《2022年高考數(shù)學(xué)考點(diǎn)分類自測(cè) 數(shù)列的概念及簡(jiǎn)單表示法 理》由會(huì)員分享,可在線閱讀,更多相關(guān)《2022年高考數(shù)學(xué)考點(diǎn)分類自測(cè) 數(shù)列的概念及簡(jiǎn)單表示法 理(4頁(yè)珍藏版)》請(qǐng)?jiān)谘b配圖網(wǎng)上搜索。
1、2022年高考數(shù)學(xué)考點(diǎn)分類自測(cè) 數(shù)列的概念及簡(jiǎn)單表示法 理一、選擇題1數(shù)列1,的一個(gè)通項(xiàng)公式an是()A. B.C. D.2已知數(shù)列an的前n項(xiàng)和為Sn,且Sn2(an1),則a2等于()A4 B2C1 D23數(shù)列an的a11,a(n,an),b(an1,n1),且ab,則a100等于()A100 B100C. D4已知數(shù)列an的前n項(xiàng)和Snkn2,若對(duì)所有的nN*,都有an1an,則實(shí)數(shù)k的取值范圍是()Ak0 Bk1Ck1 Dk05已知數(shù)列an滿足anan1(nN*),a22,Sn是數(shù)列an的前n項(xiàng)和,則S21為()A5 B.C. D.6若數(shù)列an滿足a15,an1(nN*),則其前10項(xiàng)
2、和為()A50 B100C150 D200二、填空題7數(shù)列an對(duì)任意nN*滿足an1ana2,且a36,則a10等于_8根據(jù)下圖5個(gè)圖形及相應(yīng)點(diǎn)的個(gè)數(shù)的變化規(guī)律,猜測(cè)第n個(gè)圖中有_個(gè)點(diǎn)9若數(shù)列an滿足,an1且a1,則axx_.三、解答題10數(shù)列an中,a11,對(duì)于所有的n2,nN*都有a1a2a3ann2,求a3a5的值11已知數(shù)列an的前n項(xiàng)和為Sn.(1)若Sn(1)n1n,求a5a6及an;(2)若Sn3n2n1,求an.12設(shè)函數(shù)f(x)log2xlogx2(0x1),數(shù)列an滿足f(2an)2n(nN*)(1)求數(shù)列an的通項(xiàng)公式;(2)判斷數(shù)列an的單調(diào)性詳解答案一、選擇題1解析
3、:由已知得,數(shù)列可寫成,故通項(xiàng)為.答案:B2解析:在Sn2(an1)中,令n1,得a12;令n2,得a1a22a22,所以a24.答案:A3解析:ab0,則nan1(n1) an0,100,a100100.答案:A4解析:本題考查數(shù)列中an與Sn的關(guān)系以及數(shù)列的單調(diào)性由Snkn2得ank(2n1),因?yàn)閍n1an,所以數(shù)列an是遞增的,因此k0.答案:A5解析:anan1,a22,a1,S21a1a2a20a21a1105.答案:B6解析:由an1得a2anan1a0,an1an,即an為常數(shù)列,S1010a150.答案:A二、填空題7解析:由已知,n1時(shí),a2a1a2,a10;n2時(shí),a3a
4、2a26,a23;n3時(shí),a4a3a29;n4時(shí),a5a4a212;n5時(shí),a6a5a215;n10時(shí),a10a9a227.答案:278解析:觀察圖中5個(gè)圖形點(diǎn)的個(gè)數(shù)分別為1,121,231,341,451,故第n個(gè)圖中點(diǎn)的個(gè)數(shù)為(n1)n1n2n1.答案:n2n19解析:a22a1,a3a21,a42a3, a5a41,a62a5,a72a6,此數(shù)列周期為5,axxa3.答案:三、解答題10解:由a1a2a3ann2,a1a24,a1a2a39,a3,同理a5.a3a5.11解:(1)a5a6S6S4(6)(4)2.當(dāng)n1時(shí),a1S11;當(dāng)n2時(shí),anSnSn1(1)n1n(1)n(n1)(1)n1n(n1)(1)n1(2n1)由于a1也適合于此式,所以an(1)n1(2n1)(2)當(dāng)n1時(shí),aS6;當(dāng)n2時(shí),anSnSn1(3n2n1)3n12(n1)123n12.由于a1不適合此式,所以an12解:(1)由已知得log22anlog2an22n,an2n,即a2nan10.解得ann.0x1,即02an120,an0,故ann(nN*)(2)1,而an0,an1an,即數(shù)列an是關(guān)于n的遞增數(shù)列