6.2 等差數(shù)列 課件-2025屆高三數(shù)學(xué)三輪專項(xiàng)復(fù)習(xí)

6.2等差數(shù)列考點(diǎn)1等差數(shù)列及其前n項(xiàng)和1.等差數(shù)列相關(guān)概念(1)定義:一般地,如果一個(gè)數(shù)列從第二項(xiàng)起,每一項(xiàng)與它的前一項(xiàng)的差都等于同一個(gè)常

數(shù),那么這個(gè)數(shù)列就叫做等差數(shù)列,這個(gè)常數(shù)叫做等差數(shù)列的公差,公差通常用字母d表

示.(2)等差中項(xiàng):由三個(gè)數(shù)a,A,b組成等差數(shù)列可以看成是最簡(jiǎn)單的等差數(shù)列.這時(shí)A叫做a

與b的等差中項(xiàng),根據(jù)等差數(shù)列的定義知2A=a+b.(3)通項(xiàng)公式:an=a1+(n-1)d(n∈N*).2.等差數(shù)列的前n項(xiàng)和公式(1)等差數(shù)列的前n項(xiàng)和公式:Sn=

=na1+

d.(2)等差數(shù)列的前n項(xiàng)和公式與函數(shù)的關(guān)系:由Sn=na1+

d可得Sn=

n2+

n,設(shè)a=

,b=a1-

,則Sn=an2+bn.(3)最值:若a1>0,d<0,則Sn存在最大值;若a1<0,d>0,則Sn存在最小值.考點(diǎn)2等差數(shù)列的性質(zhì)1.等差數(shù)列的常用性質(zhì)(1)通項(xiàng)公式的推廣:an=am+(n-m)d(n,m∈N*).(2)若{an}是等差數(shù)列,且k+l=m+n(k,l,m,n∈N*),則ak+al=am+an;反之,不一定成立.(3)若{an},{bn}是等差數(shù)列,則{pan+qbn}(p,q是常數(shù))也是等差數(shù)列.(4)若{an}是等差數(shù)列,則ak,ak+m,ak+2m,…(k,m∈N*)組成公差為md的等差數(shù)列.(5)當(dāng)d>0時(shí),數(shù)列{an}為遞增數(shù)列;當(dāng)d<0時(shí),數(shù)列{an}為遞減數(shù)列;當(dāng)d=0時(shí),數(shù)列{an}為常

數(shù)列.(6)若已知數(shù)列{an}的通項(xiàng)公式是an=pn+q(其中p,q為常數(shù)),則數(shù)列{an}一定是等差數(shù)列,

且公差為p.2.與等差數(shù)列前n項(xiàng)和有關(guān)的性質(zhì)(1)若{an}是等差數(shù)列,其前n項(xiàng)和為Sn,則

也是等差數(shù)列,其首項(xiàng)與{an}的首項(xiàng)相同,公差是{an}的公差的

.(2)若{an}是等差數(shù)列,Sm,S2m,S3m分別為{an}的前m項(xiàng),前2m項(xiàng),前3m項(xiàng)的和,則Sm,S2m-Sm,S3m-S2m成等差數(shù)列.(3)非零等差數(shù)列奇數(shù)項(xiàng)和與偶數(shù)項(xiàng)和的性質(zhì):①若項(xiàng)數(shù)為2n,則S偶-S奇=nd,

=

.②若項(xiàng)數(shù)為2n-1,則S偶=(n-1)an,S奇=nan,S奇-S偶=an,

=

.(4)若兩個(gè)等差數(shù)列{an}、{bn}的前n項(xiàng)和分別為Sn、Tn,則

=

.(5)若數(shù)列{an}的前n項(xiàng)和Sn=An2+Bn(A,B為常數(shù)),則數(shù)列{an}是等差數(shù)列,且公差為2A.即練即清1.判斷正誤.(正確的打“√”,錯(cuò)誤的打“?”)(1)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)的差都是常數(shù),那么這個(gè)數(shù)列是等差

數(shù)列.

(

)(2)數(shù)列{an}為等差數(shù)列的充要條件是對(duì)任意n∈N*,都有2an+1=an+an+2.

(

)(3)在等差數(shù)列{an}中,若am+an=ap+aq(m,n,p,q∈N*),則m+n=p+q.

(

)(4)等差數(shù)列的前n項(xiàng)和公式是常數(shù)項(xiàng)為0的二次函數(shù).

(

)×√××2.在等差數(shù)列{an}中,a4+a8=10,a10=6,則公差d=

(

)A.

B.

C.2

D.-

3.設(shè)Sn是等差數(shù)列{an}的前n項(xiàng)和,若a1+a3+a5=3,則S5=

.4.在等差數(shù)列{an}中,若a3+a7=10,a6=7,則數(shù)列{an}的通項(xiàng)公式為an=

,前n項(xiàng)和

為Sn=

.5.(易錯(cuò)題)在等差數(shù)列{an}中,a1=7,公差為d,前n項(xiàng)和為Sn,當(dāng)且僅當(dāng)n=8時(shí),Sn取得最大值,

則d的取值范圍為

.A52n-5n2-4n題型一等差數(shù)列基本量的計(jì)算典例1

(2025屆湖南長(zhǎng)沙六校大聯(lián)考,3)等差數(shù)列{an}(n∈N*)中,a2=10,a7-a4=2a1,則a7=

(

)A.40

B.30

C.20

D.10B解析

設(shè)等差數(shù)列{an}(n∈N*)的公差為d,由a7-a4=2a1,得3d=2a1,由a2=10,得a1+d=a1+

a1=10,解得a1=6,d=4,所以a7=a1+6d=6+24=30.故選B.歸納總結(jié)

等差數(shù)列的通項(xiàng)公式及前n項(xiàng)和公式涉及a1,an,d,n,Sn五個(gè)量,知道其中三個(gè)

就能求另外兩個(gè)(簡(jiǎn)稱“知三求二”),通常利用條件和通項(xiàng)公式、前n項(xiàng)和公式建立方

程(組)求解.變式訓(xùn)練1-1

(設(shè)問(wèn)條件變式)(2025屆廣東七校聯(lián)考,3)在等差數(shù)列{an}中,已知a1=-9,a3

+a5=-9,a2n-1=9,則n=

(

)A.7

B.8

C.9

D.10A解析

設(shè)等差數(shù)列{an}的公差為d,由a1=-9,a3+a5=2a1+6d=-9,則d=

,所以a2n-1=-9+(2n-2)×

=9,故n=7.題型二等差數(shù)列的判定與證明典例2

(2021全國(guó)甲文,18,12分)記Sn為數(shù)列{an}的前n項(xiàng)和,已知an>0,a2=3a1,且數(shù)列

{

}是等差數(shù)列.證明:{an}是等差數(shù)列.證明

設(shè)等差數(shù)列{

}的公差為d,由題意得

=

,

=

=

=2

,則d=

-

=2

-

=

,所以

=

+(n-1)

=n

,所以Sn=n2a1①,當(dāng)n≥2時(shí),有Sn-1=(n-1)2a1②.由①-②,得an=Sn-Sn-1=n2a1-(n-1)2a1=(2n-1)a1③,經(jīng)檢驗(yàn),當(dāng)n=1時(shí)也滿足③.所以an=(2n-1)a1,n∈N*,當(dāng)n≥2時(shí),an-an-1=(2n-1)a1-(2n-3)a1=2a1,所以數(shù)列{an}是首項(xiàng)為a1,公差為2a1的等差數(shù)列.歸納總結(jié)

等差數(shù)列判定與證明的方法方法解讀適合題型定義法對(duì)于任意自然數(shù)n(n≥2),an-an-1為

同一常數(shù)?{an}是等差數(shù)列解答題中的證明問(wèn)題等差中項(xiàng)法2an-1=an+an-2(n≥3,且n∈N*)成立?

{an}是等差數(shù)列通項(xiàng)公式法an=pn+q(p,q為常數(shù))對(duì)任意的正

整數(shù)n都成立?{an}是等差數(shù)列選擇題、填空題中的判定問(wèn)題前n項(xiàng)和公式法Sn=An2+Bn(A,B是常數(shù))對(duì)任意的正整數(shù)n都成立?{an}是等差數(shù)列變式訓(xùn)練2-1

(關(guān)鍵元素變式)(2025屆山東省部分學(xué)校開(kāi)學(xué)聯(lián)合教學(xué)質(zhì)量檢測(cè),15)已

知數(shù)列{an}的首項(xiàng)為a1=

,且滿足an+1+4an+1an-an=0.(1)證明:數(shù)列

為等差數(shù)列;(2)設(shè)數(shù)列

的前n項(xiàng)和為Sn,求數(shù)列{(-1)nSn}的前n項(xiàng)和.解析

(1)證明:因?yàn)閍n+1+4an+1an-an=0,所以an-an+1=4anan+1,若an+1an=0,則an=an+1=0,與a1=

矛盾,所以an+1an≠0,所以

-

=4,因?yàn)閍1=

,所以

=2,所以數(shù)列

是首項(xiàng)為2,公差為4的等差數(shù)列.(2)由(1)知

=2+(n-1)·4=4n-2,數(shù)列

的前n項(xiàng)和為Sn=

=2n2,所以(-1)nSn=2(-1)nn2,設(shè)數(shù)列{(-1)nSn}的前n項(xiàng)和為Tn,當(dāng)n為偶數(shù)時(shí),Tn=2[-12+22-32+…-(n-1)2+n2],因?yàn)閚2-(n-1)2=2n-1,所以Tn=2[3+7+…+(2n-1)]=2·

·

=n(n+1)=n2+n.當(dāng)n為奇數(shù)時(shí),n-1為偶數(shù).Tn=Tn-1+2·(-1)nn2=(n-1)n-2n2=-n2-n,所以Tn=

題型三等差數(shù)列性質(zhì)的應(yīng)用角度1項(xiàng)的性質(zhì)典例3

(2025屆重慶聯(lián)考,3)設(shè)等差數(shù)列{an}的前n項(xiàng)和為Sn,且a3+a11-a5=4,則S17=

(

)A.58

B.68

C.116

D.136B解析

設(shè)公差為d.由等差數(shù)列的性質(zhì)可得a3+a11=2a7,代入已知式子可得2a7-a5=4,∴2(a5+2d)-a5

=4,∴a5+4d=4,即a9=4,∴S17=

=

=17a9=68.故選B.歸納總結(jié)

(1)應(yīng)用等差數(shù)列的性質(zhì)解題時(shí),要靈活應(yīng)用等差數(shù)列通項(xiàng)公式及前n項(xiàng)和

公式;(2)需注意性質(zhì)成立的前提條件.變式訓(xùn)練3-1

(設(shè)問(wèn)條件變式)數(shù)列{an}是等差數(shù)列,若a3a9=8,

+

=

,則a6=

(

)A.2

B.4

C.

D.

C解析

+

=

=

=

=

,故a6=

.故選C.變式訓(xùn)練3-2已知等差數(shù)列{an}是遞增數(shù)列,且滿足a3+a5=14,a2a6=33,則a1a7等于

(

)A.33

B.16

C.13

D.12C解析

設(shè)公差為d.由等差數(shù)列的性質(zhì)得,a2+a6=a3+a5=14,又a2a6=33,解得

或

又{an}是遞增數(shù)列,所以

∴d=

=2,∴a1a7=(a2-d)(a6+d)=(3-2)×(11+2)=13.故選C.角度2和的性質(zhì)典例4

(2025屆浙江省名校第一次聯(lián)考,3)已知等差數(shù)列{an}的前n項(xiàng)和為Sn,若

=

,則

=(

)A.

B.

C.

D.

D解析

=

=

=

×

=

,故選D.典例5

(2025屆浙江省強(qiáng)基聯(lián)盟聯(lián)考,5)已知等差數(shù)列{an}的前n項(xiàng)和為Sn,

-

=2,a1=1,則a4=

(

)A.4

B.5

C.6

D.7D解析

設(shè)等差數(shù)列{an}的公差為d,則

是公差為

的等差數(shù)列.又

-

=d=2,a1=1,∴an=2n-1,∴a4=7,故選D.變式訓(xùn)練3-3

(關(guān)鍵元素變式)(2025屆河北邯鄲第一次調(diào)研,3)已知Sn為等差數(shù)列{an}

的前n項(xiàng)和,且

=

,則

=

(

)A.3

B.2

C.

D.

B解析

=

=

=

,所以

=2.變式訓(xùn)練3-4

(2024湖北武漢外國(guó)語(yǔ)學(xué)校校考,14)設(shè)等差數(shù)列{an}、{bn}的前n項(xiàng)和分

別為Sn、Tn,若對(duì)任意的n∈N*,都有

=

,則

+

=

.解析

+

=

+

=

=

=

,

=

=

=

=

=

.題型四等差數(shù)列的前n項(xiàng)和的最值典例6

(2022全國(guó)甲,理17,文18,12分)記Sn為數(shù)列{an}的前n項(xiàng)和.已知

+n=2an+1.(1)證明:{an}是等差數(shù)列;(2)若a4,a7,a9成等比數(shù)列,求Sn的最小值.解析

(1)證明:由已知條件

+n=2an+1可得,2Sn=2nan+n-n2①,當(dāng)n≥2時(shí),由①可得2Sn-1=2(n-1)·an-1+(n-1)-(n-1)2②,由an=Sn-Sn-1及①-②可得,(2n-2)an-(2n-

2)an-1=2n-2,n≥2,且n∈N*,即an-an-1=1,因此{(lán)an}是等差數(shù)列,公差為1.(2)∵a4,a7,a9成等比數(shù)列,且{an}是以1為公差的等差數(shù)列,∴(a1+3×1)(a1+8×1)=(a1+6×1)2,即(a1+3)(a1+8)=(a1+6)2,解得a1=-12.解法一:數(shù)列{an}是首項(xiàng)為-12,公差為1的等差數(shù)列,∴Sn=n×(-12)+

×1=

,∴當(dāng)n=12或n=13時(shí),Sn取最小值,為

=-78.解法二:an=-12+(n-1)×1=n-13,當(dāng)n≤12且n∈N*時(shí),an<0;當(dāng)n=13時(shí),an=0;當(dāng)n≥14且n∈N*時(shí),an>0.∴當(dāng)n=12或n=13時(shí),Sn取最小值,為-12×12+

×1=-78.歸納總結(jié)求等差數(shù)列前n項(xiàng)和Sn的最值的三種方法(1)二次函數(shù)法:利用Sn=An2+Bn(A≠0,B為常數(shù)),通過(guò)配方或借助二次函數(shù)的圖象求最

值,注意n為正整數(shù).(2)通項(xiàng)公式法:①當(dāng)a1>0,d<0時(shí),滿足

的項(xiàng)數(shù)m使得Sn取得最大值Sm;②當(dāng)a1<0,d>0時(shí),滿足

的項(xiàng)數(shù)m使得Sn取得最小值Sm.(3)不等式組法:借助當(dāng)Sn最大時(shí),有

(n≥2,n∈N*),解此不等式組確定n的范圍,進(jìn)而確定n的值和對(duì)應(yīng)Sn的值(即Sn的最大值),類似可求最小值.變式訓(xùn)練4-1

(設(shè)問(wèn)條件變式)(2025屆河北衡水檢測(cè),5)已知等差數(shù)列{an}的公差小于

0,前n項(xiàng)和為Sn