Thiết kế bài giảng đại số và giải tích lớp 11 tập 2 nâng cao

TRAN VINH
NHA XUAT BAN HA
NOI

4t_
IRAN VINH
THIET KE BAI
GIANG
DAI
SO
VA GIAI TICH
ir''-/
".V.r'
•»
It:
isiiirGCiAo
TAP
HAI
NHAXUATBANHANOI
THIET KE BAI
GIANG
DAI
sd
VA GIAI TICH 11 - NANG CAO -
TAP
HAI
TRAN VINH
NHA
XUAT
BAN
HA

NOI
Chiu trdch nhiim xudt bdn :
NGUYEN KHAC
OANH
Biin tap :
PHAM QUOC TUAN
ViBia
TAO
THANH
HUYEN
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
BAI
SOA^
§1.
Phu'cfng^
phap quy nap toan hoc
(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 GV
Cau hdi I
Hay kiem chiing khi n
=
1.
Cau hoi 2
Cd thd kiem tra dang
thiJc (1)
vdi mgi n dugc khdng?
Hoat dong cua HS
Ggi y tra Idi cau hoi 1
Vdi n
=
1 ta cd dang thiic luon
ludn diing.
Ggiy
tra ldi cau hoi 2
Khdng thd.
• 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
Xet
tinh

dung
sai
ciia cong
thiic
vdi n
=
1.
Hoat dong
cua HS
Ggi
y tra Idi cau hoi 1
Ta tha'y
n
=
1,
cdng thiic tren
ludn
ludn diing.
Cau hoi 2
Gia sii cdng thiic diing n
=
k
hay thie't lap cdng thiic.
Cau hoi 3
Hay thie't lap cdng thiic khi
n = k + 1 va chiing minh cdng
thiic
do.
Ggi y tra Idi cau hoi 2
4

Ggi y tra Idi cau hoi 3
l^
+
2^
+ +
k^
+
ik+l)^
=
(k
+
1)2
(it +
if
4 ' ;
HS tu chiing minh tie'p.
Thuc
hien
1H2J
trong 5'.
Hoat dgng cua GV
Cau hoi
1
Xet tfnh diing
sai ciia
cdng
thiic vdi n = 1.
Cau hoi
2
Gia

sii cdng thiic
diing
n = k
hay thie't lap cdng thiic.
Cau hoi 3
Hay thie't
lap
cdng thiic khi
n
=
k + 1 va Chiing minh cdng
thiic do.
Hoat dgng cua HS
Ggi y tra Idi cau hoi 1
Ta tha'y n
=
1, cdng thiic tren ludn
ludn diing.
Ggi y tra Idi cau hoi 2
HS tu thie't
lap.
Ggi y tra Idi cau hoi 3
l+3+ +
(2)t-l)
+
(2^+l)
=
(^+lf.
HS
tu chiing minh.

• Thuc hien
[H3J
trong 5'
Hoat dgng cua GV
Cau hoi 1
Xdt
tihh
diing sai cua cdng
thiic vdi n = 1.
Hoat dong cua HS
Goi y tra ldi cau hoi 1
2
Vdin=
1,
tacd
1 =1
1(4.1'-1)
3
Cau hoi 2
Gia six cdng thiic diing n
=
k
hay thie't lap cdng
thiic.
Cau hdi 3
Hay thie't lap cdng thiic khi
n
=
k + 1 va chiing minh cdng
thiic dd.

Nhu vay cdng thiic diing khi
«
= 1.
Ggi y tra ldi cau hoi 2
HS tu thie't lap.
Ggi y tra Idi cau hoi 3
l^
+
3^
+ + (2k-lf + {2(,k+l)-\f
_{k + \)[4ik + \y l)
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
Xet tfnh diing sai ciia cdng
thiic vdi n = 3.
Cau hdi 2
Gia
sir
cong
thiic
diing n
=
k
hay thidt
lap
cdng thiic theo k.
Cau hoi 3
Hay thie't lap va chiing minh

cdng thiic voi n = k +
I.
Hoat ddng cua HS
Ggi y tra ldi cau hoi 1
Ta thay n
=
1,
cdng thiic tren
ludn
ludn diing.
Ggi y tra Idi cau hdi 2
2*'
> 2/: 4- 1.
Ggi y tra Idi cau hoi 3
2''^'
> 2{k+\)+ 1.
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
L'
=
(b)2;
(c)3;
^ - •
(d)4.
Tra/^/.(b).
Cdu 2. Trong chiing minh bang phuong phap
quy
nap, gia thie't quy nap d
(a) Budc 1 ;
'
(b) Budc 2 ;
(c) Budc 3 ; (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 ; (b) Sd nguydn ;
10
(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
P(n)
Q(n)
1
2

3 4
•
5
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
P(n)
1
2
3 4
5

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)
:4VneN*;
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
P(n)
1 2
3
4
5
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*
;
Trdldi
(b).
(d) P(n)
;
6V n
G
N^*
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
Hoat ddng cua HS
Ggi y tra Idi cau hoi 1
Vdi AJ =
1, ta cd
1 = —^
. Nhu
2
cong thiic diing khi n=
I.
Ggi y tra ldi cau hdi 2

HS tu lap.
/
Ggi y tra Idi cau hdi 3
vay
12
cong
thUc
vdi n
=
k +
1.
1
+ 2 +
3
+

+
^
+
(^
+ 1)
Bai 2.
HiCdng
ddn.
Sii
dung cac budc chiing minh quy nap.
Hoat ddng ciia GV
Cau hdi 1
Xet tfnh diing sai cua cdng
ihUc

vdi n =
1.
''
Cau hdi 2
Gia sii cdng thiic diing n = k
hay thie't lap cdng thiic theo k.
Cau hdi 3
Hay
thidi
lap va chiing minh
cdng
thUc
vdi n = k +
1.
Hoat ddng cua HS
Ggi y tra Idi cau hdi 1
Vdi
n =
1, tacd
9 ^
2.1(1
+
1X2.1
+
1) ,„
2=4=
^ '^
^Nhuvay,
3
dung khi

«
= 1.
Ggi y tra Idi cau hdi 2
Cau nay GV chi
neu
van de,
HS
tu lap.
Ggi y tra ldi cau hdi 3
Ta se chiing
minh
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
thiic vdi n =
1.
cdng
Hoat ddng cua HS
Ggi y tra ldi cau hdi 1
Vdi
«
=
1,
ta cd
1
< 2
Vl.
Nhu vay, ba't
dang thiic diing khi
«
=
1.
13
Cau hdi 2
Gia
sir
cdng thiic diing n = k
hay thie't
lap
cdng thiic theo k.
Cau hdi 3
Hay thie't lap va chiing minh
cdng

thUc
vdi n = k +
1.
Ggi y tra Idi cau hdi 2
Cau nay GV chi ndu va'n de,
HS
tu lap.
Ggi y tra Idi cau hdi 3
1
1 + — — +
N/2
'"
4k
yfk^
l^k
<2V*
+
Lai cd
1
4kV\
24k
^2
+
\
4kV\
=
4/: +
4^
+
-^ <

4(/:
+
l)
4kV\
k + \
Suy ra
24k ^
VFhT
<
i4kT\
Bai 4.
Sii
dung phuong phap quy nap.
Hoat ddng cua GV
Cau hdi 1
Xet tfnh diing sai ciia cdng
thiic vdi n = 2.
Cau hdi 2
Gia
SU"
cdng thiic dung
n = k hSy thiet
lap
cdng
Hoat dgng ciia HS
Ggi y tra ldi cau hdi 1
Vdi
«
= 2, ta cd
1 _ 3 _ 2

+1
4 4 2.2
Nhu vay, (1) diing khi h - 2.
Ggi y tra ldi cau hdi 2
cau
nay GV chi ndu va'n de, HS tu lap.
14
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
1 —
9
1
k'J
1
(k +
l)'
j
k
+ l
kik

+ 2)
, k
+
2
2k
(/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.
Cau hdi 2
Gia sii cdng thiic diing n
=
k
hay thie't lap cdng thiic theo
k.
Cau hdi 3
Hay thie't lap va chung minh
cdng
thUc
vdi n = k +
1.
Hoat ddng cua HS

Ggi y tra Idi cau hdi 1
Vdi rt = 2, ta cd
1
1
_
7
_ 14 13
3 4
"^
12
"
24
^
24
•
Nhu vay, (1) diing khi
«
= 2.
Gdi y tra Idi cau hdi 2
cau
nay GV chi neu van de, HS tu lap.
Ggi y tra Idi cau hdi 3
ta se
chUng
minh
11
1 1 13
k
+ 2 k + 3
'"

2k +
\ 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
=
+
+
+ —
^+1
k+2 2k
2A:
+
1 2(it
+ l) k
+
l
HS

tur
chiing minh tiep.
Bai
6.
Sii
dung phuong phap quy nap.
Hoat ddng cua GV
Cau hdi 1
Xet tfnh diing sai cua cdng
thuc vdi n =
1.
Cau hdi 2
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.
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
nay GV chi ndu va'n de, HS tu lap.
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 ddng cua GV
Cau hdi I
Xet tfnh diing sai ciia
cdng
thiic vdi n =
1.
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;
Hoat
ddtig
cua HS
(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.
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)x
+
kx^
>l+(A:+l)x.
'• •"''•'
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
f(n)

1 2
3
•4
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
(-l)"(n-2)
1
2
3
4 5
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.
19
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 ciia GV
Cau hdi 1
Hay xac dinh so hang
thU
9.
Cau hdi 2
Hay xac dinh sd hang
thU
99.
Cau hdi 3
Hay xac dinh so hang
thii
999.
Hoat ddng cua HS
Ggi
y
tra ldi cau hdi 1
HS tu
tinh
todn.
Dap so.
lla
=
•—
^10
Ggi y tra
Idi
cau hdi 2
HS tu
tinh

todn,
Ddp sd.
Uq =
100
Ggi y tra Idi cau hdi 3
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
,
«2
••• •
"n
' ••• •
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.
• GV
neu
day sd :
•'-" """ 3«+i • :
21
Thuc hien
iH2|
trong 5
Hoat
ddiig cuai
GV
Cau hoi 1
Hay xac dinh sd hang
U33
cua
day so trdn.
Cau hdi 2
Hay xac dinh sd hang
U333
cua day so tren.
Hoat ddng cua
HS
Ggi y tra ldi cau hdi 1
HS tu
tinh
toan.

r,A
A-
8
Dap
so.
Ux\
=
—.
a 25
Ggi y tra ldi cau hdi 2
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
"n
1 2 3 4 5
6
7
8
9
• GV neu
vi
du 4 , sau dd cho
HS
dien vao bang sau:
n
Vn
1 2
3 4 5
•
6 7
8
9
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 cua GV
Cau hdi 1
Hay xac dinh
V4.
Cau hdi 2
Hay xac dinh so hang
U12
ciia
day so tren.
Hoat ddng
ciia
HS
Ggi y tra ldi cau hdi 1
Tii bang tren
HS
tu tra ldi.
Ggi y tra Idi cau hdi 2
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
Em cd nhan xet gi
giiac

BM„A.
Cau hdi 2
Hay tim cdng
thUc
hang tdng quat ciia
(Un)-
ye tam
cua
s6
day sd
Hoat dgng ciia HS
Ggi y tra ldi
cSu
hoi 1
Tam giac
BM^A
vudng tai
M^
.
Ggi y tra ldi
c:au
hdi 2
Mn
=
/4M„ =
AB.
sin
ABhl
^^\
.

AOM,,
^
.
K .
-
20
A.
sm
——^
=
2sin—.
2 n
23