6.3 等比數(shù)列 課件-2025屆高三數(shù)學(xué)三輪專項(xiàng)復(fù)習(xí)

6.3等比數(shù)列考點(diǎn)1等比數(shù)列及其前n項(xiàng)和1.等比數(shù)列的定義一般地,如果一個(gè)數(shù)列從第2項(xiàng)起,每一項(xiàng)與它的前一項(xiàng)的比都等于同一個(gè)常數(shù),那么這

個(gè)數(shù)列叫做等比數(shù)列,這個(gè)常數(shù)叫做等比數(shù)列的公比,公比通常用字母q(q≠0)表示.即:

=q(q≠0)(n∈N*).2.等比數(shù)列的通項(xiàng)公式設(shè){an}是首項(xiàng)為a1,公比為q的等比數(shù)列,則通項(xiàng)公式an=a1qn-1.推廣:an=amqn-m,其中,n,m∈N*.3.等比中項(xiàng)如果在a與b中間插入一個(gè)數(shù)G,使a,G,b成等比數(shù)列,那么G叫做a與b的等比中項(xiàng).此時(shí)G2

=ab.4.等比數(shù)列的前n項(xiàng)和公式(1)當(dāng)q=1時(shí),Sn=na1.(2)當(dāng)q≠1時(shí),Sn=

=

.考點(diǎn)2等比數(shù)列的性質(zhì)1.等比數(shù)列的單調(diào)性設(shè)等比數(shù)列{an}的首項(xiàng)為a1,公比為q.(1)當(dāng)q>1,a1>0或0<q<1,a1<0時(shí),數(shù)列{an}為遞增數(shù)列;(2)當(dāng)q>1,a1<0或0<q<1,a1>0時(shí),數(shù)列{an}為遞減數(shù)列;(3)當(dāng)q=1時(shí),數(shù)列{an}是常數(shù)列;(4)當(dāng)q<0時(shí),數(shù)列{an}是擺動(dòng)數(shù)列.2.等比數(shù)列的運(yùn)算性質(zhì)(1)若{an}是等比數(shù)列,且m+n=p+q,則aman=apaq,其中m,n,p,q∈N*.特別地,若2m=p+q,則apaq=

.反之,不一定成立.(2)若{an}是等比數(shù)列,公比為q,則ak,ak+m,ak+2m,…(k,m∈N*)是等比數(shù)列,公比為qm.(3)若數(shù)列{an},{bn}是兩個(gè)項(xiàng)數(shù)相同的等比數(shù)列,則數(shù)列{λan},{

},

,{an·bn}和

(λ≠0,n∈N*)是等比數(shù)列.(4)若三個(gè)數(shù)成等比數(shù)列,則通常設(shè)為

,x,xq;若四個(gè)符號(hào)相同的數(shù)成等比數(shù)列,則通常設(shè)為

,

,xq,xq3.3.等比數(shù)列前n項(xiàng)和(積)的性質(zhì)(1)當(dāng)q≠-1(或q=-1且k為奇數(shù))時(shí),Sk,S2k-Sk,S3k-S2k,…是等比數(shù)列,其公比為qk.(2)若a1·a2·…·an=Tn,則Tn,

,

,…成等比數(shù)列.(3)若數(shù)列{an}的項(xiàng)數(shù)為2n,S偶與S奇分別為偶數(shù)項(xiàng)與奇數(shù)項(xiàng)的和,則

=q;若項(xiàng)數(shù)為2n+1,則

=q.(4)等比數(shù)列{an}的前n項(xiàng)和Sn,可以寫成Sn=Aqn-A(A≠0,q≠1,0).即練即清1.判斷正誤.(正確的打“√”,錯(cuò)誤的打“?”)(1)等比數(shù)列的公比q是一個(gè)常數(shù),它可以是任意實(shí)數(shù).

(

)(2)三個(gè)數(shù)a,b,c成等比數(shù)列的充要條件是b2=ac.

(

)(3)若數(shù)列{an}的通項(xiàng)公式是an=an,則其前n項(xiàng)和為Sn=

.

(

)(4)若數(shù)列{an}為等比數(shù)列,則S4,S8-S4,S12-S8成等比數(shù)列.

(

)××××2.已知等比數(shù)列{an}中,a1=27,a9=

,q<0,則S8=

.3.(人教A版選擇性必修第二冊(cè)P40·T1改編)在等比數(shù)列{an}中,已知a2=6,6a1+a3=30,則an=

.4.(易錯(cuò)題)等比數(shù)列{an}中,a2=4,a6=16,則a4=

.3·2n-1或2·3n-18題型一等比數(shù)列基本量的計(jì)算典例1

(2023天津,6,5分,易)已知數(shù)列{an}的前n項(xiàng)和為Sn.若a1=2,an+1=2Sn+2(n∈N*),則a4

=

(

)A.16

B.32

C.54

D.162C解析

∵an+1=2Sn+2,∴an=2Sn-1+2(n≥2),兩式相減得an+1-an=2an,即an+1=3an(n≥2),當(dāng)n=1時(shí),由已知,得a2=2a1+2=6,則a2=3a1,∴數(shù)列{an}是等比數(shù)列,首項(xiàng)為2,公比為3.∴an=2·3n-1,∴a4=2×33=54,故選C.歸納總結(jié)

等比數(shù)列基本量的運(yùn)算是等比數(shù)列中的一類基本問(wèn)題,等比數(shù)列中有a1,n,q,

an,Sn五個(gè)基本量,一般可以“知三求二”,通過(guò)列方程(組)求基本量.變式訓(xùn)練1-1

(設(shè)問(wèn)條件變式)已知等比數(shù)列{an}的前n項(xiàng)和為Sn,且a1+a2=1,a3+a4=4,則S6

=

(

)A.9

B.16

C.21

D.25C解析

由a1+a2=1,得a3+a4=q2(a1+a2)=q2=4,所以a5+a6=q4(a1+a2)=(q2)2=42=16,所以S6=1+4+16=21,故選C.變式訓(xùn)練1-2已知等比數(shù)列{an}中,a3+a4=40,a3-a5=30,則公比q=

.解析

易知q≠1.由

得

則

題型二等比數(shù)列的判定和證明典例2已知數(shù)列{an}的首項(xiàng)a1=

,且滿足an+1=

.(1)求證:數(shù)列

為等比數(shù)列;(2)若

+

+

+…+

<100,求滿足條件的最大整數(shù)n.解析

(1)證明:

=

=

+

,

-1=

-

,而

-1=

,

=

,所以

是以

為首項(xiàng),

為公比的等比數(shù)列.(2)由(1)可得

+

+

+…+

-n=

=1-

,所以

+

+

+…+

=1-

+n.而1-

+n隨著n的增大而增大,當(dāng)n=99時(shí)滿足題意,當(dāng)n=100時(shí)不合題意,所以滿足條件的最大

整數(shù)n=99.歸納總結(jié)

等比數(shù)列的判定方法定義法若

=q(q為非零常數(shù),n∈N*)或

=q(q為非零常數(shù)且n≥2,n∈N*),則{an}是等比數(shù)列等比中項(xiàng)法若數(shù)列{an}中,an≠0且

=an·an+2(n∈N*),則數(shù)列{an}是等比數(shù)列通項(xiàng)公式法若數(shù)列的通項(xiàng)公式可寫成an=c·qn-1(c,q均是不為0的

常數(shù),n∈N*)的形式,則{an}是等比數(shù)列前n項(xiàng)和公式法若數(shù)列{an}的前n項(xiàng)和Sn=k·qn-k(k為常數(shù)且k≠0,q≠

0,1),則{an}是等比數(shù)列注意

1.只有定義法、等比中項(xiàng)法可用于證明等比數(shù)列,通項(xiàng)公式法與前n項(xiàng)和公式法

只能用于小題中等比數(shù)列的判定.2.證明一個(gè)數(shù)列{an}不是等比數(shù)列,只需要說(shuō)明前三項(xiàng)滿足

≠a1·a3,或者存在一個(gè)正整數(shù)m,使得

≠am·am+2即可.變式訓(xùn)練2-1

(關(guān)鍵元素變式)已知數(shù)列{an}和{bn}滿足a1=1,b1=0,4an+1=3an-bn+4,4bn+1=3bn-an-4.(1)證明:{an+bn}是等比數(shù)列,{an-bn}是等差數(shù)列;(2)求{an}和{bn}的通項(xiàng)公式.解析

(1)證明:由題設(shè)得4(an+1+bn+1)=2(an+bn),即an+1+bn+1=

(an+bn).又因?yàn)閍1+b1=1,所以{an+bn}是首項(xiàng)為1,公比為

的等比數(shù)列.由題設(shè)得4(an+1-bn+1)=4(an-bn)+8,即an+1-bn+1=an-bn+2.又因?yàn)閍1-b1=1,所以{an-bn}是首項(xiàng)為1,公差為2的等差數(shù)列.(2)由(1)知,an+bn=

,an-bn=2n-1.所以an=

[(an+bn)+(an-bn)]=

+n-

,bn=

[(an+bn)-(an-bn)]=

-n+

.題型三等比數(shù)列性質(zhì)的應(yīng)用角度1項(xiàng)的性質(zhì)典例3

(多選)(2025屆山東菏澤曹縣一中開學(xué)考,10)記等比數(shù)列{an}的前n項(xiàng)積為Tn,且

a5,a6∈N*,若T10=65,則a5+a6的可能取值為

(

)A.-7

B.5

C.6

D.7BD解析

∵T10=a1·a2·…·a10=(a5·a6)5=65,∴a5·a6=6,又∵a5,a6∈N*,2×3=6,1×6=6,∴a5+a6=5或7.故選BD.變式訓(xùn)練3-1

(設(shè)問(wèn)條件變式)(2024廣東江門一模,3)已知{an}是等比數(shù)列,a3a5=8a4,且a2,a6是方程x2-34x+m=0的兩根,則m=

(

)A.8

B.-8

C.64

D.-64C解析

因?yàn)閧an}是等比數(shù)列,所以a3a5=

,a2a6=

,又a3a5=8a4,所以a4=8,又a2,a6是方程x2-34x+m=0兩根,所以m=a2a6=

=64.故選C.角度2和的性質(zhì)典例4設(shè)正項(xiàng)等比數(shù)列{an}的前n項(xiàng)和為Sn,且210S30-(210+1)S20+S10=0,則公比q=

.解析

由210S30-(210+1)S20+S10=0,得210(S30-S20)=S20-S10.又正項(xiàng)等比數(shù)列{an}的前n項(xiàng)和為Sn,故S20-S10≠0,∴

=

,∵數(shù)列{an}是等比數(shù)列,∴

=

=q10,故q10=

,解得q=±

,∵{an}為正項(xiàng)等比數(shù)列,∴q>0,故q=

.典例5已知等比數(shù)列{an}共有2n項(xiàng),其和為-240,且奇數(shù)項(xiàng)的和比偶數(shù)項(xiàng)的和大80,則公

比q=

.2解析

由題意,得

解得

所以q=

=

=2.變式訓(xùn)練3-2

(關(guān)鍵元素變式)(2025屆江蘇省鎮(zhèn)江中學(xué)學(xué)情檢測(cè),12)已知{an}是公比為

的等比數(shù)列,若a1+a4+a7+…+a97=100,則a3+a6+a9+…+a99=

.25解析

因?yàn)?/p>

=q2=

,a1+a4+a7+…a97=100,所以a3+a6+a9+…+a99=25.變式訓(xùn)練3-3

(設(shè)問(wèn)條件變式)(2025屆