2.3.2等比数列的前n项和课时过关·能力提升1已知正项等比数列的前5项的和为3,前15项的和为39,则该数列前10项的和为()A.3√2B.3√13C.12D.15解析由题意可知S5,S10-S5,S15-S10成等比数列,即(S10-3)2=3(39-S10).解得S10=12或S10=-9(舍去).答案C2已知在等比数列{an}中,首项为a1,公比q≠1,它的前n项和为M,数列{2an}的前n项和为N,则MN的值为()A.2a12qnB.12a1qn-1C.12a12qn-1D.2a12qn-1解析数列{an}是公比为q的等比数列,数列{2an}是首项为2a1,公比为1q的等比数列,代入等比数列的前n项和公式得MN=12a12qn-1.答案C3已知{an}是首项为1的等比数列,Sn是{an}的前n项和,且9S3=S6,则数列{1an}的前5项和为()A.158或5B.3116或5C.3116D.158答案C4各项均为正数的等比数列{an}的前n项和为Sn,若Sn=2,S3n=14,则S4n等于()A.80B.30C.26D.16解析若q=1,由Sn=na1=2,知S3n=3na1=6≠14,故q≠1.则{Sn=a1(1-qn)1-q=2,S3n=a1(1-q3n)1-q=14.1解得qn=2,a11-q=-2.所以S4n=a11-q(1-q4n)=(-2)×(1-24)=30.答案B5设等比数列{an}的前n项和为Sn,若S6S3=3,则S9S6等于()A.2B.73C.83D.3解析设数列{an}公比为q,由已知可得S6S3=1-q61-q3=1+q3=3,即q3=2.故S9S6=1-q91-q6=1-231-22=73.答案B6已知数列{an}是等比数列,a2=2,a5=14,则a1a2+a2a3+…+anan+1=()A.16(1-4-n)B.16(1-2-n)C.323(1-4-n)D.323(1-2-n)解析由a5=a2q3,可得q=12,a1=a2q=4.又a1a2+a2a3+…+anan+1=a12q+a22q+…+an2q=q(a12+a22+…+an2)=12×16×[1-(14)n]1-14=323(1-4-n).故选C.答案C7设等比数列{an}的公比为q(q≠1),前n项和为Sn,若Sn+1,Sn,Sn+2成等差数列,则q的值为.解析设数列{an}的首项为a1,Sn=a1(1-qn)1-q,2Sn=Sn+1+Sn+2,则有2·a1(1-qn)1-q=a1(1-qn+1)1-q+a1(1-qn+2)1-q,化简得q2+q-2=0.解得q=-2.答案-28设等差数列{an}的前n项和为Sn,则S4,S8-S4,S12-S8,S16-S12成等差数列.类比以上结论有:设等比数列{bn}的前n项积为Tn,则T4,,,T16T12成等比数列.2解析∵b1b2b3b4=T4,T8T4=b5b6b7b8=b1·q4·b2·q4·b3·q4·b4·q4=T4·q16,T12T8=T4·q32,T16T12=T4·q48,故T4,T8T4,T12T8,T16T12成等比数列.答案T8T4T12T8★9在等比数列{an}中,若a1=12,a4=-4,则公比q=;|a1|+|a2|+…+|an|=.解析∵数列{an}为等比数列,且a1=12,a4=-4,∴q3=a4a1=-8,∴q=-2,∴an=12·(-2)n-1,∴|an|=2n-2.∴|a1|+|a2|+|a3|+…+|an|=12(1-2n)1-2=12(2n-1)=2n-1-12.答案-22n-1-1210已知数列{an}为等比数列,且a3+a6=36,a4+a7=18.(1)若an=12,求n;(2)设数列{an}的前n项和为Sn,求S8.解(1)设an=a1qn-1,由题意得{a1q2+a1q5=36,a1q3+a1q6=18,解得{a1=128,q=12,故an=128·(12)n-1.由an=128·(12)n-1=12,解得n=9.(2)Sn=a1(1-qn)1-q=256[1-(12)n],故S8=256×[1-(12)8]=255.11设数列{an}是公比为q的等比数列,推导数列{an}的前n项和公式.解设数列{an}的前n项和为Sn,当q=1时,Sn=a1+a1+…+a1=na1;当q≠1时,Sn=a1+a1q+a1q2+…+a1qn-1,①qSn=a1q+a1q2+…+a1qn,②由①-②得,(1-q)Sn=a1-a1qn,3∴Sn=a1(1-qn)1-q,∴Sn={na1,q=1,a1(1-qn)1-q,q≠1.★12已知a1=2,点(an,an+1)在函数f(x)=x2+2x的图象上,其中n=1,2,3,…(1)证明:数列{lg(1+an)}是等比数列;(2)设Tn=(1+a1)(1+a2)…(1+an),求Tn.(1)证明∵点(an,an+1)在函数f(x)=x2+2x的图象上,∴an+1=an2+2an,∴1+an+1=(1+an)2.又a1=2,∴1+a1=3,∴1+an≥3,lg(1+an)>0,∴lg(1+an+1)=lg(1+an)2=2lg(1+an),∴lg(1+an+1)lg(1+an)=2,∴数列{lg(1+an)}是首项为lg3,公比为2的等比数列.(2)解由(1)知lg(1+an)=2n-1lg3=lg32n-1,∴1+an=32n-1,∴Tn=(1+a1)(1+a2)…(1+an)=31×32×…×32n-1=31+2+…+2n-1=31-2n1-2=32n-1.4
