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

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

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

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

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.

+
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

;
•
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

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

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

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.

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

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

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

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

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

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

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

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

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

+ 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

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

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

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)^^^
=

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

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)

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).

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

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.

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.

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).
"-

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.

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

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