《人教a版高中數(shù)學(xué)必修5【課時(shí)作業(yè)15】等比數(shù)列前n項(xiàng)和的性質(zhì)及應(yīng)用含答案》由會(huì)員分享，可在線閱讀，更多相關(guān)《人教a版高中數(shù)學(xué)必修5【課時(shí)作業(yè)15】等比數(shù)列前n項(xiàng)和的性質(zhì)及應(yīng)用含答案（6頁(yè)珍藏版）》請(qǐng)?jiān)谘b配圖網(wǎng)上搜索。

1、人教版高中數(shù)學(xué)必修精品教學(xué)資料課時(shí)作業(yè)15等比數(shù)列前n項(xiàng)和的性質(zhì)及應(yīng)用時(shí)間：45分鐘分值：100分一、選擇題(每小題6分,共計(jì)36分)1設(shè)等比數(shù)列an的公比q2,前n項(xiàng)和為Sn,則等于()A2 B4C. D.解析：S415a1,a2a1q2a1,.答案：C2設(shè)an是由正數(shù)組成的等比數(shù)列,Sn為其前n項(xiàng)和,已知a2a41,S37,則S5等于()A. B.C. D.解析：設(shè)等比數(shù)列an的公比為q,則解得a14,q,所以S5.答案：B3一個(gè)等比數(shù)列的前7項(xiàng)和為48,前14項(xiàng)和為60,則前21項(xiàng)和為()A180 B108C75 D63解析：由題意S7,S14S7,S21S14組成等比數(shù)列48,12,3

2、,即S21S143,S2163.答案：D4在公比為整數(shù)的等比數(shù)列an中,已知a1a418,a2a312,那么a5a6a7a8等于()A480 B493C495 D498解析：已知由等比數(shù)列的通項(xiàng)公式得2q33q23q20(q1)(2q25q2)0q1或q2或q.q1,q均與已知矛盾,q2.a5a6a7a8q4(a1a2a3a4)24(1812)480.答案：A5已知數(shù)列an的前n項(xiàng)和為Sn2n1,則此數(shù)列奇數(shù)項(xiàng)的前n項(xiàng)的和是()A.(2n11) B.(2n12)C.(22n1) D.(22n2)解析：由題易知,數(shù)列an的通項(xiàng)公式為an2n1,公比q2.奇數(shù)項(xiàng)的前n項(xiàng)和為Sa1a3a2n1(22

3、n1)答案：C6一個(gè)等比數(shù)列共有3m項(xiàng),若前2m項(xiàng)和為15,后2m項(xiàng)之和為60,則中間m項(xiàng)的和為()A12 B16C20 D32解析：由已知S2m15,S3mSm60,又(S2mSm)2Sm(S3mS2m),解得Sm3,S2mSm15312.答案：A二、填空題(每小題8分,共計(jì)24分)7設(shè)Sn為等比數(shù)列an的前n項(xiàng)和,已知3S3a42,3S2a32,則公比q_.解析：由3S3a42,3S2a32兩式相減得,3(S3S2)a4a3,3a3a4a3,4a3a4,q4.答案：48已知an是首項(xiàng)為1的等比數(shù)列,Sn是an的前n項(xiàng)和,且9S3S6,則數(shù)列的前5項(xiàng)和為_(kāi)解析：顯然q1,1q39,q2,是首

4、項(xiàng)為1,公比為的等比數(shù)列,前5項(xiàng)和T5.答案：9在等比數(shù)列中,S3013S10,S10S30140,則 S20_.解析：由S3013S10,S10S30140,得S1010,S30130.再由S10,S20S10,S30S20成等比數(shù)列,得S10(S30S20)(S20S10)2,10(130S20)(S2010)2.整理得S10S2012000,解得S2040,或S2030(舍去)答案：40三、解答題(共計(jì)40分)10(10分)等比數(shù)列an的前n項(xiàng)和為Sn,已知S1,S3,S2成等差數(shù)列(1)求an的公比q；(2)若a1a33,求Sn.解：(1)依題意有a1(a1a1q)2(a1a1qa1q

5、2),由于a10,故2q2q0.又q0,從而q.(2)由已知可得a1a1()23,解得a14.從而Sn1()n11(15分)(2012山東卷)在等差數(shù)列an中,a3a4a584,a973.(1)求數(shù)列an的通項(xiàng)公式；(2)對(duì)任意mN*,將數(shù)列an中落入?yún)^(qū)間(9m,92m)內(nèi)的項(xiàng)的個(gè)數(shù)記為bm,求數(shù)列bm的前m項(xiàng)和Sm.解：(1)因?yàn)閍n是一個(gè)等差數(shù)列,所以a3a4a53a484,所以a428.設(shè)數(shù)列an的公差為d,則5da9a4732845,故d9.由a4a13d得28a139,即a11,所以ana1(n1)d19(n1)9n8(nN*)(2)對(duì)mN*,若9man92m,則9m89n92m8,因此9m11n92m1,故得bm92m19m1.于是Smb1b2b3bm(99392m1)(199m1).12(15分)給出下面的數(shù)表序列：其中表n(n1,2,3)有n行,表中每一個(gè)數(shù)“兩腳”的 兩數(shù)都是此數(shù)的2倍,記表n中所有的數(shù)之和為an,例如a25,a317,a449,試求：(1)a5；(2)數(shù)列an的通項(xiàng)an.解：(1)a5129,(2)依題意,an122322423n2n1由2得,2an12222323424n2n將得an122223242n1n2nn2n2n1n2n所以an(n1)2n1.