(福建專用)2020年高考數(shù)學(xué)總復(fù)習(xí) 第五章第1課時(shí) 數(shù)列的概念與簡單表示法課時(shí)闖關(guān)(含解析)
(福建專用)2020年高考數(shù)學(xué)總復(fù)習(xí) 第五章第1課時(shí) 數(shù)列的概念與簡單表示法課時(shí)闖關(guān)(含解析)一、選擇題1(2020·泉州質(zhì)檢)一個(gè)正整數(shù)數(shù)表如下(表中下一行中數(shù)的個(gè)數(shù)比上一行中數(shù)的個(gè)數(shù)多1個(gè)):第1行1第2行23第3行456則第11行中的第5個(gè)數(shù)是()A50B55C60 D66解析:選C.由數(shù)表知前10行數(shù)的個(gè)數(shù)共有55個(gè),故第11行中的第5個(gè)數(shù)是60.2數(shù)列1,12,12222n1,的一個(gè)通項(xiàng)an等于()A2n1 B2n1n2C2n1 D2nn解析:選A.通項(xiàng)an12222n12n1.或代入檢驗(yàn)第一項(xiàng)為1,第二項(xiàng)為3,即可排除B,C,D.3下列說法正確的是()A數(shù)列1,3,5,7可表示為1,3,5,7B數(shù)列1,0,1,2與數(shù)列2,1,0,1是相同數(shù)列C數(shù)列的第k項(xiàng)為1D數(shù)列0,2,4,6,可記為2n解析:選C.由數(shù)列定義可知A、B錯(cuò)誤;數(shù)列的第k項(xiàng)為1,故C正確;數(shù)列0,2,4,6,的通項(xiàng)公式為an2n2,故D錯(cuò),綜上可知,應(yīng)選C.4(2020·寧德質(zhì)檢)已知數(shù)列an滿足,則數(shù)列an是()A遞增數(shù)列 B遞減數(shù)列C擺動(dòng)數(shù)列 D不確定解析:選D.<1.若a1>0,則an1an,an是遞減數(shù)列;若a1<0,則an為遞增數(shù)列故數(shù)列an變化情況為不確定5已知數(shù)列an的前n項(xiàng)和Snn27n,且滿足16akak122,則正整數(shù)k的值是()A7 B8C9 D10解析:選B.由akak1Sk1Sk1(k1)27(k1)(k1)27(k1)4k14,知164k1422,所以整數(shù)k8.二、填空題6已知函數(shù)f(n),且anf(n),則a1a2a3a4a5_ .解析:a1a2a3a4a512223242521234515.答案:157已知Sn是數(shù)列an的前n項(xiàng)和,且有Snn21,則數(shù)列an的通項(xiàng)公式是_解析:當(dāng)n1時(shí),a1S1112;當(dāng)n2時(shí),anSnSn1(n21)(n1)212n1.答案:an8數(shù)列an滿足關(guān)系anan11an1(nN*),且a20202,則a2020_.解析:由anan11an1(nN*),得an1,又a20202,a20201,a20201213.答案:3三、解答題9已知數(shù)列an滿足a11,anan13n2(n2)(1)求a2,a3;(2)求數(shù)列an的通項(xiàng)公式解:(1)由已知:an滿足a11,anan13n2(n2),a2a145,a3a2712.(2)由已知:anan13n2(n2)得:anan13n2,由遞推關(guān)系,得an1an23n5,a3a27,a2a14,累加得:ana1473n2,an(n2)當(dāng)n1時(shí),1a11,數(shù)列an的通項(xiàng)公式為an.10數(shù)列an的前n項(xiàng)和為Sn,a11,an1Sn(n1,2,3,),求an.解:an1Sn,anSn1(n2),an1an(SnSn1)an(n2),an1an(n2)又a11,a2S1a1,an是從第二項(xiàng)起,公比為的等比數(shù)列,an一、選擇題1在數(shù)列an中,a11,a25,an2an1an(nN*), 則a8等于()A1 B1C5 D5解析:選C.法一:由a11,a25,an2an1an(nN*)可得該數(shù)列為1,5,4,1,5,4,1,5,4,.由此可得a85.法二:an2an1an,an3an2an1,兩式相加可得an3an,an6an,a8a25.2如圖所示的三角形數(shù)陣叫“萊布尼茲調(diào)和三角形”,它們是由整數(shù)的倒數(shù)組成的,第n行有n個(gè)數(shù)且兩端的數(shù)均為(n2),每個(gè)數(shù)是它下一行左右相鄰兩數(shù)的和,如,則第10行第4個(gè)數(shù)(從左往右數(shù))為()A. B.C. D.答案:C二、填空題3(2020·南平質(zhì)檢)已知數(shù)列an的前n項(xiàng)的乘積為Tn5n2,nN*,則數(shù)列an的通項(xiàng)公式為an_.解析:當(dāng)n1時(shí),a1T15125;當(dāng)n2時(shí),an52n1(nN*)當(dāng)n1時(shí),也適合上式,所以當(dāng)nN*時(shí),an52n1.答案:52n1(nN*)4數(shù)列an中,an,Sn9,則n_.解析:an,Sn(1)()()19,n99.答案:99三、解答題5設(shè)數(shù)列an的前n項(xiàng)和為Sn,已知(nN*)(1)求S1,S2及Sn;(2)設(shè)bnan,若對(duì)一切nN*,均有k(,m26m),求實(shí)數(shù)m的取值范圍解:(1)依題意,n1時(shí),S12,n2時(shí),S26.n2時(shí),得.Snn(n1)上式對(duì)n1也成立,Snn(n1)(nN*)(2)由(1)知,Snn(n1),當(dāng)n2時(shí),anSnSn12n.a12,an2n(nN*)bnn.,數(shù)列bn是等比數(shù)列則k.隨n的增大而增大,k.依條件,得即m0或m5.6已知二次函數(shù)f(x)x2axa(a>0,xR)有且只有一個(gè)零點(diǎn),數(shù)列an的前n項(xiàng)和Snf(n)(nN*)(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)cn1(nN*),定義所有滿足cm·cm1<0的正整數(shù)m的個(gè)數(shù),稱為這個(gè)數(shù)列cn的變號(hào)數(shù),求數(shù)列cn的變號(hào)數(shù)解:(1)依題意,a24a0,a0或a4.又由a>0得a4,f(x)x24x4.Snn24n4.當(dāng)n1時(shí),a1S11441;當(dāng)n2時(shí),anSnSn12n5.an由1可知,當(dāng)n5時(shí),恒有an>0.又c13,c25,c33,c4,c5,即c1·c2<0,c2·c3<0,c4·c5<0,數(shù)列cn的變號(hào)數(shù)為3.