式.解析(1)由a4+2是a3,a5的等差中項(xiàng)得a3+a5=2a4+4,所以a3+a4+a5=3a4+4=28,解得a4=8.由a3+a5=20得8q+1q=20,解得q=2或q=12,因?yàn)閝1,所以q=2.(2)設(shè)cn=(bn+1-bn)an,數(shù)列cn的前n項(xiàng)和為Sn.由cn=S1,n=1,Sn-Sn-1,n2,解得cn=4n-1.由(1)可知an=2n-1,所以bn+1-bn=(4n-1)12n-1,故bn-bn-1=
24、(4n-5)12n-2,n2,bn-b1=(bn-bn-1)+(bn-1-bn-2)+(b3-b2)+(b2-b1)=(4n-5)12n-2+(4n-9)12n-3+712+3.設(shè)Tn=3+712+11122+(4n-5)12n-2,n2,12Tn=312+7122+(4n-9)12n-2+(4n-5)12n-1,所以12Tn=3+412+4122+412n-2-(4n-5)12n-1,因此Tn=14-(4n+3)12n-2,n2,又b1=1,所以bn=15-(4n+3)12n-2.易錯(cuò)警示利用錯(cuò)位相減法求和時(shí),要注意以下幾點(diǎn):(1)錯(cuò)位相減法求和,只適合于數(shù)列anbn,其中an為等差數(shù)列,b
25、n為等比數(shù)列.(2)在等式兩邊所乘的數(shù)是等比數(shù)列bn的公比.(3)兩式相減時(shí),一定要錯(cuò)開一位.(4)特別要注意相減后等比數(shù)列的次數(shù).(5)進(jìn)行檢驗(yàn).2.(2018江蘇,20,16分)設(shè)an是首項(xiàng)為a1,公差為d的等差數(shù)列,bn是首項(xiàng)為b1,公比為q的等比數(shù)列.(1)設(shè)a1=0,b1=1,q=2,若|an-bn|b1對(duì)n=1,2,3,4均成立,求d的取值范圍;(2)若a1=b10,mN*,q(1,m2,證明:存在dR,使得|an-bn|b1對(duì)n=2,3,m+1均成立,并求d的取值范圍(用b1,m,q表示).解析(1)由條件知an=(n-1)d,bn=2n-1.因?yàn)閨an-bn|b1對(duì)n=1,2,
26、3,4均成立,即11,1d3,32d5,73d9,得73d52.因此,d的取值范圍為73,52.(2)由條件知:an=b1+(n-1)d,bn=b1qn-1.若存在dR,使得|an-bn|b1(n=2,3,m+1)均成立,即|b1+(n-1)d-b1qn-1|b1(n=2,3,m+1).即當(dāng)n=2,3,m+1時(shí),d滿足qn-1-2n-1b1dqn-1n-1b1.因?yàn)閝(1,m2,所以10,對(duì)n=2,3,m+1均成立.因此,取d=0時(shí),|an-bn|b1對(duì)n=2,3,m+1均成立.下面討論數(shù)列qn-1-2n-1的最大值和數(shù)列qn-1n-1的最小值(n=2,3,m+1).當(dāng)2nm時(shí),qn-2n-q
27、n-1-2n-1=nqn-qn-nqn-1+2n(n-1)=n(qn-qn-1)-qn+2n(n-1),當(dāng)10.因此,當(dāng)2nm+1時(shí),數(shù)列qn-1-2n-1單調(diào)遞增,故數(shù)列qn-1-2n-1的最大值為qm-2m.設(shè)f(x)=2x(1-x),當(dāng)x0時(shí),f(x)=(ln2-1-xln2)2x0.所以f(x)單調(diào)遞減,從而f(x)f(0)=1.當(dāng)2nm時(shí),qnnqn-1n-1=q(n-1)n21n1-1n=f1n1.因此,當(dāng)2nm+1時(shí),數(shù)列qn-1n-1單調(diào)遞減,故數(shù)列qn-1n-1的最小值為qmm.因此,d的取值范圍為b1(qm-2)m,b1qmm.疑難突破本題是數(shù)列的綜合題,考查等差數(shù)列、等比
28、數(shù)列的概念和相關(guān)性質(zhì),第(1)問主要考查絕對(duì)值不等式.第(2)問要求d的范圍,使得|an-bn|b1對(duì)n=2,3,m+1都成立,首先把d分離出來(lái),變成qn-1-2n-1b1dqn-1n-1b1,難點(diǎn)在于討論qn-1-2n-1b1的最大值和qn-1n-1b1的最小值.對(duì)于數(shù)列qn-1-2n-1,可以通過(guò)作差討論其單調(diào)性,而對(duì)于數(shù)列qn-1n-1,要作商討論單調(diào)性,因?yàn)閝nnqn-1n-1=q(n-1)n=q1-1n,當(dāng)2nm時(shí),1qn2.所以q1-1n21n1-1n,可以構(gòu)造函數(shù)f(x)=2x(1-x),通過(guò)討論f(x)在(0,+)上的單調(diào)性去證明f1n60n+800?若存在,求n的最小值;若不
29、存在,說(shuō)明理由.解析(1)設(shè)數(shù)列an的公差為d,依題意,2,2+d,2+4d成等比數(shù)列,故有(2+d)2=2(2+4d),化簡(jiǎn)得d2-4d=0,解得d=0或d=4.當(dāng)d=0時(shí),an=2;當(dāng)d=4時(shí),an=2+(n-1)4=4n-2,從而得數(shù)列an的通項(xiàng)公式為an=2或an=4n-2.(2)當(dāng)an=2時(shí),Sn=2n.顯然2n60n+800成立.當(dāng)an=4n-2時(shí),Sn=n2+(4n-2)2=2n2.令2n260n+800,即n2-30n-4000,解得n40或n60n+800成立,n的最小值為41.綜上,當(dāng)an=2時(shí),不存在滿足題意的n;當(dāng)an=4n-2時(shí),存在滿足題意的n,其最小值為41.評(píng)析
30、本題考查了數(shù)列的通項(xiàng)公式和求和公式,考查了分類討論的方法.C組教師專用題組1.(2013課標(biāo),12,5分)設(shè)AnBnCn的三邊長(zhǎng)分別為an,bn,cn,AnBnCn的面積為Sn,n=1,2,3,.若b1c1,b1+c1=2a1,an+1=an,bn+1=cn+an2,cn+1=bn+an2,則()A.Sn為遞減數(shù)列B.Sn為遞增數(shù)列C.S2n-1為遞增數(shù)列,S2n為遞減數(shù)列D.S2n-1為遞減數(shù)列,S2n為遞增數(shù)列答案B2.(2013重慶,12,5分)已知an是等差數(shù)列,a1=1,公差d0,Sn為其前n項(xiàng)和,若a1,a2,a5成等比數(shù)列,則S8=.答案643.(2012課標(biāo),16,5分)數(shù)列a
31、n滿足an+1+(-1)nan=2n-1,則an的前60項(xiàng)和為.答案18304.(2015浙江,20,15分)已知數(shù)列an滿足a1=12且an+1=an-an2(nN*).(1)證明:1anan+12(nN*);(2)設(shè)數(shù)列an2的前n項(xiàng)和為Sn,證明:12(n+2)Snn12(n+1)(nN*).證明(1)由題意得an+1-an=-an20,即an+1an,故an12.由an=(1-an-1)an-1得an=(1-an-1)(1-an-2)(1-a1)a10.由00,于是(a2n+1-a2n)+(a2n-a2n-1)0.但122n122n-1,所以|a2n+1-a2n|0,因此a2n-a2n
32、-1=122n-1=(-1)2n22n-1.因?yàn)閍2n是遞減數(shù)列,同理可得,a2n+1-a2n0,dS40B.a1d0,dS40,dS40D.a1d0答案B4.(2017天津耀華中學(xué)第二次月考,6)已知等差數(shù)列an的前n項(xiàng)和為Sn,且S21=42,若記bn=2a112-a9-a13,則數(shù)列bn()A.是等差數(shù)列但不是等比數(shù)列B.是等比數(shù)列但不是等差數(shù)列C.既是等差數(shù)列又是等比數(shù)列D.既不是等差數(shù)列又不是等比數(shù)列答案C5.(2019屆天津七校聯(lián)考,7)已知數(shù)列an的各項(xiàng)均為正數(shù),a1=1,an+1+an=1an+1-an,則數(shù)列1an+1+an的前15項(xiàng)和為()A.3B.4C.127D.128答
33、案A二、填空題(每小題5分,共10分)6.(2017天津南開中學(xué)第五次月考,12)設(shè)Sn是公差不為0的等差數(shù)列an的前n項(xiàng)和,且S1,S2,S4成等比數(shù)列,則a2a1=.答案37.(2017天津南開中學(xué)模擬,11)已知數(shù)列1,a1,a2,9是等差數(shù)列,數(shù)列1,b1,b2,b3,9是等比數(shù)列,則b2a1+a2的值為.答案310三、解答題(共75分)8.(2019屆天津南開中學(xué)第一次月考,18)設(shè)數(shù)列an的前n項(xiàng)和Sn=2n2,bn為等比數(shù)列,且a1=b1,b2(a2-a1)=b1.(1)求數(shù)列an和bn的通項(xiàng)公式;(2)設(shè)cn=anbn,求數(shù)列cn的前n項(xiàng)和Tn.解析(1)當(dāng)n2時(shí),an=Sn-
34、Sn-1=2n2-2(n-1)2=4n-2,當(dāng)n=1時(shí),a1=S1=2滿足上式,故an的通項(xiàng)公式為an=4n-2.設(shè)bn的公比為q,由a1=b1,b2(a2-a1)=b1知,b1=2,b2=12,q=14,bn=b1qn-1=214n-1,即bn=24n-1.(2)cn=anbn=4n-224n-1=(2n-1)4n-1,Tn=c1+c2+cn=1+341+542+(2n-1)4n-1,4Tn=14+342+543+(2n-3)4n-1+(2n-1)4n.兩式相減得3Tn=-1-2(41+42+43+4n-1)+(2n-1)4n=13(6n-5)4n+5.Tn=19(6n-5)4n+5.解題分
35、析本題主要考查等差數(shù)列和等比數(shù)列的相關(guān)知識(shí).(1)已知數(shù)列an的前n項(xiàng)和Sn,則通項(xiàng)為an=Sn-Sn-1(n2),a1=S1=2=b1,再根據(jù)b2(a2-a1)=b1可求得數(shù)列bn的通項(xiàng)公式.(2)根據(jù)(1)可知數(shù)列an和bn的通項(xiàng)公式,進(jìn)而可得數(shù)列cn的通項(xiàng)公式,再運(yùn)用錯(cuò)位相減法求其前n項(xiàng)和.9.(2019屆天津南開中學(xué)統(tǒng)練(2),18)已知Sn是等比數(shù)列an的前n項(xiàng)和,a10,S4,S2,S3成等差數(shù)列,16是a2和a8的等比中項(xiàng).(1)求an的通項(xiàng)公式;(2)若等差數(shù)列bn中,b1=1,前9項(xiàng)和為27,令cn=2anbn,求數(shù)列cn的前n項(xiàng)和Tn.解析(1)設(shè)an的公比為q,由S4,S
36、2,S3成等差數(shù)列可得2S2=S3+S4,即S3-S2+S4-S2=0,所以2a3+a4=0,所以q=-2,又16是a2和a8的等比中項(xiàng),所以a2a8=162,所以a1qa1q7=162,所以a1=1(a10故舍去-1),所以an=a1qn-1=(-2)n-1.(2)設(shè)bn的公差為d,因?yàn)閎1=1,S9=9b5=27,所以b5=1+4d=3,所以d=12,bn=n+12,cn=2anbn=(n+1)(-2)n-1,Tn=2(-2)0+3(-2)+4(-2)2+(n+1)(-2)n-1,-2Tn=2(-2)+3(-2)2+n(-2)n-1+(n+1)(-2)n,-得,3Tn=2+(-2)+(-2
37、)2+(-2)n-1-(n+1)(-2)n=2+-21-(-2)n-11-(-2)-(n+1)(-2)n=2+-2-(-2)n3-(n+1)(-2)n=43-n+43(-2)n,所以Tn=49-3n+49(-2)n.10.(2019屆天津耀華中學(xué)第二次月考,18)等差數(shù)列an中,a1=3,前n項(xiàng)和為Sn,數(shù)列bn是首項(xiàng)為1,公比為q(q1),且各項(xiàng)均為正數(shù)的等比數(shù)列,已知b2+S2=12,q=S2b2.(1)求an與bn;(2)證明:131S1+1S2+1Sn23.解析(1)設(shè)an的公差為d,依題意得q+6+d=12,q=6+dq,解得q=3或q=-4.因?yàn)閿?shù)列bn各項(xiàng)均為正數(shù),所以q=3,所
38、以d=3,所以an=3+3(n-1)=3n,bn=3n-1.(2)證明:因?yàn)镾n=3n(n+1)2,所以1Sn=23n(n+1)=231n-1n+1,所以1S1+1S2+1Sn=231-12+12-13+1n-1n+1=231-1n+123.又231-1n+1單調(diào)遞增,且n為正整數(shù),所以當(dāng)n=1時(shí),231-1n+1有最小值13.所以不等式成立.11.(2019屆天津耀華中學(xué)統(tǒng)練(2),20)已知數(shù)列an,a1=1,a1+2a2+3a3+nan=n+12an+1(nN*).(1)求數(shù)列an的通項(xiàng)公式;(2)求數(shù)列n2an的前n項(xiàng)和Tn;(3)若存在nN*,使得an(n+1)成立,求實(shí)數(shù)的最小值.
39、解析(1)當(dāng)n2時(shí),由題意得a1+2a2+3a3+(n-1)an-1=n2an,a1+2a2+3a3+nan=n+12an+1,-得nan=n+12an+1-n2an,3nan=(n+1)an+1(n2),即(n+1)an+1nan=3(n2),由已知得a2=a1=1,數(shù)列nan從第2項(xiàng)起,是以2為第2項(xiàng),3為公比的等比數(shù)列,n2時(shí),nan=23n-2,an=23n-2n(n2),綜上,an=1,n=123n-2n,n2.(2)由(1)知,n=1時(shí),T1=1,n2時(shí),n2an=2n3n-2.因此,當(dāng)n2時(shí),Tn=1+430+631+2n3n-2,3Tn=13+431+632+2(n-1)3n-
40、2+2n3n-1.-2Tn=1+30+231+232+23n-2-2n3n-1=2+23(1-3n-2)1-3-2n3n-1=-1-(2n-1)3n-1,Tn=12+n-123n-1,n2,當(dāng)n=1時(shí),滿足上式,Tn=12+n-123n-1,nN*.(3)an(n+1),當(dāng)n=1時(shí),12;當(dāng)n2且nN*時(shí),ann+1=23n-2n(n+1),設(shè)f(n)=n(n+1)23n-2(n2且nN*),則f(n+1)-f(n)=2(1-n)(n+1)23n-11f(n)(n2且nN*),1f(n)的最小值為1f(2)=13,即13.綜上,實(shí)數(shù)的最小值為13.12.(2018天津南開二模,18)已知數(shù)列a
41、n的前n項(xiàng)和為Sn,a1=12,且當(dāng)n2時(shí),1Sn=1Sn-1+2.(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)bn=2(1-n)an,證明:b22+b32+b42+bn223.解析(1)當(dāng)n2時(shí),1Sn=1Sn-1+2,1Sn-1Sn-1=2.又1a1=2,數(shù)列1Sn是公差為2,首項(xiàng)為2的等差數(shù)列.1Sn=2+2(n-1)=2n,Sn=12n.n2時(shí),an=Sn-Sn-1=12n-12(n-1)=-12n(n-1).an=12,n=1,-12n(n-1),n2.(2)證明:n2時(shí),bn=2(1-n)an=1n,n3時(shí),bn2=1n21n2-1=121n-1-1n+1,b22+b32+b42+bn21
42、4+1212-14+13-15+14-16+1n-2-1n+1n-1-1n+1=14+1212+13-1n-1n+1=23-121n+1n+123.13.(2018天津南開中學(xué)第五次月考,18)已知數(shù)列an滿足a1=2,an+1=2(Sn+n+1)(nN*),令bn=an+1.(1)求證:bn是等比數(shù)列;(2)求數(shù)列nbn的前n項(xiàng)和Tn;(3)求證:12-123n1a1+1a2+1a3+1an13n,1a1+1a2+1an13+132+13n=131-13n1-13=12-1213n.又由1ak=13k-1=3k+1-1(3k-1)(3k+1-1)3k+1(3k-1)(3k+1-1)=3213k-1-13k+1-1,可得1a1+1a2+1an12+32132-1-133-1+133-1-134-1+13n-1-13n+1-1=12+3218-13n+1-1=1116-3213n+1-11116.綜上可得,12-123n1a1+1a2+1a3+1an1116.
天津2020版高考數(shù)學(xué)復(fù)習(xí)6.4數(shù)列的綜合應(yīng)用精練.docx
