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1、課時知能訓(xùn)練一、選擇題1下列關(guān)于星星的圖案中星星個數(shù)構(gòu)成一個數(shù)列,該數(shù)列的一個通項(xiàng)公式是()Aann2n1BanCan Dan【解析】觀察所給圖案知,an123n.【答案】C2(2020梅州質(zhì)檢)在數(shù)列an中,a11,anan1an1(1)n(n2,nN*),則的值是()A. B.C. D.【解析】當(dāng)n2時,a2a1a1(1)2,a22,當(dāng)n3時,a3a2a2(1)3,a3,當(dāng)n4時,a4a3a3(1)4,a43,當(dāng)n5時,a5a4a4(1)5,a5,.【答案】C3已知數(shù)列an的前n項(xiàng)和為Sn,且Sn2(an1),則a2等于()A4 B2C1 D2【解析】Sn2(an1),S1a12(a11)
2、,解得a12,又S2a1a22(a21),解得a2a124.【答案】A4已知數(shù)列an滿足a11,an1an2n,則a10()A1 024 B1 023C2 048 D2 047【解析】an1an2n,anan12n1(n2),a10(a10a9)(a9a8)(a2a1)a129282121011 023.【答案】B5(2020江西高考)已知數(shù)列an的前n項(xiàng)和Sn滿足:SnSmSnm,且a11,那么a10()A1 B9C10 D55【解析】SnSmSnm,令n9,m1,即得S9S1S10,S1S10S9a101,a101.【答案】A二、填空題6(2020湛江調(diào)研)已知a12,an1an2n1(n
3、N*),則an_.【解析】an1an2n1,anan12(n1)12n1,an(anan1)(an1an2)(a2a1)a1(2n1)(2n3)322n21.【答案】n217已知數(shù)列an的前n項(xiàng)和Snn29n,第k項(xiàng)滿足5ak8,則k的值為_【解析】當(dāng)n2時,anSnSn12n10;當(dāng)n1時,a18適合上式an2n10.5ak8,52k108,k9,又kN*,k8.【答案】88數(shù)列an中,a11,對于所有的n2,nN*,都有a1a2a3ann2,則a3a5_.【解析】由題意知:a1a2a3an1(n1)2,an()2(n2),a3a5()2()2.【答案】三、解答題9已知數(shù)列an的前n項(xiàng)和為S
4、n,若S11,S22,且Sn13Sn2Sn10(nN*且n2),求該數(shù)列的通項(xiàng)公式【解】由S11得a11,又由 S22可知a21.Sn13Sn2Sn10(nN*且n2),Sn1Sn2Sn2Sn10(nN*且n2),即(Sn1Sn)2(SnSn1)0(nN*且n2)an12an(nN*且n2)故數(shù)列an從第2項(xiàng)起是以2為公比的等比數(shù)列數(shù)列an的通項(xiàng)公式為an.10(2020邯鄲模擬)已知數(shù)列an滿足前n項(xiàng)和Snn21,數(shù)列bn滿足bn,且前n項(xiàng)和為Tn,設(shè)cnT2n1Tn.(1)求數(shù)列bn的通項(xiàng)公式;(2)判斷數(shù)列cn的增減性【解】(1)a12,anSnSn12n1(n2)bn.(2)cnbn1
5、bn2b2n1,cn1cn0,cn是遞減數(shù)列11已知數(shù)列an滿足a11,a213,an22an1an2n6.(1)設(shè)bnan1an,求數(shù)列bn的通項(xiàng)公式;(2)求n為何值時an最小【解】(1)由an22an1an2n6得,(an2an1)(an1an)2n6,bn1bn2n6.當(dāng)n2時,bnbn12(n1)6,bn1bn22(n2)6,b3b2226,b2b1216,累加得bnb12(12n1)6(n1)n(n1)6n6n27n6.bnn27n8(n2),n1時,b1a2a114也適合此式故bnn27n8.(2)由bn(n8)(n1),得an1an(n8)(n1),當(dāng)n8時,an1an,當(dāng)n8時,a9a8,當(dāng)n8時,an1an.故當(dāng)n8或n9時an的值最小