2017-2018學(xué)年高中數(shù)學(xué) 第二章 數(shù)列 2.5 等比數(shù)列的前n項(xiàng)和 第1課時(shí) 等比數(shù)列的前n項(xiàng)和公式優(yōu)化練習(xí) 新人教A版必修5.doc
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2017-2018學(xué)年高中數(shù)學(xué) 第二章 數(shù)列 2.5 等比數(shù)列的前n項(xiàng)和 第1課時(shí) 等比數(shù)列的前n項(xiàng)和公式優(yōu)化練習(xí) 新人教A版必修5.doc
第1課時(shí) 等比數(shù)列的前n項(xiàng)和公式課時(shí)作業(yè)A組基礎(chǔ)鞏固1等比數(shù)列an中,an2n,則它的前n項(xiàng)和Sn()A2n1B2n2C2n11 D2n12解析:a12,q2,Sn2n12.答案:D2在等比數(shù)列an中,若a11,a4,則該數(shù)列的前10項(xiàng)和S10()A2 B2C2 D2解析:設(shè)等比數(shù)列an的公比為q,由a11,a4,得q3,解得q,于是S102.答案:B3等比數(shù)列an中,已知前4項(xiàng)之和為1,前8項(xiàng)和為17,則此等比數(shù)列的公比q為()A2 B2C2或2 D2或1解析:S41,S817,得1q417,q416.q2.答案:C4已知數(shù)列an為等比數(shù)列,Sn是它的前n項(xiàng)和,若a2a32a1,且a4與2a7的等差中項(xiàng)為,則S5()A35 B33C31 D29解析:設(shè)數(shù)列an的公比為q,a2a3aq3a1a42a1,a42.又a42a7a42a4q324q32,q.a116.S531.答案:C5等比數(shù)列an中,a33S22,a43S32,則公比q等于()A2 B.C4 D.解析:a33S22,a43S32,等式兩邊分別相減得a4a33a3,即a44a3,q4.答案:C6若數(shù)列an滿足a11,an12an,n1,2,3,則a1a2an_.解析:由2,an是以a11,q2的等比數(shù)列,故Sn2n1.答案:2n17等比數(shù)列an的前n項(xiàng)和為Sn,已知S1,2S2,3S3成等差數(shù)列,則an的公比為_解析:S1,2S2,3S3成等差數(shù)列,4S2S13S3,即4(a1a1q)a13(a1a1qa1q2),4(1q)13(1qq2),解之得q.答案:8等比數(shù)列的前n項(xiàng)和Snm3n2,則m_.解析:設(shè)等比數(shù)列為an,則a1S13m2,S2a1a29m2a26m,S3a1a2a327m2a318m,又aa1a3(6m) 2(3m2)18mm2或m0(舍去)m2.答案:29在等差數(shù)列an中,a410,且a3,a6,a10成等比數(shù)列,求數(shù)列an前20項(xiàng)的和S20.解析:設(shè)數(shù)列an的公差為d,則a3a4d10d,a6a42d102d,a10a46d106d,由a3,a6,a10成等比數(shù)列,得a3a10a,即(10d)(106d)(102d)2.整理,得10d210d0.解得d0或d1.當(dāng)d0時(shí),S2020a4200;當(dāng)d1時(shí),a1a43d10317,于是S2020a1d207190330.10已知數(shù)列an的前n項(xiàng)和Sn2nn2,anlog5bn,其中bn>0,求數(shù)列bn的前n項(xiàng)和Tn.解析:當(dāng)n2時(shí),anSnSn1(2nn2)2(n1)(n1)22n3,當(dāng)n1時(shí),a1S121121也適合上式,an的通項(xiàng)公式an2n3(nN*)又anlog5bn,log5bn2n3,于是bn52n3,bn152n1,52.因此bn是公比為的等比數(shù)列,且b15235,于是bn的前n項(xiàng)和Tn.B組能力提升1已知等比數(shù)列an的前n項(xiàng)和Sn2n1,則aaa等于()A(2n1)2 B.(2n1)C4n1 D.(4n1)解析:根據(jù)前n項(xiàng)和Sn2n1,可求出an2n1,由等比數(shù)列的性質(zhì)可得a仍為等比數(shù)列,且首項(xiàng)為a,公比為q2,aaa1222422n2(4n1)答案:D2設(shè)Sn是等比數(shù)列an的前n項(xiàng)和,若3,則()A2 B.C. D1或2解析:設(shè)S2k,則S43k,由數(shù)列an為等比數(shù)列(易知數(shù)列an的公比q1),得S2,S4S2,S6S4為等比數(shù)列,又S2k,S4S22k,S6S44k,S67k,故選B.答案:B3已知數(shù)列an是遞增的等比數(shù)列,a1a49,a2a38,則數(shù)列an的前n項(xiàng)和等于_解析:由題意,解得a11,a48或者a18,a41,而數(shù)列an是遞增的等比數(shù)列,所以a11,a48,即q38,所以q2,因而數(shù)列an的前n項(xiàng)和Sn2n1.答案:2n14設(shè)數(shù)列an(n1,2,3,)的前n項(xiàng)和Sn滿足Sna12an,且a1,a21,a3成等差數(shù)列,則a1a5_.解析:由Sna12an,得anSnSn12an2an1(n2),即an2an1(n2)從而a22a1,a32a24a1.又因?yàn)閍1,a21,a3成等差數(shù)列,所以a1a32(a21),所以a14a12(2a11),解得a12,所以數(shù)列an是首項(xiàng)為2,公比為2的等比數(shù)列,故an2n,所以a1a522534.答案:345(2016高考全國卷)已知數(shù)列an的前n項(xiàng)和Sn1an,其中0.(1)證明an是等比數(shù)列,并求其通項(xiàng)公式;(2)若S5,求.解析:(1)證明:由題意得a1S11a1,故1,a1,a10.由Sn1an,Sn11an1得an1an1an,即an1(1)an.由a10,0得an0,所以.因此an是首項(xiàng)為,公比為的等比數(shù)列,于是ann1.(2)由(1)得Sn1n.由S5得15,即5.解得1.6設(shè)an是公比大于1的等比數(shù)列,Sn為數(shù)列an的前n項(xiàng)和已知S37,且a13,3a2,a34構(gòu)成等差數(shù)列(1)求數(shù)列an的通項(xiàng);(2)令bnln a3n1,n1,2,求數(shù)列bn的前n項(xiàng)和Tn.解析:(1)由已知得解得a22.設(shè)數(shù)列an的公比為q,由a22,可得a1,a32q,又S37,可知22q7,即2q25q20.解得q12,q2.由題意得q>1,q2,a11.故數(shù)列an的通項(xiàng)為an2n1.(2)由于bnln a3n1,n1,2,由(1)得a3n123n,bnln 23n3nln 2.又bn1bn3ln 2,bn是等差數(shù)列,Tnb1b2bnln 2.故Tnln 2.