ページ1:

数列的有关概念
1831.
joj, an = ±² - 13.2"+2, ^EN * 14) 6, -66, -666,66 -..
5.
£ 4. %£
+4-13.2+2=-38
n+1
Ax+1=4 -13.2 +2
12) 450=4"-13-2"+2
-13-2-48=0
12-16) (2+3)=0
2^1^2-3 (4)
n4
∴30是数列的第4页
f. Q₁ = (-1)^-1 & (10 -1) REat
「例4.根据下列数列的递推公式,写出它的前4项,并且内其
通项公式
1) Q₁ = = = a + (2n-1)
a=0
A₁ = a₁+ (2-1)=0+1=1="
03= A₂+ (2x2-1)=1+3=4=2'
A = 2 + (2×3-1)= 9 = 3°
G3 = 04 + (2x4-1)= 16 A
An= (n-1)* REN*
例3.写点下列各数列的一个通项公式
4
3x5
3x7
A: An = 12h+1]/2x+3) hear*
7x9
3,5,9,17 32-
Q = (-1)^
2n+1
nan
=(-1)+
B) 0₁ = 10 = 0₁₁ +2
हो 0121
10.
=>
203
4 ==
30+
=
_L

ページ2:

等差数列与等比数列
Dm-1++
6. On-On- = d (132)
d可以等于0,每一项
On=
(MEN)
a.1-0.=d(d是常数)
都可以加
as
=>2)
q (q)
例1.若0.b.c成等差数列试探究方之也称为学差数:
等差、等比数列定义是证明教列等差、等比的主要依据
的充要条件
.若数列10.j为等差数列,公差为d,cro且CII,则数列
解:∵0.b.c成等差数列
2b=a+c
Ca
40cratch+24C
(6-0)=0
a=c=bt0
and
a.
A₂, 04, 0.,... Aen...
0,0,0,--...
为等差(比)数到
XQX² - 20
(*) aja.. do
(V)求证,这些方程有一个公共报
4. 等差中项与等化页
如果a.b.c房等差数列,那么我是ac的等差中项
zbz atc
如果a.b.c等比数列,那么称b是ac的学比中项
G=±36)
:>= An+Anto
tiles Anx² - (Antlitz) X+On+z = 0
(x-1) (linx-lin+2)=0
4x=1

ページ3:

2) 872 the X = Xn-f=
On+2-An
an -12 On
zd
151210 € | | aj 4 ap=q, aq=p (p+q)
#apreq=
Tap - aq + cp-qld
q=p+ cp-qld
d=-1
: Ap+q= ap + (p+q-p) x(-1)
=9-9=0
-
*j**
Они вы
5.通衣公式及推广的通项公式
Q-Q+ (n-1)d
An=am + (n-m)d
An=01944
例1.有四个数,其中前三个数成等比数列,后三个数成等差数列,
1626x9 => 9 = 27 - 9 = 3
04-02-9-6-354
丑第四冊数的和是16,第二三两数的和是,求这四个数
4 x, y, "y, 16-x
⇒
(y(34-8)(12-y)
例3.若首页为主的等差数列从巷0项起各项都他大口
比数列的公差d.的取值范围是
1 2 (12-4) = y +16=x
x=34-8
ارت
417-8
01>1
+9d>1
=>
=>
1 x=16
1,3, 9, 15 of 16., 8. 4, 0
1318+2
dsto

ページ4:

1例4.公差为0的等差數列50aj4,0,2,jon)的部分例1.若凸多边形的内角依次成等差数列,其剥等于
欢按原来的顺序由小到大床等比数列{angk
k₁ =3ks=
q
#Qk Baka
120,公差等于5°,则多边形内边数是
(n) 180° = n. 12" + (n-1)x5
2-25~+1400
Ab
X = 120 +155718 = 12" + dx = 16"
(2+2d) = 2. (2+10d)
d2-3d=0
d to
dez
032ad
== 24
(12) = 0·4^4
=2.41
== (k₁-1) x 3
2-4534 3(x-1)
kr
3-4
7. 等差数列為等比数列的性质
m+n=p+q santan = ap±aa-
#m+n=ptq of am an =ap.aq
6.等差、等比数列由前顺和
(-)
Sr = nait
n(at)
8.05
(0) 0.+++ak
akakast...+Dak,
Anker + A+++...+ask,....
为等差数列
Sn = (naq=1
1-7971
15) 1. £$2714 5100, Soo-400, BJ S;O=
Art Art - 40. O+ + + + A2, Aust
100
300
x
2·300-100+ X
=>x=500
:. 536=900

ページ5:

*
5
Art Au+-+90=2
+++Q=10
03 + x + - +30= ?
10² = 2-X
x=50
(3)_ D₁ = = Amts = 30m +1
-=3(-a)
-α-30-30
a = 30-20
=30+1
And+3(0+1)
9.由递推公式求通项公式
等差数列可以用递推公式给出:
Q₁ ==+ (NA)
==lp of 0.-9. Ann = ang CRENT)
+ ½ = (+) 3 = (+1) 3^-1
(EN)
On
a=pan+qcpq
[总可以写成: am-a+por-a),待定
in that th
(4) 0₁ =3
例,由下列递推公式术出数列的通项公式
") Q₁ = a +1=3
70-0-3
2-3
Q-2
On = 2+ (n-1) (-3)=-3n+5 VEN*)
G = Gr²+1
数列10]是等差数列
0-01
两边平台
A = A+ (n-1) = 9+ (n-1) × 1 = +8
(2) 0₁ = ± 0 =30x
03
-93
O₁ = -3 (EN)

ページ6:

A
Внос
(5) 0₁ = 0₁₂ 2011 (132)
我到
是等差数列
1= | | + (n-1) d = 2 + (n-1) = 2n
On In CnEN*)
10.05.
16) D₁ =3, O=0 (MEN*)
Igams = 19an
gamgaa
Igen = 1909 (193) - 193
On 3
(MEN)
酸压对数
| 151 2. A. J. A₁ = 2, A = Axt = ?
Al-An= *(**)
A₂ = 0, + ½
03-02 ±
+
Ap=2+
-
=3-(
=2
Cin-(S₁, R=1
()
= On = an On =
| Sn-Su-t, >, 2
Tican+)
例1.已知数列前~项为Sn=
², nent of Q₁ = Sn - Sn-1 =
[2(-1)+(-1)]=-
132
Onet)
1-111
c. Anti-Ons (nt)-폽 - (풍~-품) = 풍
1 : * {aj £d a

ページ7:

Sa., A., as
1900(b
是等差数列
11) 0₁=5₁ = 7
.: S.-S₁= (4+++4)-7=5
03-53-5₁ = (946+4)-12-7
12) 72, RENT A SA-S
D= 7, n=1
ant
anti
n°72-4-[(4-1)*+2(n-1)-4]
24+1
-2)+(24+1)=2 (32, EN)
:.. $2
=-23 ABR-12 An=13R
NS2, MEN³ / 4 BR-124=13
4B-1-12A-₁ = 13 (-1) ₤
0-4(B-B-1)-12 (An-Ax-)=13
4b-12=13
4bn = 12 (-3)+13=-12-5
bn = 124+5 (472, MEN*^)
x446-120=13x1
46₁ = 120₁+13= 12(-5)+8=-30+13=-17
b₁ = -2
• bu= -12=5 (non)
7,2 NEW* /= (O₁ + 2)²
D
-
On: 1 (0 +2) - FC +2)
-
80 m = On + 40n+ 4 - Q-, -40-1 -4
An – Am – 40-41=
(A+) (Q-Q) + (an + An-1)=0
(Anton) (On-On-1-4)=0
Anto
On-On-14
数码j是等差数列
--On-On-420

ページ8:

***
1. 裂低相法体
b-on=d, B
夜
例:知数列anj的通项公式
例:求下列数列前几天和
111, 112, 143, 1724344
ས : ཊ་ ། trk4 5+ ་ ་ 4 དར ། ། ག
Sr = ++++ An
= 10-2 ||+|+|++ (al
An = 10-2×70, Lεnss
S=101/+10/+10/
+10-09-11
36 new of Sail+10+10=+++ ||
=A+++++05-06-07-08---
>(t+tas) lata=+-+Qs) - Q₁ - Q7 ----
(2)
=2(1-1)
A = (2-1) (2143) = (-)
して一主)+(す)が..
- ½ (1 1/2 - 22 - 24 +23)
(43)
(24+2)-(2013)
= 2 0915-257-
Hon
===9x+40
Sn = 19n-n"
ISRES EX
26

ページ9:

5.分奇数,偶数
形如Darkanj的数列,其中10.j为等差数列,后」等比数例:设等差数列汤项为bn=3-2n
列,时可用错位相减求和,等比数列求和为其特例
1=3-
اسلام
203
+
2n+
。་
+++…+25
21-3 21-1
2n-1
0-1
2h-1
++
2n-1
4.例序相加法:如等差数列前n项和公式的推导方法
例:该开区,利用课本中推导等差数列前n项和公式
+(-8) + (-7) +-+for+-+f(8) + f(9)61
PS= fr - 9) + (-1) + fi-6) + + fix. + fist fig)
S = f(9) +f13) +f(7) + ... + (1) + ... + √(-7) + -√(-8)
fort+x+ + + + = + + + +
25=18-1=95
| #Snabbe-b2b3+ b3bq --- + (-1) - baby
| S₁ = bibe - beb3+bshq-~~+br
= bacb₁-b₂+ba (bs-bs)+-+be cban-bat
Alba+be++b)
4 bath)
=(1+3-2)
=-20²+2n
b-b1 [3-21-1)]-[3-2(n-1)]=-4
S₁ = SAY +baba
-2(-1)+2(-1)+(3-2) [3-24
=2m²-2n-1
]an -k4-1
1
奇

ページ10:

Y
6.分段
-232+1 1510
1675
JES, ENS=Art Act an=
-4)+3+3+3
C1-
--100 of 1-
3-310
-100+
(-121-34)
ds, an
300 t, 2
x=-2 o
A-13+(-1)(-2)30
A = 13+ (n-1+1) (-2)5
**
Qn+170
(2(1-4)5- 35 115112
(-)
例2.在数列{0.j中、ニュ(ベーに)取最小時,
12. 等差数列前n项的最值
-dna-din
• 150
dard予定数
当do时,Sa是关于几的二次函数,且常数项为零
San`t bn.
11. 241 19.147, 9₁ = 13, S2 = S₁ #4=
大值时,入的值
30+ d=ait d
80+d=
x13=-2
Sn = 1413+ ^(n-1) (-2) = -n+14^
=-11-71749
(S.) max=49
M=112
公差d20,10.j是递增数列
Ja, soat, nithe ar net

ページ11:

3.一道题
√15]
Sony = Art Art
SF Son = (2n-1) and 10
+ Alan-2 +2m-1
S-1 = 21+ +-+ A₂+A₁
Sony = (Aon-1+Q₁1 (2n-1) = (2n-1) (20n)
1. Sn-1 = (2n-1)Q
差数列値ib前の坂本和Ta
n
S-1 = (2n-1) On_
1. Tany = (2n-1) bn.
On
S
2(2n-1)
bx Tax-t
=
2/27-1)+1
3-1
Skn zn In = km (3n+1), kt
Sn-Soak-2k(-)"
24-1
= 2 (133)
| bij b₂ = 0 -1 (new)
1)术数到100中值最大的和最小的友
| bus - bu = ax = -1 - α==
A1 = 2-
| 2) b₁ = α = = -
b= - +-4-1=^- ^EN") ===
I
An = +1 cment
¥15n53, NENO V
NA, NEN
1=3 (an)=-1
(a) =
Qn.