1、等比数列的前n项和(一)一、基础过关1已知等比数列an的前n项和为Sn,且a11,a464,则S4等于()A48 B49 C50 D512在等比数列an中,公比q是整数,a1a418,a2a312,则此数列的前8项和为()A513 B512 C511 D5103设Sn为等比数列an的前n项和,8a2a50,则等于()A11 B5 C8 D114设等比数列an的公比q2,前n项和为Sn,则等于 ()A2 B4 C. D.5等比数列an的前n项和为Sn,已知S1,2S2,3S3成等差数列,则an的公比为_6设等比数列an的前n项和为Sn,若a11,S64S3,则a4_.7若等比数列an中,a11,
2、an512,前n项和为Sn341,则n的值是_8设等比数列an的前n项和为Sn,已知a26,6a1a330,求an和Sn.二、能力提升9已知an是等比数列,a22,a5,则a1a2a2a3anan1等于()A16(14n) B16(12n)C.(14n) D.(12n)10设an是由正数组成的等比数列,Sn为其前n项和,已知a2a41,S37,则S5等于()A. B.C. D.11在等比数列an中,已知Sn48,S2n60,求S3n.12已知等比数列an中,a12,a32是a2和a4的等差中项(1)求数列an的通项公式;(2)记bnanlog2an,求数列bn的前n项和Sn.三、探究与拓展13
3、已知等差数列an满足a20,a6a810.(1)求数列an的通项公式;(2)求数列的前n项和答案1D2.D3.D4.C5.6.37.108解设an的公比为q,由题设得解得或当a13,q2时,an32n1,Sn3(2n1);当a12,q3时,an23n1,Sn3n1.9Cq3,q,a14.ana1qn14()n1,anan116()n1()n32()n.a1a2a2a3anan1(14n)10B11解因为S2n2Sn,所以q1,由已知得得1qn,即qn.将代入得64,所以S3n6463.12解(1)设数列an的公比为q,由题知:2(a32)a2a4,q32q2q20,即(q2)(q21)0.q2,即an22n12n.(2)bnn2n,Sn12222323n2n.2Sn122223324(n1)2nn2n1.得Sn212223242nn2n12(n1)2n1.Sn2(n1)2n1.13解(1)设等差数列an的公差为d,由已知条件可得解得.故数列an的通项公式为an2n.(2)设数列的前n项和为Sn,即Sna1,故S11,.所以,当n1时,得a11()1(1).所以Sn.当n1时也成立综上,数列的前n项和Sn.