Đại số và Giải tích 11 nâng cao và hướng dẫn thiết kế bài giảng (Tập 2): Phần 1

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THIET KE BAI GIANG

DAI SO VA GIAI TICH
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It:

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NHAXUATBANHANOI


THIET KE BAI GIANG
DAI s d VA GIAI TICH 11 - NANG CAO - TAP HAI
TRAN VINH

N H A XUAT B A N H A NOI

Chiu trdch nhiim xudt bdn :
NGUYEN K H A C O A N H
Biin tap :
PHAM QUOC TUAN
ViBia .TAO

THANH H U Y E N
Trinh bdy :

QUYNH TRANG
Sica bdn in :
PHAM QUOC TUAN

In 1000 cuon, tai Xf nghiep In ACS Viet Nann.
Km 10 - Dudng Pham Van Dong - Kien Thuy - [Hai Phong.
Giay phep xuat ban so: 208 -2007/CXB/46 m TK - 47/HN.
In xong va nop luu chieu qu^ I nam 2008.


Cki/ONq III

DAY SO.
CAP SO CONG VA CAP SO NHAIN
Phan 1

^mUrn^ VAX D^ CUA CHUWSTG
I. NOI DUNG
Noi dung chinh cua chuung III:
Phuong phap quy nap toan hoc: Dinh nghia, cac bu6c chiing minh bang phuong phap
quy nap.

Day so: Dinh nghla day so la gi ? Day sd hihi han va day s6' yo han, cong thiic
tdng quat ciia day sd, cac phuong phap cho day sd, day sd tang, day sd giam va
day sd hi chan.
• Qip sd cdng : Dinh nghia, cdng sai, sd hang tdng qiiat, tinh ch&'t cac sd hang,
tdng n sd hang dSu tien cua cap sd cdng.
• Cap sd nhan : Dinh nghTa, cdng bdi, sd hang tdng quat, tinh chat cac sd hang,
tdng n sd hang dau tien ciia ca'p sd nhan.
II. MUC TIEU
1. Kien thiirc
Nam duoc toan bd Icien thiic co ban trong chuong da neu tren, cu the :
- Biet chiing minh va nhan bie't khi nao sit dung phuang phap quy nap toan hoc.
Bidt tim cac sd hang tdng quat cua day sd; Chiing minh duoc day sd la day sd
tang, giam, day sd bi chan.
• Nam duoc Ichai niem va each nhan bie't mot day sd la ca'p sd cdng ; tim duoc
sd hang tdng quat va tinh tdng n sd hang dau tien cua mot ca'p sd cdng.


Nam duoc khai niem va each nhan bidt mot day sd la ca'p sd nhan ; tim duoc
sd hang tdng quat va tmh tdng n sd hang dSu tien cua mot ca'p sd nhan.
2. Kinang
Vun dung cac budc quy nap de chiing minh bai toan bang phuong phap quy
nap toan hoc.
van dung thanh thao c^c tinh chat cua ca'p sd cdng va cap sd nhan trong giai
toan.
Bie't each cho mot day sd, each khao sat tmh tang, giam ciia cac day sd dcfn gian.
Nhan bie't duoc cap sd cdng, cap sd nhan ; bie't each tim so hang tdng quat va
tdng n sd hang d^u tien ciia cac ca'p sd do trong cac trudng hop khdng phiic tap ;
Bie't van dung nhiing kie'n thiic trong chuong de giai quye't cac bai toan cd lien
quan dugc dat ra d cac mdn hoc khac, cung nhu trong thuc tiln cude sd'ng.
3. Thai do

• Tu giac, tfch cue, ddc lap va chii ddng phat hien cung nhu ITnh hdi kie'n thUc
trong qua trinh hoat ddng.
• Can than chinh xac trong lap liian va tinh toan.
Cam nhan dugc thuc teciia toan hoc, nha't la dd'i vdi day sd.


Pban 2
CAC B A I S O A ^

§1. Phu'cfng^ phap q u y n a p t o a n h o c
(tie't 1, 2)
I. MUC TIEU
1. Kien thtic
HS nam dugc :
• Phuong phap va cac budc chiing minh quy nap.
Khi nao thi van dung phuong phap quy nap.
Giai thich dugc phuong phap quy nap.
2. KT nang
Van dung thanh thao phuong phap quy nap trong giai toan.
Bie't them mdt phuong phap chiing minh dd'i vdi bai toan cd lien quan
den sd tu nhien.
3. Thai dp
Tu giac, tich cue trong hgC tap.
« Biet phan biet rd cac Idiai niem co ban va van dung trong timg trudng hop cu
the.
.
- Tu duy cac va'n de ciia toan hgc mdl each logic va he thd'ng.
II. CHUAN BI CUA GVVA HS
1. Chuan bj ciia GV
Chuan bi cac cau hdi ggi nid.

Chuan bi cdc vi du sinh dgng.
• Chudn bi pha'n mau, va mdt sd dd dung khac.
2. Chuan bi ciia HS


- Can dn lai mgt sd kie'n thiic da hgc ve sd tu nhien d ldp dudi.
IIL PHAN PHOI

TH6I

LUONG

Bai nay chia lam 2 tie't :
Tii't 1 : Tix ddu din hit vi du 1.
Tii't 2 : Tiep theo den het phdn bdi tap.
IV- TIEN TRINH DAY - HOC

A. DAT VANDE
Cau hdi 1
Xet tmh diing - sal cua cac cau sau day :
a)Ne'ua>bthia" >'b"
b)Ne'ua>b> 1 thia" > b",
Cau hoi 2
Cho cac menh 6i sau :
a) Sd nguyen duong le ldn hon 1 la so nguyen to.
b) 1 + 2 + 3 +,.. + n = "^""^^\ n e N.
2
Hay xem xet tinh diing sai cua cac menh de tren vdi 3 sd hang dSu tien.

B. BAIMdl

, HOATDONCl
1. Phuong phap quy nap toan hoc
« GV dat va'n de :
HI. Hay phat bi^u mdt vai menh de chiia bie'n tu nhien A(n).
• Sau dd GV neu bai toan trong SGK.
Chdng minh rang ydi mgi so nguyen duang n, ta ludn cd
1.2 + 2.3 + ... t n(n + 1) =

x

:


• Thuc hien [HI] trong 4'
Hoat dong cua HS

Hoat dong cua GV

Ggi y tra Idi cau hoi 1

Cau hdi I
Hay kiem chiing khi n = 1.
Cau hoi 2

Vdi n = 1 ta cd dang thiic luon
ludn diing.
G g i y tra ldi cau hoi 2

Cd thd kiem tra dang thiJc (1) Khdng thd.
vdi mgi n dugc khdng?

• GV neU cac budc quy nap ;
• Bu&c 1 (budc ca sd, hay budc khdi ddu). Chieng minh A(n) la mpt
menh de diing khi n = 1.
• Bu&c 2 (budc quy ngp, hay budc "di truyen"). Vdi k Id mpt sd
nguyen duang tuy y, xudt phdt td gid thii't A(n) Id mpt menh de diing
khi n = k, chimg minh A(n) cdng Id mpt minh de diing khi n = k + 1.
Ngudi ta ggi phuang phdp chimg minh viCa niu tren Id phuang phdp
quy ngp todn hgc (hay cdn ggi tdt Id phuang phdp quy ngp). Gid thie't
dugc noi tai a budc 2 ggi Id gid thie't quy ngp.
• GV dua ra mdt sd cau hdi ciing cd :
H2. Hay giai thich tai saO phep chiing minh bang quy nap la diing.
H3. Phep chiing minh bang quy nap cd ap dung cho bai toan chiing minh menh de
A(x)ba'tki.
HOATDONC 2
2. Vi du ap dung
• Thuc hien vi du 1 trong 5 phiit.
Hoat ddng cua GV
Cau hoi 1

Hoat dong cua H S
Ggi y tra Idi cau hoi 1

Xet tinh dung sai ciia cong Ta tha'y n = 1, cdng thiic tren ludn
thiic vdi n = 1.
ludn diing.


Cau hoi 2

Ggi y tra Idi cau hoi 2

Gia sii cdng thiic diing n = k
hay thie't lap cdng thiic.
Cau hoi 3

4
Ggi y tra Idi cau hoi 3

l^ + 2^ + ... + k^ + ik+l)^ =
Hay thie't lap cdng thiic khi
n = k + 1 va chiing minh cdng
(k + 1)2 (it + if
thiic do.
4 ';
HS tu chiing minh tie'p.
Thuc hien 1H2J trong 5'.
Hoat dgng cua GV
Cau hoi 1

Hoat dgng cua HS
Ggi y tra Idi cau hoi 1

Xet tfnh diing sai ciia cdng Ta tha'y n = 1, cdng thiic tren ludn
thiic vdi n = 1.
ludn diing.
Cau hoi 2
Ggi y tra Idi cau hoi 2
Gia sii cdng thiic diing n = k
HS tu thie't lap.
hay thie't lap cdng thiic.

Cau hoi 3
Ggi y tra Idi cau hoi 3
Hay thie't lap cdng thiic khi
l + 3 + . . . + (2)t-l) + (2^+l)
n = k + 1 va Chiing minh cdng
thiic do.
= (^+lf.
HS tu chiing minh.
• Thuc hien [H3J trong 5'
Hoat dgng cua GV
Cau hoi 1

Hoat dong cua HS
Goi y tra ldi cau hoi 1

Xdt tihh diing sai cua cdng Vdin= 1, tacd 12 = 1
thiic vdi n = 1.
1(4.1'-1)
3


Nhu vay cdng thiic diing khi « = 1.
Cau hoi 2

Ggi y tra ldi cau hoi 2

Gia six cdng thiic diing n = k HS tu thie't lap.
hay thie't lap cdng thiic.
Ggi y tra Idi cau hoi 3
Cau hdi 3

l^ + 3^ + ... + (2k-lf + {2(,k+l)-\f
Hay thie't lap cdng thiic khi
n = k + 1 va chiing minh cdng
_{k + \)[4ik + \y--l)
thiic dd.
3
HS tu chiing minh.
• GV neu chu y trong SGK:

Trong thuc te, ta cdn gap cdc bdi todn vdi yiu cdu chieng minh minh
de chica bie'n A(n) Id mpt minh de dung vdi mgi gid tri nguyen dicang
n >p, trong dd p Id mpt sd nguyin duang cho tricdc. Trong triCdng hgp
ndy, de gidi quye't bdi todn ddt ra bang phuang phdp quy ngp, a budc
I ta cdri chimg minh A(n) Id menh de ddng khi n = pvd a budc 2, cdn
xet gid thiet quy ngp vdi k Id so nguyin duang tuy y ldn han hodc bdng p
' Thuc hien vf du 2 trong 5 phiit;
Hoat ddng cua GV
Cau hoi 1

Hoat ddng cua HS
Ggi y tra ldi cau hoi 1

Xet tfnh diing sai ciia cdng Ta thay n = 1, cdng thiic tren ludn
thiic vdi n = 3.
ludn diing.
Cau hdi 2
Ggi y tra Idi cau hdi 2
Gia sir cong thiic diing n = k
hay thidt lap cdng thiic theo k. 2*' > 2/: 4- 1.
Cau hoi 3

Ggi y tra Idi cau hoi 3

Hay thie't lap va chiing minh
2''^' > 2{k+\)+ 1.
cdng thiic voi n = k + I.
HS chiing minh tiep.


nOATDONCB

TOM TfiT Bfll HOC
1. Cac budc quy nap toan hgc.
• Budc 1 {budc CO so, hay bicdc khdi ddu). ChUng minh A{n) la mdt menh de diing
khi

n-\.

• Budc 2 {budc quy ngp, hay buac "di truyen"). Vdi k la mdt sd nguydn duong tuy
y, xua't phat tir gia thie't A{n) la mdt menh de diing khi « = A:, chiing minh A{n)
cung la mdt menh di diing khi n = k+ i.
Ngudi ta ggi phuong phap chiing minh viia neu trdn la phuang phdp cpiy ngp todn
hgc (hay cdn ggi tat la phuang phdp quy ngp). Gia thiet dugc ndi tdi d budc 2 ggi
la gid thie't quy ngp.
HOATDONC 4

MOT SO Cfia HOI TRfiC NGHIEM ON TfiP Bfil 1
Cdu I. Thdng thudng, trong phuong phap quy nap toan hgc ngudi ta chia sd budc
la:
(a)l

(c)3;

L'

^ -

=

(b)2;
•

(d)4.

Tra/^/.(b).
Cdu 2. Trong chiing minh bang phuong phap quy nap, gia thie't quy nap d
(a) Budc 1 ;
(c) Budc 3 ;

'

(b) Budc 2 ;
(d) Budc 4.

Trd ldi. (b).
Cdu 3. Phuong phap quy nap toan hgc thudng chi van dung cho bai toan cd lien
quan den :
(a) Sd tu nhien ;

10

(b) Sd nguydn ;


(c)Sdhi}uti;

(d) So thuc.

Trd ldi. (a).
Cdu 4. Cho bai toan : Chiing minh rang n (n + 1) chia he't cho 2 vdi mgi n e N*
(a) Menh de dd diing vdi n = 1.
(b) Menh de dd diing vdi n = 2.
(c) Menh de do diing vdi n = 3.
(d) Ca ba ke't luan tren deu sai.
Hay chgn cau tra ldi sai.
Trd ldi. (d).
Cdu 5. Cho P(n) = (1+h)", Q(n) = 1 + nh. Hay dien vao d trdng trong bang sau

n

1

2

3

4

• 5

P(n)

Q(n)
Cdu 6. Cho bai toan nhu cau 5, vdi h > (). Hay chgn phuofng an diing trong cac
phuong an sau day:
(a) P(n) > Q(n) V n e N* ;

(b) P(n) = Q(n) V n e N* ;

(c) P(n) < Q(n) V n e N*

(d) P(n) > Q(n) V n

G

N*

Trd ldi (a).
Cdu 7. Cho P(n) = n(n + l)(n + 2). Hay dien vao d trdng trong bang sau

n

1

2

3

4

5

P(n)
Cdu 8. Cho bai toan nhu cau 7. Hay chgn phuong an diing trong cac phuong an
sau day:
(a) P(n) • 3 V n e N* ;

(b)P(n) : 4 V n e N * ;
11


(c) P(n) : 5 V n e N* ;

(d) P(n) ; 6V n e N*

Trd ldi (a).
Cdu 9. Cho P(n) = n(n + l)(n + 2)(n + 3). Hay dien vao d trdng trong bang sau :
n

1

2

3

5

4

P(n)
Cdu 10. Cho bai toan nhu cau 9. Hay chgn phuong an diing trong cac phuong an
sau day:

(a) P(n) : 3 V n e N* ;

(b) P(n) ; 4 V n 6 N* ;

(c) P(n) ; 5 V n e N* ;

(d) P(n) ; 6V n G N^*

Trdldi (b).
HOATDONCL

naCFNG DfiN Bfil TfiP SfiCH GIfiO KHOfi
Bai 1. Hudng dan. Su dung cac budc chiing minh quy nap.
Hoat ddng cua GV
Cau hoi 1
Xet tfnh dung sai cua cong
thUe vdi n = 1.

Cau hoi 2
Gia sii cong thUc diing n = k
hay thie't lap cong thiic theo k.
Cau hdi 3
Hay thie't lap va chiing minh
12

Hoat ddng cua HS
Ggi y tra Idi cau hoi 1
Vdi AJ = 1, ta cd 1 = —^
2
cong thiic diing khi n= I.

. Nhuvay

Ggi y tra ldi cau hdi 2
HS tu lap.

Ggi y tra Idi cau hdi 3

/


1 + 2 + 3 + ... + ^ + (^ + 1)

cong thUc vdi n = k + 1.

Bai 2. HiCdng ddn. Sii dung cac budc chiing minh quy nap.
Hoat ddng cua HS

Hoat ddng ciia GV

Ggi y tra Idi cau hdi 1

Cau hdi 1

Vdi n = 1, tacd

Xet tfnh diing sai cua cdng
ihUc vdi n = 1.

9

^
2 = 4 =
''

Cau hdi 2

2.1(1 + 1X2.1 + 1) , „
^
'^
^Nhuvay,
3

dung khi « = 1.
Ggi y tra Idi cau hdi 2

Gia sii cdng thiic diing n = k Cau nay GV chi neu van de, HS tu lap.
hay thie't lap cdng thiic theo k.
Cau hdi 3

Ggi y tra ldi cau hdi 3

Hay thidi lap va chiing minh Ta se chiing minh
cdng thUc vdi n = k + 1.
2^ + A^ +.... + {2kf + {2k + 2f,
2{k+\)(k + 2){2k + 2,)
3
Bang phuong pha'p quy nap. HS tu
chiing minh.
Bai 3. Ihcdng ddn. Sir dung cac budc chiing minh quy nap.
Hoat ddng ciia GV

Cau hdi 1
Xet tfnh diing sai cua cdng
thiic vdi n = 1.

Hoat ddng cua HS
Ggi y tra ldi cau hdi 1
Vdi « = 1, ta cd 1 < 2 V l . Nhu vay, ba't
dang thiic diing khi « = 1.

13


Cau hdi 2

Ggi y tra Idi cau hdi 2

Gia sir cdng thiic diing n = k Cau nay GV chi ndu va'n de, HS tu lap.
hay thie't lap cdng thiic theo k.
Cau hdi 3

Ggi y tra Idi cau hdi 3

Hay thie't lap va chiing minh
cdng thUc vdi n = k + 1.

1 + —
'"

N/2
< 2l^k

V*

— +
4k

+

1
yfk^

1

4kV\

Lai cd
^2

24k +
4kV\
\
= 4/: + 4 ^ + - ^ < 4(/: + l)
4kV\ k + \
Suy ra

24k ^

VFhT

<

i4kT\

Bai 4. Sii dung phuong phap quy nap.
Hoat ddng cua G V
C a u hdi 1

H o a t dgng ciia H S
Ggi y t r a ldi cau hdi 1

Xet tfnh diing sai ciia cdng Vdi « = 2, ta cd
thiic vdi n = 2.
1 _ 3 _ 2 +1
4

4

2.2

Nhu vay, (1) diing khi h - 2.
C a u hdi 2
Gia SU" cdng thiic dung
n = k hSy thiet lap cdng

14

Ggi y t r a ldi cau hdi 2
cau nay GV chi ndu va'n de, HS tu lap.


thuc theo k.

Cau hdi 3
Hay thie't lap va chiing
minh cong thUc vdi
n = k + 1.

Ggi y tra ldi cau hdi 3
ta cd
/

1

1

k+l
2k

1—
9

1

1

k'J

(k + l)' j

kik + 2) , k + 2
(/t + l)2
2(k + \)

Tii caC chiing minh tren suy ra (1) diing vdi
mgi so nguyen « > 2.
Bai 5. Sii dung phuong phap quy nap
Hoat ddng ciia GV
Cau hdi 1
Xet tfnh diing sai cua cdng
thiic vdi n = 2.

Hoat ddng cua HS
Ggi y tra Idi cau hdi 1
Vdi rt = 2, ta cd
1
3

1 _ 7 _ 14
13
4 "^ 12 " 24 ^ 24 •

Nhu vay, (1) diing khi « = 2.
Cau hdi 2

Gdi y tra Idi cau hdi 2

Gia sii cdng thiic diing n = k cau nay GV chi neu van de, HS tu lap.
hay thie't lap cdng thiic theo
k.
Cau hdi 3

Ggi y tra Idi cau hdi 3

ta se chUng minh

Hay thie't lap va chung minh
cdng thUc vdi n = k + 1.

1 1
k + 2 k + 3 '"

1
2k + \

1
13
2{k + \) 24

Bang cac phan tfch :

15


1
1
-;
+
+ ...
k+2
k+3
^ 1
1
1

2k ' 2k + l
2ik + l)
1
=

1
+

^+1

k+2

2A: + 1

1
+ ... + —
2k

2(it + l)

k+l

HS tur chiing minh tiep.

Bai 6. Sii dung phuong phap quy nap.
Hoat ddng cua G V
Cau hdi 1
Xet tfnh diing sai cua cdng
thuc vdi n = 1.

Hoat ddng cua HS
Ggi y tra ldi cau hdi 1
Vdi « = 1, ta cd
Hj = 7.2^' ~^ + 3^-' ~ ' = 7 + 3 = 10,
chia he't cho 5.
Suy ra menh de tren diing khi « = 1.
Ggi y tra Idi cau hdi 2

Cau hdi 2

cau nay GV chi ndu va'n de, HS tu lap.

Gia sir cdng thiic dung n = k
hay thie't lap cdng thiic theo k.
Cau hdi 3
Hay thie't lap va chiing minh
cong thUc vdi n = k + 1.

Ggi y tra Idi cau hdi 3
tacd
_792(k+l)-2

-2(k+l)-l

= 4 . 7 . 2 ^ ^ ' ^ + 9.3^''"*
= 4(7.2^*'^^ + 3^''"S + 5 . 3 ^ ' ' ' '
= 4.Mk + 5.3^''"''

16

Vi u^^ : 5 (theo gia thie't quy nap), nen
tUdd ta dugc didu can chiing minh.
Bai 7. Sir dung phuong phap quy nap.
Hoat ddtig cua H S

Hoat ddng cua GV
Cau hdi I
Xet tfnh diing sai ciia cdng
thiic vdi n = 1.

(Jgi y tra ldi cau hdi I
Vdi n =1, ta eg t
( l + x ) ^ = l + x = 1 + 1.X.' ^^
Nhu vay, ta cd (1) diing khi « =1.
Ggi y tra ldi caii hdi 2
cau nay GV chi neu va'n de, HS tu lap.

Cau hdi 2
Gia sii cong thdc diing n = k
hay thie't lap cdng thuc theo k.
Cau hdi 3
Hay thie't lap va chdng minh
cdng thuc vdi n = k + I;

Ggi y tra ldi cau hdi 3
That vay, lU gia thie't x > - 1 va gia
thie't quy nap, ta cd
il+x)^^^

= {l+x){l+x)^'

>{l+x)il+kx)

= l+ik+

>l+(A:+l)x.

'
l)x + kx^
'• •"''•'

TU cac chiing minh tren suy ra (1)
diing vdi mgi n & N*
Bai 8. Sii dung phuang phap quy nap.
HS tu giai.

2-TKB(.DSVGTl iNCT2

17


§2. Day so
(tiet 3, 4)
1. MUC TIEI)
L Kien thiirc
HS nam dugc :
Dinh nghla day so : Sd hang tdng quat ciia day so, day sd hiiu han, sd
hang dau va sd hang cud'i ciia day so hiiu han.
• Cac phuong phap cho day so : Day sd cho bdi cdng thiic, day sd cho bdi

md ta. day sd cho bdi truy hdi. - Bieu dien hinh hgc cua day sd tren he true toa do,
• Day sd tang, day sd giam va day so bichan.
2. KT nang
Sau khi hgc xong bai nay, HS can giai thanh thao cac dang toan vi day
''SO.

• Tim dugc sd hang tdng quat Ciia day so, sd hang dau, sd h^ng cudj giia
day sd hiiu han.
Chiing minh mdt day so bi chan tren, mdt day so bi chan dudi, day .0' bi
chan.
3. Thai do
- Tu giac, tfch cue trong hgc tap.
Bie't phan biet rd cac khai niem co ban va van dung trong timg truenig hop cy
the.
•• Tu duy cac va'n de ciia toan hgc mgt each Idgic va he thdng.
IL CHUAN BI CUA GV VA HS
1. Chuan hi ciia GV
• Chuain bi cac cau hdi ggi md.
Chuan bi phan mau va mgt sd do dung khac.

18


2. Chuan bi ciia HS
• Can on lai mgt sd kie'n thiic da hgc ve day sd da hgc, da biet.
IIL PHAN PHOI

TH6I LUONG

Bai nay chia lam 2 tie't:

Tii't I -: Td ddu den hit phdn 2.
Tiet 2 : Tiep theo den het phdn bdi tap.
IV. TIEN TRiNH DAY - HOC

A. OAT VANDE
Cau hdiT
2

Cho f(n) = n Hay dien vao cac d trdng sau day:
n

1

3

2

•4

f(n)

Cau hdi 2
Hay nhan xdt ve tfnh tang, giam ciia nam so hang tren.

B. BAI Mdl
HOATDONCl
1. Dinh nghTa va vi du
• GV dat va'n de nhu sau
Hay dien vao d trdng.
n

1

2

3

4

5

(-l)"(n-2)
H1. Nhan xet gi ve dau ciia day sd tren.
112. Ta cd the tim dugc mgt sd hang nao dd vdi n bat ki hay khong?
H3. Hay xac dinh sd hang d vi tri thii 100.
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H4. Cdng thiic tren cho ta mdt day sd. Em hay neu dinh nghTa day so theo quan
diem ciia minh.
• Neu dinh nghla day so.
Mpt hdm sou xdc dinh tren tap hgp cdc sd nguyen duang N * dicgc
ggi Id mpt ddy sd vd hgn (hay con ggi tdt Id ddy so).
MSi gid tri cita hdm sd u dicgc ggi Id mpt sdhgng cua day so ; ii(l)
dugc ggi Id sd hgng thif nlid't (hay sdhgng ddu); H(2) dugc ggi Id so
hgng thic hai : ...
• GV neu vf du 1, sau dd ihuc hien [jHIJ trong 5'
Hoat ddng cua H S

Hoat ddng ciia G V

C a u hdi 1
Hay xac dinh so hang thU 9.

Ggi y t r a ldi cau hdi 1
HS tu tinh todn.
Dap so. lla = •—
^ 1 0

C a u hdi 2
Hay xac dinh sd hang thU 99.

Ggi y tra Idi cau hdi 2
HS tu tinh todn,
Ddp sd. Uq =

C a u hdi 3

100

Hay xac dinh so hang thii Ggi y tra Idi cau hdi 3
999.
HS tu tfjih toan,
Ddp sd.

MQ

= —~.
1000

• Tie'p theo GV dua ra cac kf hieu ciia day sd

Ngicdi ta tlncdiig ki hieu ddy sd u = u(n) bai (u^), vd ggi «„ la sdhgng
tong qudt ciia day sd do.
NgiCdi la ciing thudng viel ddy sd(uj dicdi dgng khai trien :
20


Ul , ô 2

ããã ã " n ' ããã ã

H5. Hay la'y vf du mgt vai day sd cho dudi dang cdng thiie tdng quat va chi ra sd
hang thU 10, 100 ciia day sddp-

'

'"•••'

116. Hay lay vf du mdt vai day sd cho dudi dang khai trien va chi ra sd hang thii
10, 100 ciia day sodd.
H7. Hay so sanh trong each chd nao de tim cac sd hang cua day sd hon.
• GV neu chii y trong SGK:

;

NgiCdi ta ciing ggi mpt hdm sdu xdc dinh tren tap hgp gom m sd
nguyen duang ddu tien (m tuy y thupc N*) la mpt ddy sd. Ro rdng, ddy sd
trong trudng hgp ndy chi co hicii hgn so hgng (m so hgng : u^, 112, •••• u,,,
) ; vi the, ngiCdi ta con ggi no Id ddy sd hdu hgn ; Wj ggi Id sd hgng ddu
vd i(,„ ggi Id sdhgng cudi.
H8. hay neu su khac biet giiia day sd hiru han va day sd vo han.

• GV neu vf du 2 va dua ra cau hdi :
H9. Hay neu sd hang dau va so hang cudi ciia day sd tren.
HOATDONC 2
2. Cac each cho day so
• GV dat va'n de nhu sau:
HIO; Mgt day sd dugc xac dinh khi nao ?
• GV neu each 1 (each cho day so).
Cdch 1 : Cho ddy sdhai cong thicc ciia sd hgng tong qudt.
HI 1. Em hay la'y vf du cho each cho day so nay.
ã GV neu day sd :
ã'-"

"""

3ô+i

ã

:

21


Thuc hien iH2| trong 5
Hoat ddiig cuai GV

Hoat ddng cua HS

Cau hoi 1

Ggi y tra ldi cau hdi 1

Hay xac dinh sd hang U33 cua HS tu tinh toan.
day so trdn.
r,A
A8
Dap so. Ux\ = — .
a 25
Ggi y tra ldi cau hdi 2

Cau hdi 2
Hay xac dinh sd hang U333
cua day so tren.

HS tu tfnh toan.
r,A

83

.

Dapso. "333=225"

• GV ndu each 2:
Cho ddy sdbdi hi thitc truy hoi (hay cdn ndi: Cho ddy so bdng
quy ngp).
"• GV neu vl du 3 , sau do cho HS dien vao bang sau:
n

1

2

3

4

5

6

7

8

9

7

8

9

"n

• GV neu vi du 4 , sau dd cho HS dien vao bang sau:
n

1

2

3

4

5 •

6

Vn

HI2. Ne'u chi bie't mdt sd hang ciia (v^ ) thi cd bie't dugc tat ca cac sd hang cua Vn
hay khdng?
Hi3 Hay cho mdt vai vi du khac v^ phudng phap cho day sd bang quy nap.
Thuc hien [H3j trong 3'.

22


Hoat ddng ciia HS

Hoat ddng cua GV

Ggi y tra ldi cau hdi 1

Cau hdi 1

Tii bang tren HS tu tra ldi.

Hay xac dinh V4.

Ggi y tra Idi cau hdi 2

Cau hdi 2
Hay xac dinh so hang U12 ciia
day so tren.

HS tu tfnh toan.

• GV neu each 3 :
Diin dgt bang ldi each xdc dinhmSi sdhgng ciia ddy sd.
• GV neu vf du 5 va dua ra cac cau hdi nhu sau :

'

H14. Hay la'y vf du ve each cho day- so bang ldi.
1115. Neu su khac nhau giiia: ba each cho day so.
• GV neu chii y :
M-pt day sdcd the cho bang niiieu cdch. Chdng hgn, ddy sd(\^J^) d
vi du 3 CO the cho bdi cong thicc ciia sdhgng tong qudt nhusau :
u„ = 2 " - / , n

em.

• Thuc hien |H4| trong 3 .
Hoat ddng cua GV
C^u hdi 1

Hoat dgng ciia HS

Ggi y tra ldi cSu hoi 1

Em cd nhan xet gi ye tam Tam giac BM^A vudng tai M^ .
giiac BM„A.
Ggi y tra ldi c:au hdi 2
Cau hdi 2
Mn = /4M„ = AB. sin ABhl
Hay tim cdng thUc cua s6
hang tdng quat ciia day sd
^^\ . AOM,, ^ . K .
- 20A. sm ——^ = 2sin—.
2
n
(Un)-

23