高考數(shù)學(xué) 考點(diǎn)匯總 考點(diǎn)23 等比數(shù)列及其前n項(xiàng)和(含解析)
高考數(shù)學(xué) 考點(diǎn)匯總 考點(diǎn)23 等比數(shù)列及其前n項(xiàng)和(含解析)一、 選擇題1. (xx·天津高考文科·5)設(shè)是首項(xiàng)為,公差為的等差數(shù)列,為其前n項(xiàng)和,若成等比數(shù)列,則=( )A.2 B.-2 C. D.【解析】選D.因?yàn)槌傻缺葦?shù)列,所以即,解得2. (xx·新課標(biāo)全國卷高考文科數(shù)學(xué)·T5)等差數(shù)列的公差為2,若a2,a4,a8成等比數(shù)列,則的前n項(xiàng)和Sn= ()A.n(n+1) B.n(n-1) C. D. 【解題提示】利用a2,a4,a8成等比數(shù)列求得公差,然后利用等差數(shù)列求和公式求和.【解析】選A.因?yàn)閐=2,a2,a4,a8成等比,所以=a2a8,即(a2+2d)2=a2(a2+6d),解得a2=4,a1=2.所以利用等差數(shù)列的求和公式可求得Sn=n(n+1).二、填空題3.(xx·廣東高考文科·T13)等比數(shù)列an的各項(xiàng)均為正數(shù),且a1a5=4,則log2a1+log2a2+log2a3+log2a4+log2a5=.【解析】方法一:各項(xiàng)均為正數(shù)的等比數(shù)列an中a1a5=a2a4=4,則a1a2a3a4a5=25,log2a1+log2a2+log2a3+log2a4+log2a5=log2a1a2a3a4a5=log225=5.方法二:各項(xiàng)均為正數(shù)的等比數(shù)列an中a1a5=a2a4=4,設(shè)log2a1+log2a2+log2a3+log2a4+log2a5=S,則log2a5+log2a4+log2a3+log2a2+log2a1=S,2S=5log2(a1a5)=10,S=5.答案:54.(xx·廣東高考理科)若等比數(shù)列an的各項(xiàng)均為正數(shù),且a10a11+a9a12=2e5,則lna1+lna2+lna20=.【解析】各項(xiàng)均為正數(shù)的等比數(shù)列an中a10a11=a9a12=a1a20,則a1a20=e5,lna1+lna2+lna20=ln(a1a20)10=lne50=50.方法二:各項(xiàng)均為正數(shù)的等比數(shù)列an中a10a11=a9a12=a1a20,則a1a20=e5,設(shè)lna1+lna2+lna20=S,則lna20+lna19+lna1=S,2S=20ln(a1a20)=100,S=50.答案:50【誤區(qū)警示】易算錯(cuò)項(xiàng)數(shù)和冪次,要充分利用等比數(shù)列的性質(zhì).5. (xx·天津高考理科·11)設(shè)是首項(xiàng)為,公差為-1的等差數(shù)列,為其前項(xiàng)和.若成等比數(shù)列,則的值為_.【解析】因?yàn)樗?,即,解?【答案】6.(xx·安徽高考理科·12)數(shù)列是等差數(shù)列,若,構(gòu)成公比為的等比數(shù)列,則_.【解題提示】 求出等差數(shù)列的公差即可用表示出等比數(shù)列的三項(xiàng),即可計(jì)算出公比?!窘馕觥吭O(shè)等差數(shù)列的公差為d,則,即,解得d=-1,所以,所以.答案:1三、解答題7.(xx·福建高考文科·17)17.(本小題滿分12分)在等比數(shù)列中,.(1) 求;(2) 設(shè),求數(shù)列的前項(xiàng)和.【解題指南】(1)利用等比數(shù)列通項(xiàng)公式求出首項(xiàng)和公比(2)由求出的通項(xiàng)公式,為等差數(shù)列,利用等差數(shù)列前n項(xiàng)和公式求前n項(xiàng)和【解析】(1)設(shè)的公比為q,依題意得, 解得,因此,.(2)因?yàn)?,所以?shù)列的前n項(xiàng)和.8. (xx·天津高考文科·20)(xx·天津高考理科·19)(本小題滿分14分)已知和均為給定的大于1的自然數(shù),設(shè)集合,集合,(1) 當(dāng)時(shí),用列舉法表示集合A;設(shè)其中證明:若則.【解析】(1)當(dāng)q=2,n=3時(shí),M=0,1,A=x|x=x1+x2·2+x3·22,xiM,i=1,2,3.可得,A=0,1,2,3,4,5,6,7.(2)由s,tA,s=a1+a2q+anqn-1,t=b1+b2q+bnqn-1,ai,biM,i=1,2,n及an<bn,可得s-t=(a1-b1)+(a2-b2)q+(an-1-bn-1)qn-2+(an-bn)qn-1(q-1)+(q-1)q+(q-1)qn-2-qn-1= - qn-1=-1<0.所以s<t.