Đang tải... (xem toàn văn)
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 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 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 ^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 Kien thiirc Nam duoc toan bd Icien thiic co ban chuong da neu tren, cu the : - Biet chiing minh va nhan bie't 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 Kinang Vun dung cac budc quy nap de chiing minh bai toan bang phuong phap quy nap toan hoc van dung thao c^c tinh chat cua ca'p sd cdng va cap sd nhan 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 cac trudng hop khdng phiic tap ; Bie't van dung nhiing kie'n thiic chuong de giai quye't cac bai toan cd lien quan dugc dat d cac mdn hoc khac, cung nhu thuc tiln cude sd'ng Thai • Tu giac, tfch cue, ddc lap va chii ddng phat hien cung nhu ITnh hdi kie'n thUc qua trinh hoat ddng • Can than chinh xac lap liian va tinh toan Cam nhan dugc thuc teciia toan hoc, nha't la dd'i vdi day sd Pban 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 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 KT nang Van dung thao phuong phap quy nap giai toan Bie't them mdt phuong phap chiing minh dd'i vdi bai toan cd lien quan den sd tu nhien Thai dp Tu giac, tich cue hgC tap « Biet phan biet rd cac Idiai niem co ban va van dung timg trudng hop cu the - Tu cac va'n de ciia toan hgc mdl each logic va he thd'ng II CHUAN BI CUA GVVA HS 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 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 chia lam tie't : Tii't : Tix ddu din hit vi du Tii't : Tiep theo den het phdn bdi tap IV- TIEN TRINH DAY - HOC A DAT VANDE Cau hdi Xet tmh diing - sal cua cac cau sau day : a)Ne'ua>bthia" >'b" b)Ne'ua>b> thia" > b", Cau hoi Cho cac menh 6i sau : a) Sd nguyen duong le ldn hon la so nguyen to b) + + +, + n = "^""^^\ n e N Hay xem xet tinh diing sai cua cac menh de tren vdi sd hang dSu tien B BAIMdl , HOATDONCl 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 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] 4' Hoat dong cua HS Hoat dong cua GV Ggi y tra Idi cau hoi Cau hdi I Hay kiem chiing n = Cau hoi Vdi n = ta cd dang thiic luon ludn diing G g i y tra ldi cau hoi Cd thd kiem tra dang thiJc (1) Khdng thd vdi mgi n dugc khdng? • GV neU cac budc quy nap ; • Bu&c (budc ca sd, hay budc khdi ddu) Chieng minh A(n) la mpt menh de diing n = • Bu&c (budc quy ngp, hay budc "di truyen") Vdi k Id mpt sd nguyen duang y, xudt phdt td gid thii't A(n) Id mpt menh de diing n = k, chimg minh A(n) cdng Id mpt minh de diing n = k + 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 ggi Id gid thie't quy ngp • GV dua 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 Vi du ap dung • Thuc hien vi du phiit Hoat ddng cua GV Cau hoi Hoat dong cua H S Ggi y tra Idi cau hoi Xet tinh dung sai ciia cong Ta tha'y n = 1, cdng thiic tren ludn thiic vdi n = ludn diing Cau hoi Ggi y tra Idi cau hoi Gia sii cdng thiic diing n = k hay thie't lap cdng thiic Cau hoi Ggi y tra Idi cau hoi l^ + 2^ + + k^ + ik+l)^ = Hay thie't lap cdng thiic n = k + va chiing minh cdng (k + 1)2 (it + if thiic '; HS tu chiing minh tie'p Thuc hien 1H2J 5' Hoat dgng cua GV Cau hoi Hoat dgng cua HS Ggi y tra Idi cau hoi Xet tfnh diing sai ciia cdng Ta tha'y n = 1, cdng thiic tren ludn thiic vdi n = ludn diing Cau hoi Ggi y tra Idi cau hoi Gia sii cdng thiic diing n = k HS tu thie't lap hay thie't lap cdng thiic Cau hoi Ggi y tra Idi cau hoi Hay thie't lap cdng thiic l + + + (2)t-l) + (2^+l) n = k + va Chiing minh cdng thiic = (^+lf HS tu chiing minh • Thuc hien [H3J 5' Hoat dgng cua GV Cau hoi Hoat dong cua HS Goi y tra ldi cau hoi Xdt tihh diing sai cua cdng Vdin= 1, tacd 12 = thiic vdi n = 1(4.1'-1) HOATDONC 3 Mdt vai quy tdc tfnh gidi han tai vd cue • GV neu quy tdc 1: Ne'u limMn = ±00 va limv„ = ±oo thi lim(t7nVn) dugc cho bang sau : limHn limi'n lim.(MnV„) +00 +00 +00 +00 —00 -00 -00 +00 —00 -^ -IrOO GV cho HS thuc hanh bang each ldy cac vi du cu the va dien vao bang sau "n lim(MnV„) ^n +00 • • • —00 -00 +00 • GV neii Vl du Cd th^ thay bang cac vf du khac • GV ndu quy tac 2: Neu lim«„ = ± oo va Iimv„ = L^0 thi lim(M„v„) dugc cho bang sau : limMn Da'u cua L Um(M„Vn) +00 + +00 +00 —00 123 GV cho HS thuc hanh bang each ldy cae vf du cu the va dien vao bang sau : "n Ddu eiia L lim(77nVn) + +00 —00 + —00 - +00 • GV neu vf du Cd the thay bang cac vf du khac Thuc hien [HIJ 5' Hoat ddng ciia HS Hoat ddng ciia GV Cau hdi Ggi y tra Idi cau hdi Tim lim(/7 sin/7 - 2/7 ) nsmn - 2/7 = /? sin/7 ^ , - + —r— VOl moi n^ ) I' n Vi Um lim ^ n sin/7"' = = - 2< + t V '- 77^ y , lim(/7sin/7 -2n ) = -oo V +oo va nen Cau hdi Ggi y tra ldi cau hoi Tim r Vi Iim nsinn — 2n \ = +oo nen •2 nsinn-2n GV neu quy tdc 3: 124 hm = 77 sin n - 2/? Ni'u IimM„ = L^ 0,limv„ = va v„ > hodc v„ < ki tic mpt sd hgng ndo dd trd.diM Xim -^ dugc cho bang sau : lim^ +00 -00 + —00 +00 1 + + Ddu cua Vn + Dau cua L GV cho HS thuc hanh bang each ldy cac yf du cu the va dien vao bang sau : Dd'u ciia L limi^ Vn + +00 + —00 —00 - +00 • GV neu vf du Cd the thay bang cac yf du khac • Thuc hien [H2J 5' Hoat dghg ciia GV Hoat ddng ciia HS Cau hdi Ggi y tra ldi cau hdi Chia ca ttr va mdu cho so Chia ca tir va mau cho n nao? Ggi y tra Idi cau hdi Cau hdi _ 3/7^+2/7-1 Tim hm r-^ 2/7^/7 , -2n^+n Iim — = -co 3/7-2 125 HOATDONC MOT SO C^a HOI TR^C NGHIEM ON T^P Bfll 3-n L.au 11 m —— Dang: 2n2 ( a ) CO (b) +00 ; (0 ; -1 7ra7^V/ (b Cau h m r— bang: 2n^ (a) 00 (b) + « ; (o-i; -i T/-a uyi (a) ^ , - n - l Cfl77 J llm — p = - bang: 2\ U' (a) 00 < • ib) +oo; -i- Trd ldi (a) ' 3n^+2 , C(77/ 11 m ,, ; — bang: V n^ + n +1 (a) + 00 (b) -00 ; (c, ^ • (d) I Trd ldi (a) 126 Cdu + 00 la gidi han cua day sd nao sau day: 3n + (a) lim 3n + (b) lim Vn^-n + : Vn^ +n + l 3n+2 (c) lim (d) lim vn" +n + 3n + VnTl Trd ldi (d) Cdu _ 00 la gidi ban cua day so nao sau day -3n + (a) lim (b) lim Vn^ + n + -Vn^ + n + l -3n + (c) lim (d) lim V n-^ + n+ Trd Idi (d) „ , V3 + n^ , bang: Cau lim n+ (a) + 00 (b) (c) ; -f- 00 Trd.ldi (a) ^ , Cau im V3+n^ -n +n+l 3n + hano- ( a ) + 00 (c,i: Trd ldi (b) ^ n , v3 + n^ ,^ Cau l\m — bang: 3n^+n + l ( b ) 00 -3n + 4n + l (a) + 00 (b)_a) | Trd ldi ia) Cdu yo lim -\/3 + n- bang: -3n^+n + l (b) 00 (a) + 00 (0^; Trd ldi (b) • HOATDONC HCrO^NG D^N Bfil T^P SfiCH GIfiO KHOfi Bai 11 Hicdng ddn Sir dung dinh nghTa va quy tdc tfnh gidi ban tai vd cue Ddp so a) -00 ; b) +oo , Bai 12 Hicdng ddn Sir dung dinh nghTa va quy tdc tfnh gidi ban tai vd cue a) Hoat ddng ciia GV Cau hdi I I Hoat ddng ciia HS Ggi y tra ldi cau hdi Chia ca tir va mdu so cho so Chia ca> tir•> va mau cho n nao ? Cau hdi Ggi y tra ldi cau hoi Tim gidi ban ciia cac day sd (t