2018_2019学年高中数学第二章数列2.5等比数列的前n项和课后作业新人教版.docx

2.5等比数列的前n项和1.等比数列an的前n项和为Sn,且4a1,2a2,a3成等差数列.若a1=1,则S4等于()A.7B.8C.15D.16解析:设等比数列的公比为q,则由4a1,2a2,a3成等差数列,得4a2=4a1+a3,4a1q=4a1+a1q2.q2-4q+4=0.q=2.S4=15.答案:C2.设Sn为等比数列an的前n项和,已知3S3=a4-2,3S2=a3-2,则公比q等于()A.3B.4C.5D.6解析:由题意,得3S3-3S2=(a4-2)-(a3-2),则3a3=a4-a3,即a4=4a3,故q=4.答案:B3.设Sn为等比数列an的前n项和,8a2+a5=0,则等于()A.11B.5C.-8D.-11解析:设an的公比为q.an为等比数列,且8a2+a5=0,8a2+a2q3=0a20,q3=-8.q=-2,=-11.答案:D4.设等比数列an的前n项和为Sn,若=3,则等于()A.B.C.2D.3解析:=3,设S3=k,则S6=3k.又数列an是等比数列,S3=k,S6-S3=2k,S9-S6=4k.S9=k+2k+4k=7k.答案:B5.在等比数列an中,已知a1+a2+an=2n-1,则+等于()A.(2n-1)2B.(2n-1)2C.4n-1D.(4n-1)解析:设等比数列an的前n项和为Sn,则Sn=2n-1.易知等比数列an的公比q=2,首项a1=1,an=2n-1,于是=4n-1,+=1+4+42+4n-1=(4n-1).答案:D6.等比数列an的前n项和为Sn,若S3+3S2=0,则公比q=.解析:由S3=-3S2,可得a1+a2+a3=-3(a1+a2),即a1(1+q+q2)=-3a1(1+q),化简整理得q2+4q+4=0,解得q=-2.答案:-27.已知等比数列an共有2n项,其和为-240,且奇数项的和比偶数项的和大80,则公比q=.解析:根据题意,得S奇=-80,S偶=-160.q=2.答案:28.如果数列an满足a1,a2-a1,a3-a2,an-an-1,是首项为1,公比为2的等比数列,那么an=.解析:an-an-1=a1qn-1=2n-1,即相加得an-a1=2+22+2n-1=2n-2,故an=a1+2n-2=2n-1.答案:2n-19.已知数列an是等比数列,其中a7=1,且a4,a5+1,a6成等差数列.(1)求数列an的通项公式;(2)数列an的前n项和记为Sn,证明:Sn128(n=1,2,3,).(1)解:设等比数列an的公比为q(qR),由a7=a1q6=1,得a1=q-6,从而a4=a1q3=q-3,a5=a1q4=q-2,a6=a1q5=q-1.因为a4,a5+1,a6成等差数列,所以a4+a6=2(a5+1),即q-3+q-1=2(q-2+1),q-1(q-2+1)=2(q-2+1).所以q=.故an=a1qn-1=q-6qn-1=64.(2)证明:Sn=128128.10.数列an的前n项和为Sn,且a1=1,an+1=Sn,nN*,求:(1)a2,a3,a4的值,及数列an的通项公式;(2)a2+a4+a6+a2n的值.解:(1)由a1=1,an+1=Sn,n=1,2,3,得a2=S1=a1=,a3=S2=(a1+a2)=,a4=S3=(a1+a2+a3)=.由an+1-an=(Sn-Sn-1)=an(n2),得an+1=an(n2),又a