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1、(江蘇專(zhuān)用)2013年高考數(shù)學(xué)總復(fù)習(xí) 第五章第4課時(shí) 數(shù)列通項(xiàng)與求和 課時(shí)闖關(guān)(含解析)A級(jí)雙基鞏固一、填空題1數(shù)列1,的前n項(xiàng)和為_(kāi)解析:2(),12(1)()().答案:2數(shù)列1,2,3,4,的前n項(xiàng)和為_(kāi)解析:ann,Sn12n(123n)(),Snn(n1)1(n2n2).答案:(n2n2)3已知數(shù)列an的前n項(xiàng)和Sn159131721(1)n1(4n3). 則S15S22S31的值為_(kāi)解析:S15(4)7a15285729,S2211(4)44,S3115(4)a316012161,S15S22S3129446176.答案:764數(shù)列an滿(mǎn)足anan1(nN*),且a11,Sn是數(shù)列
2、an的前n項(xiàng)和,則S21_.解析:依題意得anan1an1an2,則an2an,即數(shù)列an中的奇數(shù)項(xiàng)、偶數(shù)項(xiàng)分別相等,則a21a11.S21(a1a2)(a3a4)(a19a20)a2110(a1a2)a211016.答案:65(2012鎮(zhèn)江質(zhì)檢)已知數(shù)列an的前n項(xiàng)和為Sn,且a11,an1Sn(nN*),則Sn_.解析:由已知得an1SnanSn1,且n2時(shí)有:anSn1,兩式聯(lián)立得Sn2Sn1.Sn是以2為公比,S11為首項(xiàng)的等比數(shù)列,Sn2n1.答案:2n16已知數(shù)列an的通項(xiàng)公式ann2cos n,Sn為它的前n項(xiàng)和,則_.解析:注意到cosn(1)n(nN*),an(1)nn2,因
3、此S2010(1222)(3242)(2009220102)12342009201010052011.1005.答案:10057數(shù)列an滿(mǎn)足:an1an(1an1),a11,數(shù)列bn滿(mǎn)足:bnanan1,則數(shù)列bn的前10項(xiàng)和S10_.解析:由題可知an1an(1an1),整理可得1,則1(n1)n,所以an,bnanan1,故S10b1b2b101.答案:8已知數(shù)列an中,Sn是其前n項(xiàng)和,若a11,a22,anan1an2anan1an2,且an1an21,則a1a2a3_,S2010_.解析:由12a312a3,得a33,a1a2a36.繼續(xù)依據(jù)遞推關(guān)系得到a41,a52,a63,故該數(shù)
4、列是周期為3的數(shù)列,S201064020.答案:64020二、解答題9在等差數(shù)列an中,a16a17a18a936,其前n項(xiàng)的和為Sn.(1)求Sn的最小值,并求出Sn取最小值時(shí)n的值;(2)求Tn|a1|a2|an|.解:a16a17a183a1736a1712.又a936,公差d3.首項(xiàng)a1a98d60,an3n63.(1)法一:設(shè)前n項(xiàng)的和Sn最小,則即n20或21.這表明:當(dāng)n20或21時(shí),Sn取最小值,最小值為S20S21630.法二:Sn60n3(n241n)(n)2.nN*,當(dāng)n20或21時(shí),Sn取最小值(2024120)630.(2)由an3n630n21,當(dāng)n21時(shí),TnSn
5、(41nn2);當(dāng)n21時(shí),Tna1a2a21a22anSn2S21(n241n)1260.綜上可知,Tn.10已知數(shù)列anbn滿(mǎn)足a12,b11且anan1bn11(n2,nN*),bnan1bn11(n2,nN*)(1)令cnanbn,求數(shù)列cn的通項(xiàng)公式;(2)求數(shù)列an的通項(xiàng)公式及前n項(xiàng)和解:(1)已知兩式相加得,anbnan1bn12,即cncn12,故cn為等差數(shù)列,其公差為2,首項(xiàng)為2n1.(2)已知兩式相減得,anbn(an1bn1),設(shè)dnanbn,dndn1,故dn成等比數(shù)列,其公比為,首項(xiàng)為1,故dn.于是有兩式相加得2an2n1.ann.設(shè)數(shù)列an前n項(xiàng)和為Sn,則Sn
6、()(123n)1n.B級(jí)能力提升一、填空題1(2012南京調(diào)研)在數(shù)列an中,對(duì)任意自然數(shù)nN*,a1a2a3an2n1,則aaa_.解析:設(shè)Sna1a2an2n1,anSnSn1(2n1)(2n11)2n1(n2)當(dāng)n1時(shí),a12111滿(mǎn)足上式an2n1,a4n1,aaa14424n1(4n1)答案:(4n1)2將奇數(shù)數(shù)列如下分組:1,(3,5),(7,9,11),(13,15,17,19),使得第n組中含有n個(gè)數(shù),那么第n組中的n個(gè)奇數(shù)的和為_(kāi)解析:113,35823,79112733,131517196443.故猜想第n組和為n3, 另外,第n組數(shù)共n個(gè)數(shù),首項(xiàng)為n2n1,公差為2,所
7、以第n組各數(shù)之和為n3.答案:n33(2012徐州質(zhì)檢)在數(shù)列an中,若對(duì)任意的n均有anan1an2為定值(nN*),且a72,a93,a984,則此數(shù)列an的前100項(xiàng)的和S100_.解析:由題設(shè)得anan1an2an1an2an3,anan3,a3k12(kN),a3k24(kN),a3k3(kN*),S100342334333299.答案:2994點(diǎn)P在直徑為2的球面上,過(guò)P作兩兩垂直的三條弦,若這三條弦長(zhǎng)成等差數(shù)列,則這三條弦長(zhǎng)和的最大值是_解析:由題意,設(shè)三條弦長(zhǎng)分別為ad,a,ad,三條弦長(zhǎng)之和為l3a,由球的幾何特征及三弦兩兩垂直可知,只有在四點(diǎn)共球且構(gòu)成的對(duì)角線為直徑時(shí)l最大
8、,即(ad)2a2(ad)24,3a22d24時(shí),a .l2.答案:2二、解答題5數(shù)列an滿(mǎn)足a11,a22,an2(1cos2)ansin2,nN*.(1)求a3,a4并求數(shù)列an的通項(xiàng);(2)設(shè)bn,Snb1b2bn,求Sn.解:(1)a11,a22,a3(1cos2)a1sin2a112,a4(1cos2)a2sin22a24.一般地當(dāng)n2k1(kN*)時(shí),a2k11cos2a2k1sin2a2k11,即a2k1a2k11,a2k1是首項(xiàng)為1,公差為1的等差數(shù)列因此a2k1k.當(dāng)n2k(kN*)時(shí),a2k2(1cos2)a2ksin22a2k,因此a2k是首項(xiàng)為2,公比為2的等比數(shù)列,a
9、2k2k.故數(shù)列an的通項(xiàng)公式為an.(2)由(1)知bn,Sn,故Sn,得Sn1.Sn22.6已知數(shù)列an的前n項(xiàng)和為Sn,且Snn2n.數(shù)列bn滿(mǎn)足bn22bn1bn0(nN*),且b311,b1b2b9153.(1)求數(shù)列an,bn的通項(xiàng)公式;(2)設(shè)cn,數(shù)列cn的前n項(xiàng)和為T(mén)n,求使不等式Tn對(duì)一切nN*都成立的最大正整數(shù)k的值;(3)設(shè)f(n),是否存在mN*,使得f(m15)5f(m)成立?若存在,求出m的值;若不存在,請(qǐng)說(shuō)明理由解:(1)當(dāng)n1時(shí),a1S16,當(dāng)n2時(shí),anSnSn1(n2n)(n1)2(n1)n5,而當(dāng)n1時(shí),n56,ann5(nN*),又bn22bn1bn0,即bn2bn1bn1bn.數(shù)列bn是等差數(shù)列,又b311,b1b2b9153,解得b15,公差d3.bn3n2(nN*)(2)由(1)可得,cn(),Tnc1c2cn(1)()(),Tn1Tn0,Tn單調(diào)遞增,故TnminT1,令,得k19,所以kmax18.(3)f(n),當(dāng)m為奇數(shù)時(shí),m15為偶數(shù),3m475m25,m11.當(dāng)m為偶數(shù)時(shí),m15為奇數(shù),m2015m10,mN*(舍去)綜上所述,存在惟一正整數(shù)m11,使得f(m15)5f(m)成立