2020高考數(shù)學(xué)刷題首選卷 第四章 數(shù)列 考點(diǎn)測試31 數(shù)列求和 文(含解析).docx
《2020高考數(shù)學(xué)刷題首選卷 第四章 數(shù)列 考點(diǎn)測試31 數(shù)列求和 文(含解析).docx》由會(huì)員分享,可在線閱讀,更多相關(guān)《2020高考數(shù)學(xué)刷題首選卷 第四章 數(shù)列 考點(diǎn)測試31 數(shù)列求和 文(含解析).docx(11頁珍藏版)》請(qǐng)?jiān)谘b配圖網(wǎng)上搜索。
考點(diǎn)測試31數(shù)列求和一、基礎(chǔ)小題1若數(shù)列an的通項(xiàng)公式為an2n2n1,則數(shù)列an的前n項(xiàng)和為()A2nn21 B2n1n21C2n1n22 D2nn2答案C解析Sn2n12n2故選C2數(shù)列an的前n項(xiàng)和為Sn,若an,則S5等于()A1 B C D答案B解析an,S51故選B3等差數(shù)列an的前n項(xiàng)和為Sn,若S410a1,則()A B1 C D2答案B解析由S410a1得10a1,即da1所以1故選B4已知數(shù)列an滿足a1a2a3an2a2,則()Aa10 Ca1a2 Da20答案D解析a1a2a3an2a2,當(dāng)n1時(shí),a12a2,當(dāng)n2時(shí),a1a22a2,a20故選D5設(shè)數(shù)列an的前n項(xiàng)和為Sn,且Sn,若a38,則a1()A B C64 D128答案B解析S3S2a3,8,a1,故選B6已知數(shù)列an的前n項(xiàng)和為Sn,a11,當(dāng)n2時(shí),an2Sn1n,則S11()A5 B6 C7 D8答案B解析由當(dāng)n2時(shí),an2Sn1n得an12Snn1,上面兩式相減得an1an2an1,即an1an1,所以S11a1(a2a3)(a4a5)(a10a11)5116故選B7設(shè)Sn1234(1)n1n,則S4mS2m1S2m3(mN*)的值為()A0 B3C4 D隨m的變化而變化答案B解析容易求得S2kk,S2k1k1,所以S4mS2m1S2m32mm1m23故選B8在等比數(shù)列an中,前7項(xiàng)的和S716,且aaa128,則a1a2a3a4a5a6a7()A8 B C6 D答案A解析設(shè)數(shù)列an的公比為q,則a1a2a3a4a5a6a7,a1a2a3a4a5a6a716,aaa128,a1a2a3a4a5a6a78故選A二、高考小題9(2017全國卷)幾位大學(xué)生響應(yīng)國家的創(chuàng)業(yè)號(hào)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(dòng)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:N100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()A440 B330 C220 D110答案A解析設(shè)首項(xiàng)為第1組,接下來的兩項(xiàng)為第2組,再接下來的三項(xiàng)為第3組,依此類推,則第n組的項(xiàng)數(shù)為n,前n組的項(xiàng)數(shù)和為由題意知,N100,令100,解得n14且nN*,即N出現(xiàn)在第13組之后第n組的各項(xiàng)和為2n1,前n組所有項(xiàng)的和為n2n12n設(shè)N是第n1組的第k項(xiàng),若要使前N項(xiàng)和為2的整數(shù)冪,則N項(xiàng)的和即第n1組的前k項(xiàng)的和2k1應(yīng)與2n互為相反數(shù),即2k12n(kN*,n14),klog2(n3),n最小為29,此時(shí)k5則N5440故選A10(2016北京高考)已知an為等差數(shù)列,Sn為其前n項(xiàng)和若a16,a3a50,則S6_答案6解析設(shè)等差數(shù)列an的公差為d,a16,a3a50,62d64d0,d2,S666(2)611(2017全國卷)等差數(shù)列an的前n項(xiàng)和為Sn,a33,S410,則_答案解析設(shè)公差為d,則ann前n項(xiàng)和Sn12n,2,2121212(2015全國卷)設(shè)Sn是數(shù)列an的前n項(xiàng)和,且a11,an1SnSn1,則Sn_答案解析an1Sn1Sn,Sn1SnSnSn1,又由a11,知Sn0,1,是等差數(shù)列,且公差為1,而1,1(n1)(1)n,Sn13(2018江蘇高考)已知集合Ax|x2n1,nN*,Bx|x2n,nN*將AB的所有元素從小到大依次排列構(gòu)成一個(gè)數(shù)列an記Sn為數(shù)列an的前n項(xiàng)和,則使得Sn12an1成立的n的最小值為_答案27解析設(shè)An2n1,Bn2n,nN*,當(dāng)AkBlAk1(k,lN*)時(shí),2k12l2k1,有k2l1k,則k2l1,設(shè)TlA1A2A2l1B1B2Bl,則共有kl2l1l個(gè)數(shù),即TlS2l1l,而A1A2A2l12l122l2,B1B2Bl2l12則Tl22l22l12,則l,Tl,n,an1的對(duì)應(yīng)關(guān)系為lTlnan112an1132336210456033079108494121720453182133396611503865780觀察到l5時(shí),TlS2112a39,則n22,38),nN*時(shí),存在n,使Sn12an1,此時(shí)T5A1A2A16B1B2B3B4B5,則當(dāng)n22,38),nN*時(shí),SnT5n210n87an1An15An4,12an1122(n4)124n108,Sn12an1n234n195(n17)294,則n27時(shí),Sn12an10,即nmin27三、模擬小題14(2018福建廈門第一學(xué)期期末)已知數(shù)列an滿足an1(1)n1an2,則其前100項(xiàng)和為()A250 B200 C150 D100答案D解析n2k(kN*)時(shí),a2k1a2k2,n2k1(kN*)時(shí),a2ka2k12,n2k1(kN*)時(shí),a2k2a2k12,a2k1a2k14,a2k2a2k0,an的前100項(xiàng)和(a1a3)(a97a99)(a2a4)(a98a100)254250100故選D15(2018浙江模擬)已知數(shù)列an的通項(xiàng)公式為an則數(shù)列3ann7的前2n項(xiàng)和的最小值為()A B C D答案D解析設(shè)bn3ann7,3ann7的前2n項(xiàng)和為S2n,則S2nb1b2b3b2n3(1232n)14n91n2n213n,又2n213n2n2,當(dāng)n4時(shí),f(n)2n2是關(guān)于n的增函數(shù),又g(n)91n也是關(guān)于n的增函數(shù),S8S10S12,S8,S6,S4,S2,S6S8S41024的最小n的值為_答案9解析當(dāng)n1時(shí),a14,當(dāng)n2時(shí),anSnSn12n12n2n,所以an所以bn所以Tn當(dāng)n9時(shí),T9210910211161024;當(dāng)n8時(shí),T8298925861024的最小n的值為917(2018江西南昌蓮塘一中質(zhì)檢)函數(shù)f(x),g(x)f(x1)1,angggg,nN*,則數(shù)列an的通項(xiàng)公式為_答案an2n1解析由題意知f(x)的定義域?yàn)镽,又f(x)f(x),函數(shù)f(x)為奇函數(shù),g(x)g(2x)f(x1)1f(2x1)1f(x1)f(1x)2,由f(x)為奇函數(shù),知f(x1)f(1x)0,g(x)g(2x)2angggg,nN*,angggg,nN*,由得2angggggg(2n1)2,則數(shù)列an的通項(xiàng)公式為an2n118(2018洛陽質(zhì)檢)已知正項(xiàng)數(shù)列an滿足a6aan1an若a12,則數(shù)列an的前n項(xiàng)和Sn為_答案3n1解析a6aan1an,(an13an)(an12an)0,an0,an13an,an是公比為3的等比數(shù)列,Sn3n119(2018石家莊質(zhì)檢二)已知數(shù)列an的前n項(xiàng)和Snn,如果存在正整數(shù)n,使得(man)(man1)0成立,那么實(shí)數(shù)m的取值范圍是_答案,解析易得a1,n2時(shí),有anSnSn1nn13n則有a1a3a2k10a2ka4a2(kN*)若存在正整數(shù)n,使得(man)(man1)0成立,則只需滿足a1ma2即可,即m1,且a3a4a528,a42是a3,a5的等差中項(xiàng)數(shù)列bn滿足b11,數(shù)列(bn1bn)an的前n項(xiàng)和為2n2n(1)求q的值;(2)求數(shù)列bn的通項(xiàng)公式解(1)由a42是a3,a5的等差中項(xiàng)得a3a52a44,所以a3a4a53a4428,解得a48由a3a520得8q20,解得q2或q,因?yàn)閝1,所以q2(2)設(shè)cn(bn1bn)an,數(shù)列cn的前n項(xiàng)和為Sn由cn解得cn4n1由(1)可知an2n1,所以bn1bn(4n1)n1,故bnbn1(4n5)n2,n2,bnb1(bnbn1)(bn1bn2)(b3b2)(b2b1)(4n5)n2(4n9)n373設(shè)Tn37112(4n5)n2,n2,Tn372(4n9)n2(4n5)n1,所以Tn34424n2(4n5)n1,因此Tn14(4n3)n2,n2,又b11,所以bn15(4n3)n2,n2經(jīng)檢驗(yàn),當(dāng)n1時(shí),bn也成立故bn15(4n3)n22(2018天津高考)設(shè)an是等差數(shù)列,其前n項(xiàng)和為Sn(nN*);bn是等比數(shù)列,公比大于0,其前n項(xiàng)和為Tn(nN*)已知b11,b3b22,b4a3a5,b5a42a6(1)求Sn和Tn;(2)若Sn(T1T2Tn)an4bn,求正整數(shù)n的值解(1)設(shè)等比數(shù)列bn的公比為q由b11,b3b22,可得q2q20因?yàn)閝0,可得q2,故bn2n1所以Tn2n1設(shè)等差數(shù)列an的公差為d由b4a3a5,可得a13d4由b5a42a6,可得3a113d16,從而a11,d1,故ann所以Sn(2)由(1),有T1T2Tn(21222n)nn2n1n2由Sn(T1T2Tn)an4bn可得2n1n2n2n1,整理得n23n40,解得n1(舍去)或n4所以n的值為43(2017天津高考)已知an為等差數(shù)列,前n項(xiàng)和為Sn(nN*),bn是首項(xiàng)為2的等比數(shù)列,且公比大于0,b2b312,b3a42a1,S1111b4(1)求an和bn的通項(xiàng)公式;(2)求數(shù)列a2nb2n1的前n項(xiàng)和(nN*)解(1)設(shè)等差數(shù)列an的公差為d,等比數(shù)列bn的公比為q由已知b2b312,得b1(qq2)12,而b12,所以q2q60,解得q2或q3,又因?yàn)閝0,所以q2所以bn2n由b3a42a1,可得3da18由S1111b4,可得a15d16,聯(lián)立,解得a11,d3,由此可得an3n2所以,數(shù)列an的通項(xiàng)公式為an3n2,數(shù)列bn的通項(xiàng)公式為bn2n(2)設(shè)數(shù)列a2nb2n1的前n項(xiàng)和為Tn,由a2n6n2,b2n124n1,有a2nb2n1(3n1)4n,故Tn24542843(3n1)4n,4Tn242543844(3n4)4n(3n1)4n1,上述兩式相減,得3Tn2434234334n(3n1)4n14(3n1)4n1(3n2)4n18得Tn4n1所以數(shù)列a2nb2n1的前n項(xiàng)和為4n1二、模擬大題4(2018山西太原模擬)已知數(shù)列an的前n項(xiàng)和為Sn,數(shù)列bn滿足bnanan1(nN*)(1)求數(shù)列bn的通項(xiàng)公式;(2)若cn2an(bn1)(nN*),求數(shù)列cn的前n項(xiàng)和Tn解(1)當(dāng)n1時(shí),a1S11;當(dāng)n2時(shí),anSnSn1n,又a11符合上式,ann(nN*),bnanan12n1(2)由(1)得cn2an(bn1)n2n1,Tn122223324(n1)2nn2n1,2Tn123224325(n1)2n1n2n2,得,Tn2223242n1n2n2n2n2(1n)2n24,Tn(n1)2n245(2018沈陽質(zhì)檢)已知數(shù)列an是遞增的等比數(shù)列,且a1a49,a2a38(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)Sn為數(shù)列an的前n項(xiàng)和,bn,求數(shù)列bn的前n項(xiàng)和Tn解(1)由題設(shè)知a1a4a2a38,又a1a49,可解得或(舍去)設(shè)等比數(shù)列an的公比為q,由a4a1q3得q2,故ana1qn12n1,nN*(2)Sn2n1,又bn,所以Tnb1b2bn1,nN*6(2018安徽馬鞍山第二次教學(xué)質(zhì)量監(jiān)測)已知數(shù)列an是等差數(shù)列,其前n項(xiàng)和為Sn,a237,S4152(1)求數(shù)列an的通項(xiàng)公式;(2)求數(shù)列|an2n|的前n項(xiàng)和Tn解(1)設(shè)數(shù)列an的首項(xiàng)為a1,公差為d,則解得所以數(shù)列an的通項(xiàng)公式為an2n33(nN*)(2)由(1)知,|an2n|2n332n|當(dāng)1n5時(shí),Tnn234n2n12;當(dāng)n6時(shí),T5133,|2n332n|2n(2n33),TnT52n1n234n131,Tn2n1n234n264綜上所述,Tn- 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