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1、2020年高考數(shù)學(xué)總復(fù)習(xí) 第五章 第4課時(shí) 數(shù)列求和課時(shí)闖關(guān)(含解析) 新人教版一、選擇題1(2020遼陽(yáng)質(zhì)檢)已知數(shù)列an的前n項(xiàng)和Snan2bn(a、bR),且S25100,則a12a14等于()A16B8C4 D不確定解析:選B.由數(shù)列an的前n項(xiàng)和Snan2bn(a、bR),可得數(shù)列an是等差數(shù)列,S25100,解得a1a258,所以a12a14a1a258.2數(shù)列1,3,5,7,(2n1),的前n項(xiàng)和Sn的值為()An21 B2n2n1Cn21 Dn2n1解析:選A.該數(shù)列的通項(xiàng)公式為an(2n1),則Sn135(2n1)()n21.故選A.3若數(shù)列an的前n項(xiàng)和為Sn,且滿足Sna
2、n3,則數(shù)列an的前n項(xiàng)和Sn為()A3n13 B3n3C3n13 D3n3解析:選A.Snan3,Sn1an13,兩式相減得:Sn1Sn(an1an)即an1(an1an),3,即q3.又S1a13,即a1a13,a16.ana1qn163n123n.Snan323n33n13,故應(yīng)選A.4已知函數(shù)f(x)x2bx的圖象在點(diǎn)A(1,f(1)處的切線的斜率為3,數(shù)列的前n項(xiàng)和為Sn,則S2020的值為()A. B.C. D.解析:選D.f(x)2xb,f(1)2b3,b1,f(x)x2x,S202011.5設(shè)數(shù)列an是首項(xiàng)為1,公比為3的等比數(shù)列,把a(bǔ)n中的每一項(xiàng)都減去2后,得到一個(gè)新數(shù)列bn
3、,bn的前n項(xiàng)和為Sn,則對(duì)任意的nN*,下列結(jié)論正確的是()Abn13bn2,且Sn(3n1)Bbn13bn2,且Sn(3n1)Cbn13bn4,且Sn(3n1)2nDbn13bn4,且Sn(3n1)2n解析:選C.因?yàn)閿?shù)列an是首項(xiàng)為1,公比為3的等比數(shù)列,所以數(shù)列an的通項(xiàng)公式為an3n1,則依題意得,數(shù)列bn的通項(xiàng)公式為bn3n12,bn13n2,3bn3(3n12)3n6,bn13bn4.bn的前n項(xiàng)和為:Sn(12)(312)(322)(332)(3n12)(13132333n1)2n2n(3n1)2n.二、填空題6數(shù)列1,的前n項(xiàng)和Sn_.解析:由于an2(),Sn2(1)2(1
4、).答案:7對(duì)于數(shù)列an,定義數(shù)列an1an為數(shù)列an的“差數(shù)列”,若a12,an的“差數(shù)列”的通項(xiàng)為2n,則數(shù)列an的前n項(xiàng)和Sn_.解析:an1an2n,an(anan1)(an1an2)(a2a1)a12n12n2222222n222n.Sn2n12.答案:2n128若數(shù)列an是正項(xiàng)數(shù)列,且n23n(nN*),則_.解析:令n1,得4,a116.當(dāng)n2時(shí),(n1)23(n1),與已知式相減,得(n23n)(n1)23(n1)2n2,an4(n1)2,n1時(shí),a1也適合an.an4(n1)2,4n4,2n26n.答案:2n26n三、解答題9(2020高考重慶卷)設(shè)an是公比為正數(shù)的等比數(shù)列
5、,a12,a3a24.求an的通項(xiàng)公式;設(shè)bn是首項(xiàng)為1,公差為2的等差數(shù)列,求數(shù)列anbn的前n項(xiàng)和Sn.解:設(shè)q為等比數(shù)列an的公比,則由a12,a3a24得2q22q4,即q2q20,解得q2或q1,因此q2.所以an的通項(xiàng)公式為an22n12n.Snn122n1n22.10數(shù)列an中a13,已知點(diǎn)(an,an1)在直線yx2上,(1)求數(shù)列an的通項(xiàng)公式;(2)若bnan3n,求數(shù)列bn的前n項(xiàng)和Tn.解:(1)點(diǎn)(an,an1)在直線yx2上,an1an2,即an1an2.數(shù)列an是以3為首項(xiàng),2為公差的等差數(shù)列,an32(n1)2n1.(2)bnan3n,bn(2n1)3n,Tn3
6、3532(2n1)3n1(2n1)3n,3Tn332533(2n1)3n(2n1)3n1,由得2Tn332(32333n)(2n1)3n192(2n1)3n1.Tnn3n1.11(探究選做)已知函數(shù)f(x)ax2bx(a0)的導(dǎo)函數(shù)f(x)2x7,數(shù)列an的前n項(xiàng)和為Sn,點(diǎn)Pn(n,Sn)(nN*)均在函數(shù)yf(x)的圖象上(1)求數(shù)列an的通項(xiàng)公式及Sn的最大值;(2)令bn,其中nN*,求數(shù)列nbn的前n項(xiàng)和Tn.解:(1)f(x)ax2bx(a0),f(x)2axb,由f(x)2x7,得a1,b7,f(x)x27x,又點(diǎn)Pn(n,Sn)(nN*)均在函數(shù)yf(x)的圖象上,Snn27n.當(dāng)n1時(shí),a1S16;當(dāng)n2時(shí),anSnSn12n8,an2n8(nN*)令an2n80,得n4,當(dāng)n3或n4時(shí),Sn取得最大值12.綜上,an2n8(nN*),當(dāng)n3或n4時(shí),Sn取得最大值12.(2)由題意得b18,bn2n4,即數(shù)列bn是首項(xiàng)為8,公比是的等比數(shù)列,故nbn的前n項(xiàng)和Tn123222n2n4,Tn12222(n1)2n4n2n3,得:Tn23222n4n2n3,Tnn24n32(2n)24n.