高中數(shù)學(xué)新人教A版必修5習(xí)題 2.5 等比數(shù)列的前n項(xiàng)和2

1、人教版高中數(shù)學(xué)必修精品教學(xué)資料等比數(shù)列的綜合應(yīng)用a組基礎(chǔ)鞏固1已知等比數(shù)列的公比為2,且前5項(xiàng)和為1,那么前10項(xiàng)和等于()a31 b33c35 d37解析：根據(jù)等比數(shù)列性質(zhì)得q5,25,s1033.答案：b2在等比數(shù)列an中,s41,s83,則a17a18a19a20的值是()a14 b16c18 d20解析：s41,s83.s8s42,s4,s8s4,s12s8,s16s12,s20s16,成等比數(shù)列,且公比為2,a17a18a19a20s20s161·2416.答案：b3如果一個(gè)項(xiàng)數(shù)為偶數(shù)的等比數(shù)列,它的偶數(shù)項(xiàng)和是奇數(shù)項(xiàng)和的2倍,又它的首項(xiàng)為1,且中間兩項(xiàng)的和為96,則此等比數(shù)

2、列的項(xiàng)數(shù)為()a12 b10c8 d6解析：設(shè)等比數(shù)列為an,其項(xiàng)數(shù)為2n,公比為q,則a11,anan196.s偶,s奇.由s偶2s奇,得a22a12.q2.由anan196,得2n12n96.3·2n196,即2n132,n6,2n12.答案：a4在等比數(shù)列中,已知a1aa15243,則的值為()a3 b9c27 d81解析：a1a15a,a24335,a83,a9·a7a9.答案：b51的值為()a18 b20c22 d18解析：該數(shù)列的通項(xiàng)為an12,可以看作11項(xiàng)求和,則前11項(xiàng)的和為s112222×112×20,所以b正確答案：b6已知數(shù)列a

3、n的前n項(xiàng)和為sn,且ansn·sn1(n2),a1,則a10等于()a. b.c. d.解析：由ansn·sn1(n2),得snsn1sn·sn1(n2),1,n1n,sn,a10s10s9.答案：c7設(shè)等差數(shù)列an的前n項(xiàng)和為sn,則s4,s8s4,s12s8,s16s12成等差數(shù)列類比以上結(jié)論有：設(shè)等比數(shù)列bn的前n項(xiàng)積為tn,則t4,_,_,成等比數(shù)列解析：對(duì)于等比數(shù)列,通過類比,若等比數(shù)列bn的前n項(xiàng)積為tn,則t4,成等比數(shù)列答案：8在等比數(shù)列an中,a13,a481,若數(shù)列bn滿足bnlog3an,則數(shù)列的前n項(xiàng)和sn_.解析：a13,a481,3q

4、381,q3,an3×3n13n,bnlog33nn,sn1.答案：9已知等差數(shù)列an滿足a20,.a6a810.(1)求數(shù)列an的通項(xiàng)公式；(2)求數(shù)列的前n項(xiàng)和解：(1)設(shè)等差數(shù)列an的公差為d,由已知條件,得解得故數(shù)列an的通項(xiàng)公式為an2n.(2)設(shè)數(shù)列的前n項(xiàng)和為sn,即sna1.故s11,.所以,當(dāng)n>1時(shí),a111.所以sn.當(dāng)n1時(shí),s1符合此式綜上,數(shù)列的前n項(xiàng)和sn.10已知數(shù)列an的前n項(xiàng)和為sn,且sn2n2n,nn*,數(shù)列bn滿足an4log2bn3,nn*.(1)求an,bn；(2)求數(shù)列an·bn的前n項(xiàng)和tn.解：(1)由sn2n2n,

5、得當(dāng)n1時(shí),a1s13；當(dāng)n2時(shí),ansnsn14n1,當(dāng)n1時(shí)也適合,所以an4n1,nn*.由4n1an4log2bn3,得bn2n1,nn*.(2)由(1)知an·bn(4n1)·2n1,nn*,所以tn37×211×22(4n1)·2n1,2tn3×27×22(4n5)·2n1(4n1)·2n,所以2tntn(4n1)·2n(4n5)·2n5.故tn(4n5)·2n5,nn*.b組能力提升11設(shè)等比數(shù)列an的前n項(xiàng)和為sn,若8a2a50,則下列式子中數(shù)值不能確定的是

6、()a. b.c. d.解析：設(shè)等比數(shù)列an的公比為q,則8a1qa1q40,得q2,q24；q2；而,由于n未知,故無法確定其值答案：d12等比數(shù)列an的前n項(xiàng)和為sn,公比不為1.若a11,且對(duì)任意的nn*都有an2an12an0,則s5_.解析：由an2an12an0,得anq2anq2an0,顯然an0,所以q2q20.又q1,解得q2.又a11,所以s511.答案：1113設(shè)數(shù)列an的前n項(xiàng)和為sn2n2,數(shù)列bn為等比數(shù)列,且a1b1,b2(a2a1)b1.(1)求數(shù)列an和bn的通項(xiàng)公式；(2)設(shè)cn,求數(shù)列cn的前n項(xiàng)和tn.解：(1)當(dāng)n1時(shí),a1s12；當(dāng)n2時(shí),ansns

7、n12n22(n1)24n2,故an的通項(xiàng)公式為an4n2,即an是a12,公差d4的等差數(shù)列設(shè)bn的公比為q,則b1qdb1,q.故bnb1qn12×.即bn的通項(xiàng)公式為bn.(2)cn(2n1)4n1,tnc1c2cn13×415×42(2n1)4n1,4tn1×43×425×43(2n3)4n1(2n1)4n.兩式相減,得3tn12(4142434n1)(2n1)4n,tn14將各項(xiàng)均為正數(shù)的數(shù)列an中的所有項(xiàng)按每一行比上一行多一項(xiàng)的規(guī)律排成數(shù)表,如下表記表中各行的第一個(gè)數(shù)a1,a2,a4,a7,構(gòu)成的數(shù)列為bn,各行的最后一個(gè)數(shù)a1,a3,a6,a10,構(gòu)成的數(shù)列為cn,第n行所有數(shù)的和為sn(n1,2,3,4,)已知數(shù)列bn是公差為d的等差數(shù)列,從第二行起,每一行中的數(shù)按照從左到右的順序每一個(gè)數(shù)與它前面一個(gè)數(shù)的比是常數(shù)q,且a1a131,a31.a1a2a3a4a5a6a7a8a9a10 (1)求數(shù)列cn,sn的通項(xiàng)公式；(2)記dn(nn*),求證：d1d2d3dn>.解析：(1)由題意得bndnd1.前n行共有123n個(gè)數(shù)133,a