1第一章数列§3等比数列3.2等比数列的前n项和第1课时等比数列前n项和的推导及初步应用课后篇巩固提升必备知识基础练1.等比数列{an}中,a1=2,a2=1,则S100等于()A.4-2100B.4+2100C.4-2-98D.4-2-100答案C解析公比q=a2a1=12.S100=a1\(1-q100\)1-q=2[1-(12)100]1-12=4(1-2-100)=4-2-98.故选C.2.在等比数列{an}中,已知a1=3,an=48,Sn=93,则n的值为()A.4B.5C.6D.7答案B解析显然公比q≠1,由Sn=a1-anq1-q,得93=3-48q1-q,解得q=2.由an=a1qn-1,得48=3×2n-1,解得n=5.故选B.3.设Sn为等比数列{an}的前n项和,8a2+a5=0,则S5S2等于()A.11B.5C.-8D.-11答案D解析设数列{an}的公比是q,由8a2+a5=0得8a1q+a1q4=0, a1≠0,q≠0,∴q=-2,则S5S2=a1\(1+25\)a1\(1-22\)=-11.4.等比数列{an}的前n项和为Sn,已知S3=a2+10a1,a5=9,则a1等于()A.13B.-13C.19D.-192答案C解析设等比数列{an}的公比为q,由S3=a2+10a1,得a1+a2+a3=a2+10a1,即a3=9a1,q2=9,又a5=a1q4=9,所以a1=19.5.设公比为q(q>0)的等比数列{an}的前n项和为Sn.若S2=3a2+2,S4=3a4+2,则a1等于()A.-2B.-1C.12D.23答案B解析由S2=3a2+2,S4=3a4+2,得a3+a4=3a4-3a2,即q+q2=3q2-3,解得q=-1(舍去)或q=32,将q=32代入S2=3a2+2中得a1+32a1=3×32a1+2,解得a1=-1,故选B.6.已知数列{an}满足3an+1+an=0,a2=-43,则{an}的前10项和等于()A.-6(1-3-10)B.19(1-3-10)C.3(1-3-10)D.3(1+3-10)答案C解析由3an+1+an=0,得an+1an=-13,故数列{an}是公比q=-13的等比数列.又a2=-43,可得a1=4.所以S10=4[1-(-13)10]1-(-13)=3(1-3-10).故选C.7.设等比数列{an}的前n项和为Sn,若a1=1,S6=4S3,则a4=.答案3解析 S6=4S3,∴q≠1,∴a1\(1-q6\)1-q=4·a1\(1-q3\)1-q,∴q3=3,∴a4=a1·q3=1×3=3.8.记Sn为等比数列{an}的前n项和.设S3=6,S4=a1-3,则公比q=,S4=.答案-1253解析由S3=6,S4=a1-3,得{a1+a1q+a1q2=6,a1q+a1q2+a1q3=-3,解得q=-12,a1=8,则S4=a1-3=8-3=5.9.已知等差数列{an}满足a2=0,a6+a8=-10.(1)求数列{an}的通项公式;(2)求数列an2n-1的前n项和.解(1)设等差数列{an}的公差为d,由已知条件可得{a1+d=0,2a1+12d=-10,解得{a1=1,d=-1.故数列{an}的通项公式为an=2-n,n∈N+.(2)设数列an2n-1的前n项和为Sn,即Sn=a1+a22+…+an2n-1,①Sn2=a12+a24+…+an-12n-1+an2n.②①-②得Sn2=a1+a2-a12+…+an-an-12n-1−an2n=1-12+14+…+12n-1-2-n2n=1-1-12n-1-2-n2n=n2n.所以Sn=n2n-1,n∈N+.关键能力提升练10.在等比数列{an}中,对任意n∈N+,a1+a2+…+an=2n-1,则a12+a22+…+an2等于()A.(2n-...
