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

Phần 1 Tài liệu Thiết kế bài giảng Đại số và Giải tích 11 nâng cao (Tập 2) các phương pháp thiết kế bài giảng về dãy số, cấp số cộng và cấp số nhân, giới hạn, giới hạn của dãy số. Mời các bạn cùng tham khảo nội dung chi tiết.

TRAN VINH

NHA XUAT BAN HA NOI

IRAN VINH

THIET KE BAI GIANG

DAI SO VA GIAI TICH

ir''-/ ".V.r' - •» It:

isiiirGCiAo

TAP HAI

NHAXUATBANHANOI

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

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

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

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

• 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

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

IIL PHAN PHOI T H 6 I 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

Cho cac menh 6i sau :

a) Sd nguyen duong le ldn hon 1 la so nguyen to

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

G g i y 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

Cau hoi 2

Gia sii cdng thiic diing n = k

hay thie't lap cdng thiic

l^ + 2^ + + k^ + ik+l)^ = (k + 1)2 (it + if

4 ' ;

HS tu chiing minh tie'p

Thuc hien 1H2J trong 5'

Gia sii cdng thiic diing n = k

hay thie't lap cdng thiic

Ta tha'y n = 1, cdng thiic tren ludn ludn diing

Ggi y tra Idi cau hoi 2

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;

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

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:

( 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)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

Cdu 6 Cho bai toan nhu cau 5, vdi h > () Hay chgn phuofng an diing trong cac

phuong an sau day:

naCFNG DfiN Bfil TfiP SfiCH GIfiO KHOfi

Bai 1 Hudng dan Su dung cac budc chiing minh quy nap

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

cong thUc vdi n = k + 1 1 + 2 + 3 + + ^ + (^ + 1)

Bai 2 HiCdng ddn Sii dung cac budc chiing minh quy nap

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 = ^ '^ ^ N h u v a y ,

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

Vdi « = 1, ta cd 1 < 2 V l 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

Gia SU" cdng thiic dung

n = k hSy thiet lap cdng

H o a t d g n g ciia H S Ggi y t r a 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 t r a ldi c a u 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

Tii caC chiing minh tren suy ra (1) diing vdi mgi so nguyen « > 2

Bai 5 Sii dung phuong phap quy nap

Gia sii cdng thiic diing n = k

hay thie't lap cdng thiic theo

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

1 1 -; + +

HS tur chiing minh tiep

Bai 6 Sii dung phuong phap quy nap

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

G g i 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

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

Gia sii cong thdc diing n = k

hay thie't lap cdng thuc theo k

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)

• 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

- 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

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 T H 6 I L U O N G

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

• GV dat va'n de nhu sau

Hay dien vao d trdng

n

(-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

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'

HS tu tinh todn

Dap so lla = •—

^ 1 0 Ggi y t r a 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 :

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

Hi3 Hay cho mdt vai vi du khac v^ phudng phap cho day sd bang quy nap

Thuc hien [H3j trong 3'

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 :

Hay tim cdng thUc

hang tdng quat ciia

Mn = /4M„ = AB sin ABhl

^^\ AOM,, ^ K

- 20 A sm — — ^ = 2sin—

2 n

23

HOATDONC 3

3 Day sd tang, day so giam

• GV dat va'n de nhu sau:

+ Cho day sd {u„) = n + 1

Hay dien vao bang sau':

HI7 Hay nhan xet ve su tang,, giam ciia cac sd hang

• GV neu dinh hghIa 2:

Day sd(it^.^) dugc ggi Id ddy sdtdng ne'u vdi mgi n ta co »„ <

»n+l-Day sd(Uj^) dugc ggi la ddy sg gidm ne'u vdi mgi n ta co M„ > «(„+i

• GV neu vf du 6 -va dat ra cac cau hdi nhu sau

HIS Ne'u day sd khong tang thi day sd giam Diing hay.sai ? ;

HI9 Ne'u day sd khong gi^m thi day sd tang! Dung Hay sai ?

Thirc hien [HSJ trong 3 '

Hay tim cdng thUc ciia so

hang tdng quat ciia day so

- runj

-J

Hoat ddng cua HS Ggi y tra Idi cau hdi 1

GV ggi HS tra ldi

Ggi y tra ldi cau hdi 2

Mn = AMn = AB sin AfiF,

^,, / AOM-, ^ K:

=• 20A sin ^ = 2sin —

2 n HOATDONC 4

4 Day sd bi chan

• GV dat va'n de nhu sau:

+ Cho day sd (Ujj) = 1

Hay dien vao bang sau :

H21 Tim sd hang nhd nha't ciia day

• Sau dd GV neu dinh nghTa

a) Ddy sd(uj dugc ggi la ddy sd bi chgn trin ne'u ton tgi mpt so M

c) Day sd(iij dicgc ggi la ddy sdbi chdn ne'u no vica bi chdn tren, vda

hi chgn dudi ; nghia la, ton tgi mpt sdM vd mpt sdm sao cho

Vn e N^, m < u^ < M

GV neu vf du 7, cho HS la'y them mdt so vf du khac

• Thuc hien |H6| trong 4

GV ggi HS tra Idi

Ggi y tra ldi cau hdi 2

Cac khang dinh diing : b), c), d) va e)

• Mdt sd cau hdi on tap

H22 Day so la mdt ham so

(a) Diing ; (b) Sai

H23 Mgi day sd deu la day sd tang hac giam

(a) Diing ; (b) Sai

H24 Day sdbi chart la day sd tang,hac giam

(a) Diing; < (b) Sai

H25 Day sd giam la day sd bj chan

(a) Diing ; (b) Sai

H26 Day so tang la day so bj chan

(a) Diing; (b) Sai

H27 Ne'u day sd khdng tang thi giam

(a) Diing ; (b) Sai

H28 Day sd khdng giam la day so tang

(a) Diing ; (b) Sai

H29 Cd nhihig day sd khdng tang ciing khong giam

(a)Diing; (b) Sai

HOATDONC 5

TOM T^T Bfil HOC

1 Mdt ham so u xac djnh tren tap hgp cac so nguyen duong N * dugc ggi la mdt

day sd vd han (hay cdn ggi tat la day so')

Mdi gia trj ciia ham so u dugc ggi la mdt sd hgng ciia day sd ; M(1) duge ggi la sd hgng thic nhdt (hay sdhgng ddu); M(2) dugc ggi ^sdhgng thic hai;

Ngudi ta thudng kf hieu cac gia trj M(1), M(2), tuong iing bdi MJ , 112 ,

2 Ngudi ta thudng kf hieu day sd M = «(«) bdi (Mn), va ggi Mn la sdhgng tdng qudt

cd hiru han sd hang (m sd hang : MJ, M2, , M„, ) ; vi the, ngudi ta cdn ggi nd la ddy sdlncu hgn ; MJ ggi la sdhgng ddu vh ii,„ gglla sdhgng cudi

4 Cd cac each cho day so sau :

Cho ddy sdbdi cong thicc ciia sdhgng tong qudt

Cho ddy sdhai he thdc truy hoi (hay cdn ndi: Cho ddy sdbdng quy ngp)

Dien dgt hdng ldi cdch xdc dinh moi sdhgng ciia day sd

5 Day sd (»„) dugc ggi la day so tang ne'u vdi mgi n ta cd u^ < M^+I

Day sd (M„) dugc ggi la day sd giam nd'u vdi mgi n ta cd Mn > M„^.-J

6 a) Day sd (M„) dugc ggi la day sd bj chan tren ne'u ton tai mgt sd M.sao cho

V/7 G N*, Mn < M

b) Day sd (M,,) dugc ggi la day sd bj chan dudi neu tdn tai mdt sd m sao cho

Vn G N*, Mn > m

c) Day sd (M^) dugc ggi la day sd bj chan ne'u nd vUa bi chan trdn, vUa bj chan dudi;

nghTa la, tdn tai mgt sd M ya nidt sd m sao cho :

Vn G N*, m < u^ < M

HOAT DONG'6

MOT SO CfiU HOI TR^C NGHIEM KHfiCH QUfiN

Hdy dien diing sai vdo 6 tr.dng sau

Cdu I Cho day sd Uj, = —

n (a) Day so da cho la day sd tang

(b) Day sd da cho la day sq giam

D

D

(c) Day sd da cho la day sd khdng ddi

(d) Day sd da cho la day so khong tang, khong giam

(b) Day so da cho la day sd giam

(c) Day sd da cho la day sd khdng doi

(d) Day sd da cho la day sd khong tang, khong giam

(b) Day so da cho la day sd giam

(c) Day sd da cho la day sd khdng ddi

(d) Day sd da cho la day sd khong tang, khdng giam

C<3M ^ Cho day so u^ = —•

:-n^ + 1

(a) Day sd da cho la day so tang

(b) Day sd da cho la day so giam

(c) Day so da cho la day so khdng ddi

(d) Day so da cho la day sd khong tang, khdng giam

(b) I3ay sd da cho la day so giam

(c) Day so da cho la day so bj chan

(d) Day sd da cho la day so khong tang, khdng giam

Cdu 7 Cho day sd Uj - 2, u,^ = u„_j + 3

(a) Day sd da eho la day so tang

(b) Day sd da cho la day sd giam

(c) Day sd da cho la day so bj chan dudi

(d) Day sd da cho la day sd khong tang, khong giam

Cdu 8 Cho day so U| = 2, Uj, = Uj^_i - 3

(a) Day sd da cho la day sd tang U (b) Day sd da cho la day sd giam |_J (c) Day sd da cho la day sd bj chan tren |_|

(d) Day sd da cho la day sd khdng tang, khdng giam |_|

Hdy chgn khdng dinh diing trong cdc cdu sau

Cdu 9 Cho day so

U i i l , u , / = U n _ i - 2 Trong cac sd sau day sd nao la sd hang ciia day sd:

Cdu J1 Cho day sd

U i = l , U n = U n _ , + 2 Trong cac sd sau day sd nao la so hang ciia day sd:

( a ) - 1 6 2 ; (b) 162

(c) 81; ( d ) - 8 1

Trd uyi (a)

HOATDQNC 7

HCrd^NG DfiN Bfil TfiP SGK

Bai 9 Huang ddn Six dung cac cdng thiic sd hang tdng quat cua day sd:

GV ggi ba HS len bang va cho dien vao d trdng sau :

ChUng minh cdng thiic tong

quat ciia day so ddng vdi

Cdng thiic diing idii « = 1

Ggi y tra Idi cau hdi 2

«k + l = 2Mk + 3=2.(2'' + * - 3 ) + 3

_ T ( k + l)-l- 1 _ T

Tir cac chiing minh trdn suy ra

* diing vdi mgi « G N

2

Ta cd M„ + J - Mn = 3n - 4n + 3 > 0

Cau hdi 2

Ket luan:

Ggi y tra Idi cau hdi 2

Day so tang

Cau b, cau c, HS giai tuang tu

Bai 14 Hifdng ddn Dua vao djnh nghTa day sd tang, day sd giam va day sd bj

Luyen tap (tiet 5)

I MUC TIJEIJ

1 Kien thiirc

HS on tap lai kien thuc ciia hai bai da hgc ; PhUOng phip quy nap toin hgc va day sd Dac biet la bai day sd, thdng qua cac bai tap nham ciing cd lai cac kie'n thUc :

• Djnh nghla va cac phuong phap cho day so

- Day sd tang va day so giam

Day sd bj chan

2 KT nang

Khac sau them viec xet hieu Uj^^j-Uj, va thuong "'"^ de xet tfnh

Un

tang, giam ciia day so

Phuong phap phan tfch thanh long de chiing minh dSy sd bj chan

3 Thai do

- Tu giac, tfch cue trong hgc tap.,

Biei plian biet rd cac khai niem co ban va van dung trong tiing trudng hgp cy the

- Tu duy cac va'n de ciia toan hgc mdt each Idgic va he thd'ng

Bie't phan biet hinh thiic va ndi dung bai toan

n CHUAN BI CUA GV VA HS

1 Chuan bj ciia GV

Chuan bj cac cau hdi ggi md

Chuan bj pha'n mau va mgt dung dd ddng kh^c

2 Chuan bi ciia HS

Can on lai mot sd kien thiic hai bai da hoc

m PHAN PHOI THCJI LU0NG

Bai nay chia lam 1 tie't

IV. T I £ ' N TRINH DAY - HOC

Bai 15 Bai nay, nham hudng HS de'n bai sau, tuy nhien GV nen cho HS phan tich

kT bai tap ve : Cdng thiic tdng quat va truy hdi cua day sd

a) GV ggi HS dien vao cac d trdng sau :

37

khdng?

Cau hdi 2

Hay chiing minh cdng thu'c

bang quy nap

Cdng thiic dung khi « = 1

Ggi y tra ldi cau hdi 2

.= 1.+ ( 1 - 1).2^