. CHAPTER 0 SEQUENCE, SERIES, d INDUCTION Secto 0. Seqece A lst of mers specfc order. E / Fd the frst terms : of the gve seqece: Aswer: Frst, I ll show how to fd the terms ltcll the I ll show how to se the TI to fd them. 3 Let, Let, 4 3 4 4 Let 3 3, Let 4 4, 3 9 4 6 6 3 4 6 Let, The frst terms of the seqece re,,,, 4 9 6 Tr / Fd the frst terms of the gve seqece: where the dom of s {,,3, } Seqece mode: 4= 9z=( D =3 94= 94=44 ) =7 s=zcs ) =9 ' Fcto mode: Isert seqece here, (seqece, vrle, strt, ed, step)
.. of Focc Seqece, gve tht d Fd the frst terms of the seqece 3 3 3 4 4 4 3 3 43 3 The frst terms of the seqece re,,, 3,. E /Fd the frst terms of the seqece 3, gve tht,=z 948+3= +3+3= s=t3 z, 9 ]= +3=8 E / Fd the frst terms of the seqece (), gve tht.t#iith :*, 's" 3 = 8 E 3/ Fd the geerl term of seqece whose frst for terms re,6,7,8,... I... +4 ={,/3,4 3 9= 9h, + 9= E 4/ Fd the geerl term of seqece whose frst for terms re,4,6,8,.. h={,, }.. =#h 9,=z 9=Zg, E / Fd the geerl term of seqece whose frst for terms re,3,,7,... { 4,3 E 6/ Fd the geerl term of seqece whose frst for terms re,4,8,6,... z E 7/ Fd the geerl term of seqece whose frst for terms re 3,9,7,8,... Ht: ( ) + fht! " }
Seres The reslt of ddg the terms of seqece. Smmto otto (codesed form of seres): E 8/ Fd the sm: Aswer: The emple sks for the sm of the frst terms of the seqece whose th term s. The de the smmto otto strts t d eds t. Replce the th term wth,, 3, 4, d, d the fd the sm of the fve terms. 3 4 076 60 d LIST MATH : sm ( d LIST OPS : seq (,,,,) ENTER seqece, vrle, strt, ed, step * Or o cold se the seqece mode d jst dd the mers eeded. Tr 8/ j 6 j Tr 9/ = '=+s+ +st=7@ =fd '. +4.+43>+44.4+4+4,96 Tr 0 / = 9+6+.+36=,0 =#+#+#'+±Pt#, +.4 + KIH ) 4+ ) 36+ ) 4C4+ ) SCSTD +.tt#tot+3ot 3
4 c c 6 4 3 3 3 0 3 Evlte the sm wthot g clcltor. 00 0 00 40 00 000 4 ( ) ( ) oiteg3ois.87ot63okoe.t3ei.qjs870zooezetezzocztt_zocs3@toczeitte.oeyoextoidlockd68970ls4oeielooc.e.t?e4
. Arthmetc d Geometrc Seqeces Secto 03 A Arthmetc Seqece s seqece whch the dfferece etwee two sccessve terms s lws the sme costt. The costt s clled the commo dfferece d. ' Tr/ Wht s the commo dfferece, d, for the followg: 8, 4, 8,, 6, 30,... W d=4 The th Term of Arthmetc Seqece: 0 3 3 3 3 4 3 3 4 3 6 6 4 9 6 0 zr = 3 4 4 3 4 4 4 6 4 =3 6 co 4 9=3 4 z GGF =3C 4 DHO th The term of rthmetc seqece s gve d E / I the seqece, 4, 7,0,3,6, fd 0 (tht s, fd the tweteth term).,= D= 7 4=3 =,+( Dd A= test ) } 0+46 ) } 90=+(0) } = +967=8 @ E / I the seqece9,,, 3, 7,, fd (tht s, fd the twetffth term)., = 9 D= 9=4. q=d(ht)t9, 9=Stk. 9zs=, +46 4) zs=+zzf4) =t(h 49 90=0+(04) } }) = qs=9+464 ) 87 87
',. ' 8D 9 d 83. E 3/ Fd the frst terms the rthmetc seqece f, t.c : 9 ( ) d, 83 ( ) d,t4d,t0d 9=9/+44 ) 9=449 9=9+6 9. 9 3=9 Sm of Arthmetc Seqeces: I the rthmetc seqece,, 8,,4,..., f we deote the sm of the frst terms the rthmetc seqece S the 8 S. S d, se ths forml f d d re gve. S, se ths forml f d re gve. 3+4,. S=zt[,+D=EE3+, E 4/ Fd the sm of the frst 7 terms, f = 3 d 7 = 4 : h= 83=4 64,6=4 6 4=d )4,=3 z= > 93= =l 9=9 +4 Dd D= ' z E / Fd the sm of the frst 00 postve tegers the rthmetc seqece, 3,, 7, 9,. %o=toe[z(d+godd, D= 3= o%o=t99() 0,000 Tr 6/ Fd the sm of the frst 0 terms, f 9 = d 3 = 3. Ht: look t emple 3.,+8G. >s)= g=qt8d= ±}0== 9,3=9,+d=3, = + 's: =,= 4 S,o=t [ zetdo DHD =. D= 3,7g 6
' Geometrc Seqeces, Seres, d ther Applctos A Geometrc Seqece s seqece whch the rto etwee two sccessve terms s lws the sme costt. The costt s clled the commo rto r. 6 8 4 If, 6, 8, 4,...s seqece, the 3, 3, 3,, r 6 8 The th Term of Geometrc Seqece:, r 3 r ()3 8 3 4 3 4 r 8 3 4 r 4 3 6 r 6 3 486 3 4, r () ( ) EX / I the seqece, 3, 9, 7,8, 43, fd 0 (tht s, fd the teth term) Ht: r = 3 = 9 3 = 7 9 r =,o= = = 3 (3) h,= (39 = k$683 Therefore, 3 3 4 3 3 4 4 4 6 r r r r r r r r r r r r r r r r r r r r r EX / I the seqece6, 8, 4,,, /, fd 0 (tht s, fd the teth term). Ht: r = 8 6 = = = 0. = 6 fsj 9,0=6 C 0.9=3 's ', = 6 7
/3 d 8. EX 3/ Fd the frst terms the geometrc seqece f 3 ) Use the forml r r, 8 r 3 0 3 ) Dvde the secod eqto the frst eqto to elmte. 8 6 : 9 3 3 ) solve for r g "chge of se." terse 6 49 't v) Fd. e 0 rts 8 z= v) Use d r to fd the reqested terms.,= to r= 9=Fs(D " ",=Ks z=t,, = stge 9 to 8
. 36=9. Sm of Geometrc Seqeces: I the geometrc seqece, 8, 3,8,..., f we deote the sm of the frst terms the rthmetc seqece S the 8 3 S. r r S or S r r EX 4/ Fd the sm of the frst 0 terms the geometrc seqece, f =, r = 3,0=4 3,= 34,88 EX / Fd the sm of the frst 7 terms the geometrc seqece,6,8,4,6, S>=C 33 #= 86 9=, r = =s g=z EX 6/ Fd the sm of the frst 7 terms the geometrc seqece, f 3 = 36, 4 = 08. Ht: r },r 36=9,( 3 =o8=, r3 9 T ' =3= I " ' 3=, 4*333=437 9
Ifte Geometrc Seres: K If r, r proper frcto, the the fte geometrc seres S 3 4 coverges d ts sm s gve the forml S, r =9 f If r, the the fte geometrc seres S 34 dverges d hs o sm. Emple : Fd the sm of the geometrc seres,,,,, 3 9 7 8 d r., 3 3 S r ± =3 : Emple 6: Chge the repetg decml to rtol mer. 0.34 0.34343434 The repetg decml mer c e wrtte s 0.34 0.0034 0.000034 0.00000034 0.0034 We see tht ths s fte seres wth 0.34 d r 0.34 00 0.34 0.34 34 0.34 0.34343434 99 99 00 00 or 34 99 Sometmes o hve to thk ot wht the d r re. Emple 7: Fd Ht: Wrte t ot. zs ' +64 +C tsp 3 + ts + Is, =3 r= g=t #4 so = s=t 0
Comto Notto: The Boml Theorem Secto 0.6! C, r Cr Cr :red s " choose r". r r r!! r Grphg Clcltor Solto: To evlte 0 C press 0 MATH PRB 3 : Cr ENTER Emple : Evlte Itrodcto: The epso of the oml 0 0 6 ) ) c) 3 4 wth dfferet vles of s gve elow: whe the epoet s 0 the epso wll e whe the epoet s the epso wll e 3 whe the epoet s the epso wll e whe the epoet s 3 the epso wll e 3 3 whe the epoet s 4 the epso wll e 3 3 4 4 3 3 4 4 6 4 whe the epoet s the epso wll e 4 3 0 0 3 4 If we ol wrte the coeffcets of ech epso we shll get the followg (Pscl's trgle): 0 whe the epoet s 0 the coeffcets of the epso wll e whe the epoet s the coeffcets of the epso wll e 3 whe the epoet s the coeffcets of the epso wll e 4 whe the epoet s 3 the coeffcets of the epso wll e 3 3 whe the epoet s 4 the coeffcets of the epso wll e 464 whe the epoet s the coeffcets of the epso wll e 00
0 whe the epoet s 0 the coeffcets of the epso wll e 0 0 whe the epoet s the coeffcets of the epso wll e 0 whe the epoet s the coeffcets of the epso wll e 0 3 whe the epoet s 3 the coeffcets of the epso wll e 3 3 3 3 0 3 4 whe the epoet s 4 the coeffcets of the epso wll e 4 4 4 4 4 0 3 4 whe the epoet s the coeffcets of the epso wll e 0 3 4 All of the ove comto mers re clled the oml coeffcets. The Boml Theorem: For ech postve teger, 3 4 r 3 3 4 4 r r Term Term Term 3 Term 4 Term Term (r) Term (+) Note : I ech term, the epoet of d the epoet of the dd p to. Note : The k th term the epso s, k th the mer of the term. EX/ The 6th term for ( + ) s Tr/ Fd the 8th term for ( + ) Tr/ Fd the th term for ( + ) 7 79 7 7 Tr/ Fd the term of the oml epso of ( + ) 7 wth power of 4.. Notce tht the epoet of s oe less CY)Y ' = 6,80 447 ( I, ) 34 3 44 = (3) 4 ' 3 = 3 4
To fd ll of the oml coeffcets of the oml epso ( + ) oe step: ) Go to the fcto edtor presg Y ) Net to Y=, eter Cr, where s the epoet of the oml to e epded. 3) Press TABLE TBLST, move crsor et to TlStrt d eter zero. Net to Tl=, eter. Move crsor to to d hghlghted t presg ENTER. 4) Press TABLE ) All coeffcets wll e show the colm leled Y. Igore vle greter th. The screes elow show how to fd the oml coeffcets of oml of degree. EX/ Epd the oml XGYY Yz) 3GP YzP s + + 0 + 0 + + XS +0 44 +40 3+80 43 44 +80 44+3 (zjs EX/ Epd the oml 4 3 Emple 3: Fd the 8 th term the epso of oml 4. Emple 4: I the oml3, fd the coeffcet of the term whose vrle prt s Emple : I the oml 0K 644*44*+434 4 34+44 4 'D ' 6 4 96 3/+6 4 ( 3 0, fd the term whose vrle prt s 6 6 43+8 4 )C4Xj kt 8,008 7 CYKXKH'.t#oE*m.ssE*dEeed3 6. 946447,66 4
Mthemtcl Idcto Secto 0.3 Coteremple To show cojectre fls we jst eed oe emple of the cojectre to fl. Ths s clled coteremple. Fd coter emple for the followg cojectres: ) + < 0 ) 3 + = 6 + 6 N TRUE N TRUE 3 4 6 7 8 9 3 4 6 7 8 9 Lookg For Ptter: P : 4 + 8 + + + 4 = ( + ) N TRUE 3 4 6 7 8 9 Let s s tht t s tre for =. If we c show tht t s tre for the =k, the t s tre for =k+. Wht does ths tell s? Dd we jst prove the eqto s tre? 4
Theorem : Prcple of Mthemtcl Idcto ) Show tht t s tre for the frst oe. ) Show tht f t s tre for k, the t s tre for k+. Wrm ps: Crete p k d p k+ P : 4 + 8 + + + 4 = ( + ) p k: p k+: P : + 8 + 6 + + 8( ) = ( ) ; > p k: p k+: P : ( ) = p k: p k+: P : + 4 + 8 + = ( ) p k: p k+:
Now we c eg P : 4 + 8 + + + 4 = ( + ) ) Show tht t s tre for =. ) Show tht f t s tre for k, the t s tre for k+. p k: p k+: Ssttte: + 4 + 8 + = ( ) ) Show tht t s tre for =. ) Show tht f t s tre for k, the t s tre for k+. p k: p k+: Ssttte: 6
P : + 8 + 6 + + 8( ) = ( ) ; > ) Show tht t s tre for =. ) Show tht f t s tre for k, the t s tre for k+. p k: p k+: Ssttte: P : ( ) = ) Show tht t s tre for =. ) Show tht f t s tre for k, the t s tre for k+. p k: p k+: Ssttte: 7
P : + + 3 + 4 + + + = ) Show tht t s tre for =. ( + ) ) Show tht f t s tre for k, the t s tre for k+. p k: p k+: Ssttte: P : + + 3 + 4 + + + = ) Show tht t s tre for =. ( + )( + ) 6 ) Show tht f t s tre for k, the t s tre for k+. p k: p k+: Ssttte: 8
TRIGONOMETRIC IDENTITIES Chpter 6 revew, pge 60 #3 Techqes for smplfg oce o hve plced the epresso terms of e d e. Wrte ech epresso terms of e d e d smplf so tht o qotets pper the fl epresso. ) θ sec θ ) Wrte terms of e or e. ) Smplf d se the qotet dettes f eeded. ) t θ θ ) Wrte terms of e or e. ) Smplf d se the qotet dettes f eeded. Techqes for smplfg oce o hve plced the epresso terms of e d e. c) Tr ddg d) Comple frcto. 9
e) Tr Mltplg f) Tr fctorg θ θ ht: = ( )( + ) g) Tr fctorg 3 θ 3 θ ht: 3 3, 3 3 h) Tr fctorg θ + θ θ ) Tr to fctor kll. θ θθ θ or tr to wrte s two seprte frctos. θ θθ θ 0
Wrte terms of e or e d/or Smplf s e. ) θ θ θθ ) θ(cscθ θ) c) tθ secθ θ
Prove the dett sec Left sde (more complestrt here) sec Rght sde Prove the dett Left sde tsec t sec Rght sde(more complestrt here)
3 SIN TAN t t t t ) t( t t t t t COS t cot t sec csc sec csc IDENTITIES RECIPROCAL csc cot sec t IDENTITIES PYTHAGOREAN ANGLE DOUBLE t t t t ANGLE HALF t ) ( ) ( ) ( ) ( ) ( ) ( ) ( v v v v v v v v v v v v SUM TO PRODUCT PRODUCT TO SUM Prove the dett sec cot csc Ct the followg ot:
Use the dettes to prove the followg: ( )( + ) = θ = t θ + t θ θ = θ θ 4