BINOMIAL THEOREM. Mathematics is a most exact science and its conclusions are capable of absolute proofs. C.P. STEINMETZ

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1 Chapte 8 BINOMIAL THEOREM Maematics is a most eact sciece ad its coclusios ae capable of absolute poofs. C.P. STEINMETZ 8. Itoductio I ealie classes, we have leat how to fid e squaes ad cubes of biomials like a + b ad a b. Usig em, we could evaluate e umeical values of umbes like (98) 2 (00 2) 2, (999) 3 (000 ) 3, etc. Howeve, fo highe powes like (98) 5, (0) 6, etc., e calculatios become difficult by usig epeated multiplicatio. This difficulty was ovecome by a eoem kow as biomial eoem. It gives a easie way to epad (a + b), whee is a itege o a atioal umbe. I is Chapte, we study biomial eoem fo positive itegal idices oly. 8.2 Biomial Theoem fo Positive Itegal Idices Let us have a look at e followig idetities doe ealie: (a+ b) 0 a + b 0 (a+ b) a + b (a+ b) 2 a 2 + 2ab + b 2 (a+ b) 3 a 3 + 3a 2 b + 3ab 2 + b 3 (a+ b) (a + b) 3 (a + b) a + a 3 b + 6a 2 b 2 + ab 3 + b I ese epasios, we obseve at (i) (ii) (iii) Blaise Pascal ( ) The total umbe of tems i e epasio is oe moe a e ide. Fo eample, i e epasio of (a + b) 2, umbe of tems is 3 wheeas e ide of (a + b) 2 is 2. Powes of e fist quatity a go o deceasig by wheeas e powes of e secod quatity b icease by, i e successive tems. I each tem of e epasio, e sum of e idices of a ad b is e same ad is equal to e ide of a + b.

2 BINOMIAL THEOREM 6 We ow aage e coefficiets i ese epasios as follows (Fig 8.): Fig 8. Do we obseve ay patte i is table at will help us to wite e et ow? Yes we do. It ca be see at e additio of s i e ow fo ide gives ise to 2 i e ow fo ide 2. The additio of, 2 ad 2, i e ow fo ide 2, gives ise to 3 ad 3 i e ow fo ide 3 ad so o. Also, is peset at e begiig ad at e ed of each ow. This ca be cotiued till ay ide of ou iteest. We ca eted e patte give i Fig 8.2 by witig a few moe ows. Fig 8.2 Pascal s Tiagle The stuctue give i Fig 8.2 looks like a tiagle wi at e top vete ad uig dow e two slatig sides. This aay of umbes is kow as Pascal s tiagle, afte e ame of Fech maematicia Blaise Pascal. It is also kow as Meu Pastaa by Pigla. Epasios fo e highe powes of a biomial ae also possible by usig Pascal s tiagle. Let us epad (2 + 3y) 5 by usig Pascal s tiagle. The ow fo ide 5 is Usig is ow ad ou obsevatios (i), (ii) ad (iii), we get (2 + 3y) 5 (2) 5 + 5(2) (3y) + 0(2) 3 (3y) 2 +0 (2) 2 (3y) 3 + 5(2)(3y) +(3y) y y y y + 23y 5.

3 62 MATHEMATICS Now, if we wat to fid e epasio of (2 + 3y) 2, we ae fist equied to get e ow fo ide 2. This ca be doe by witig all e ows of e Pascal s tiagle till ide 2. This is a slightly legy pocess. The pocess, as you obseve, will become moe difficult, if we eed e epasios ivolvig still lage powes. We us ty to fid a ule at will help us to fid e epasio of e biomial fo ay powe wiout witig all e ows of e Pascal s tiagle, at come befoe e ow of e desied ide. Fo is, we make use of e cocept of combiatios studied ealie to ewite! e umbes i e Pascal s tiagle. We kow at C, 0 ad!( )! is a o-egative itege. Also, C 0 C The Pascal s tiagle ca ow be ewitte as (Fig 8.3) Fig 8.3 Pascal s tiagle Obsevig is patte, we ca ow wite e ow of e Pascal s tiagle fo ay ide wiout witig e ealie ows. Fo eample, fo e ide e ow would be C 0 C C 2 C 3 C C 5 C 6 C. Thus, usig is ow ad e obsevatios (i), (ii) ad (iii), we have (a + b) C 0 a + C a 6 b + C 2 a 5 b 2 + C 3 a b 3 + C a 3 b + C 5 a 2 b 5 + C 6 ab 6 + C b A epasio of a biomial to ay positive itegal ide say ca ow be visualised usig ese obsevatios. We ae ow i a positio to wite e epasio of a biomial to ay positive itegal ide.

4 BINOMIAL THEOREM Biomial eoem fo ay positive itege, (a + b) C 0 a + C a b + C 2 a 2 b C a.b + C b Poof The poof is obtaied by applyig piciple of maematical iductio. Let e give statemet be P() : (a + b) C 0 a + C a b + C 2 a 2 b C a.b + C b Fo, we have P () : (a + b) C 0 a + C b a + b Thus, P () is tue. Suppose P (k) is tue fo some positive itege k, i.e. (a + b) k k C 0 a k + k C a k b + k C 2 a k 2 b k C k b k... () We shall pove at P(k + ) is also tue, i.e., (a + b) k + k + C 0 a k + + k + C a k b + k + C 2 a k b k + C k+ b k + Now, (a + b) k + (a + b) (a + b) k (a + b) ( k C 0 a k + k C a k b + k C 2 a k 2 b k C k ab k + k C k b k ) [fom ()] k C 0 a k + + k C a k b + k C 2 a k b k C k a 2 b k + k C k ab k + k C 0 a k b + k C a k b 2 + k C 2 a k 2 b k C k- ab k + k C k b k + [by actual multiplicatio] k C 0 a k + + ( k C + k C 0 ) a k b + ( k C 2 + k C )a k b ( k C k + k C k ) ab k + k C k b k + [goupig like tems] k + C 0 a k + + k + C a k b + k + C 2 a k b k + C k ab k + k + C k + b k + (by usig k + C 0, k C + k C k + C ad k C k k + C k + ) Thus, it has bee poved at P (k + ) is tue wheeve P(k) is tue. Theefoe, by piciple of maematical iductio, P() is tue fo evey positive itege. We illustate is eoem by epadig ( + 2) 6 : ( + 2) 6 6 C C C C C C C Thus ( + 2)

5 6 MATHEMATICS Obsevatios. The otatio k 0 C k a k b k stads fo C 0 a b 0 + C a b C a b C a b, whee b 0 a. Hece e eoem ca also be stated as k k ( a b) C k a b. k 0 2. The coefficiets C occuig i e biomial eoem ae kow as biomial coefficiets. 3. Thee ae (+) tems i e epasio of (a+b), i.e., oe moe a e ide.. I e successive tems of e epasio e ide of a goes o deceasig by uity. It is i e fist tem, ( ) i e secod tem, ad so o edig wi zeo i e last tem. At e same time e ide of b iceases by uity, statig wi zeo i e fist tem, i e secod ad so o edig wi i e last tem. 5. I e epasio of (a+b), e sum of e idices of a ad b is + 0 i e fist tem, ( ) + i e secod tem ad so o 0 + i e last tem. Thus, it ca be see at e sum of e idices of a ad b is i evey tem of e epasio Some special cases I e epasio of (a + b), (i) Takig a ad b y, we obtai ( y) [ + ( y)] C 0 + C ( y) + C 2 2 ( y) 2 + C 3 3 ( y) C ( y) C 0 C y + C 2 2 y 2 C 3 3 y ( ) C y Thus ( y) C 0 C y + C 2 2 y ( ) C y Usig is, we have ( 2y) 5 5 C C (2y) + 5 C 2 3 (2y) 2 5 C 3 2 (2y) 3 + (ii) Thus 5 C (2y) 5 C 5 (2y) y y y y 32y 5. Takig a, b, we obtai ( + ) C 0 () + C () + C 2 () C C 0 + C + C C C ( + ) C 0 + C + C C C

6 BINOMIAL THEOREM 65 I paticula, fo, we have 2 C 0 + C + C C. (iii) Takig a, b, we obtai ( ) C 0 C + C ( ) C I paticula, fo, we get 0 C 0 C + C ( ) C Eample Epad 2 3, 0 Solutio By usig biomial eoem, we have C 0 ( 2 ) + C ( 2 ) 3 + C 2 ( 2 ) Eample 2 Compute (98) 5. + C 3 ( 2 ) C 3 Solutio We epess 98 as e sum o diffeece of two umbes whose powes ae easie to calculate, ad e use Biomial Theoem. Wite Theefoe, (98) 5 (00 2) 5 5 C 0 (00) 5 5 C (00) C 2 (00) C 3 (00) 2 (2) C (00) (2) 5 C 5 (2) Eample 3 Which is lage (.0) o 0,000? Solutio Splittig.0 ad usig biomial eoem to wite e fist few tems we have

7 66 MATHEMATICS (.0) ( + 0.0) C C (0.0) + oe positive tems oe positive tems oe positive tems > 0000 Hece (.0) > 0000 Eample Usig biomial eoem, pove at 6 5 always leaves emaide whe divided by 25. Solutio Fo two umbes a ad b if we ca fid umbes q ad such at a bq +, e we say at b divides a wi q as quotiet ad as emaide. Thus, i ode to show at 6 5 leaves emaide whe divided by 25, we pove at k +, whee k is some atual umbe. We have Fo a 5, we get ( + a) C 0 + C a + C 2 a C a ( + 5) C 0 + C 5 + C C 5 i.e. (6) C C i.e ( C 2 + C ) o ( C C ) o k+ whee k C C This shows at whe divided by 25, 6 5 leaves emaide. EXERCISE 8. Epad each of e epessios i Eecises to 5.. ( 2) (2 3) 6

8 BINOMIAL THEOREM Usig biomial eoem, evaluate each of e followig: 6. (96) 3. (02) 5 8. (0) 9. (99) 5 0. Usig Biomial Theoem, idicate which umbe is lage (.) 0000 o Fid (a + b) (a b). Hece, evaluate ( 3 2) ( 3 2). 2. Fid ( + ) 6 + ( ) 6. Hece o oewise evaluate ( 2 + )6 + ( 2 )6. 3. Show at is divisible by 6, wheeve is a positive itege.. Pove at 3 C Geeal ad Middle Tems. I e biomial epasio fo (a + b), we obseve at e fist tem is C 0 a, e secod tem is C a b, e id tem is C 2 a 2 b 2, ad so o. Lookig at e patte of e successive tems we ca say at e ( + ) tem is C a b. The ( + ) tem is also called e geeal tem of e epasio (a + b). It is deoted by T +. Thus T + C a b. 2. Regadig e middle tem i e epasio (a + b), we have (i) If is eve, e e umbe of tems i e epasio will be +. Sice is eve so + is odd. Theefoe, e middle tem is, i.e., 2 2 tem. 8 Fo eample, i e epasio of ( + 2y) 8, e middle tem is 2 5 tem. i.e., (ii) If is odd, e + is eve, so ee will be two middle tems i e

9 68 MATHEMATICS epasio, amely, 2 tem ad 2 tem. So i e epasio (2 y), e middle tems ae, i.e., 2 ad 2 2, i.e., 5 tem I e epasio of, whee 0, e middle tem is 2, i.e., ( + ) tem, as 2 is eve. It is give by 2 C 2 C (costat). This tem is called e tem idepedet of o e costat tem. Eample 5 Fid a if e ad 8 tems of e epasio (2 + a) 50 ae equal. Solutio The ( + ) tem of e epasio ( + y) is give by T + C y. Fo e tem, we have, +, i.e., 6 Theefoe, T T C 6 (2) 50 6 a 6 50 C a 6. Similaly, T 8 50 C 2 33 a Give at T T 8 So 50 C 6 (2) 3 a 6 50 C (2) 33 a C 6.2 a Theefoe C.2 a 50 3 i.e., a 50 C C 50!!. 33! 2 6!3! 50! Eample 6 Show at e middle tem i e epasio of (+) 2 is 35...(2) 2, whee is a positive itege.!

10 BINOMIAL THEOREM 69 Solutio As 2 is eve, e middle tem of e epasio ( + ) 2 is i.e., ( + ) tem which is give by, T + 2 C () 2 () 2 C (2 )!!! 2 2, 2 (2) (2 2) !! (2 2) (2 )(2 )!! [ (2 )][ (2)]!! [ (2 )]2!! [ ] [ (2 )]! 2.!! (2 ) 2! Eample Fid e coefficiet of 6 y 3 i e epasio of ( + 2y) 9. Solutio Suppose 6 y 3 occus i e ( + ) tem of e epasio ( + 2y) 9. Now T + 9 C 9 (2y) 9 C y. Compaig e idices of as well as y i 6 y 3 ad i T +, we get 3. Thus, e coefficiet of 6 y 3 is 9 C ! !6! Eample 8 The secod, id ad fou tems i e biomial epasio ( + a) ae 20, 20 ad 080, espectively. Fid, a ad. Solutio Give at secod tem T 2 20

11 0 MATHEMATICS We have T 2 C. a So C. a () Similaly C 2 2 a (2) ad C 3 3 a (3) Dividig (2) by (), we get 2 2 C2 a 20 ( )! a i.e.,. 6 C a 20 ( 2)! a 6 o ( ) Dividig (3) by (2), we have a 9 2( 2) Fom () ad (5), 6 9. Thus, 5 2( 2) a 3 Hece, fom (), 5 a 20, ad fom (), 2 Solvig ese equatios fo a ad, we get 2 ad a ()... (5) Eample 9 The coefficiets of ee cosecutive tems i e epasio of ( + a) ae i e atio: : 2. Fid. Solutio Suppose e ee cosecutive tems i e epasio of ( + a) ae ( ), ad ( + ) tems. The ( ) tem is C 2 a 2, ad its coefficiet is C 2. Similaly, e coefficiets of ad ( + ) tems ae C ad C, espectively. Sice e coefficiets ae i e atio : : 2, so we have, C 2, i.e., () C C ad, i.e., (2) C 2 Solvig equatios() ad (2), we get, 55.

12 BINOMIAL THEOREM Fid e coefficiet of EXERCISE i ( + 3) 8 2. a 5 b i (a 2b) 2. Wite e geeal tem i e epasio of 3. ( 2 y) 6. ( 2 y) 2, Fid e tem i e epasio of ( 2y) Fid e 3 tem i e epasio of 9 3 Fid e middle tems i e epasios of y , I e epasio of ( + a) m+, pove at coefficiets of a m ad a ae equal. 0. The coefficiets of e ( ), ad ( + ) tems i e epasio of ( + ) ae i e atio : 3 : 5. Fid ad.. Pove at e coefficiet of i e epasio of ( + ) 2 is twice e coefficiet of i e epasio of ( + ) Fid a positive value of m fo which e coefficiet of 2 i e epasio ( + ) m is 6. Miscellaeous Eamples Eample 0 Fid e tem idepedet of i e epasio of Solutio We have T C C

13 2 MATHEMATICS (3) ( ) C (2) The tem will be idepedet of if e ide of is zeo, i.e., Thus, 68 (3) 5 Hece 5 tem is idepedet of ad is give by ( ) 6 C 6. (2) 2 Eample If e coefficiets of a, a ad a + i e epasio of ( + a) ae i aimetic pogessio, pove at 2 ( + ) Solutio The ( + ) tem i e epasio is C a. Thus it ca be see at a occus i e ( + ) tem, ad its coefficiet is C. Hece e coefficiets of a, a ad a + ae C, C ad C +, espectively. Sice ese coefficiets ae i aimetic pogessio, so we have, C + C + 2. C. This gives!!! 2 ( )!( )! ( )!( )!!( )! i.e. ( )!( )( ) ( )! ( )( )( )!( )! 2 ( )!( )( )! o ( )! ( )! ( ) ( ) ( ) ( ) 2 ( )! ( )![ ( )] i.e. o 2, ( )( ) ( ) ( ) ( ) ( )( ) 2 ( )( ) ( ) ( ) o ( + ) + ( ) ( + ) 2 ( + ) ( + ) o ( )

14 BINOMIAL THEOREM 3 o i.e., 2 ( + ) Eample 2 Show at e coefficiet of e middle tem i e epasio of ( + ) 2 is equal to e sum of e coefficiets of two middle tems i e epasio of ( + ) 2. Solutio As 2 is eve so e epasio ( + ) 2 has oly oe middle tem which is 2 2 i.e., ( + ) tem. The ( + ) tem is 2 C. The coefficiet of is 2 C Similaly, (2 ) beig odd, e oe epasio has two middle tems, 2 2 ad i.e., 2 ad ( + ) tems. The coefficiets of 2 ese tems ae 2 C ad 2 C, espectively. Now 2 C + 2 C 2 C [As C + C + C ]. as equied. Eample 3 Fid e coefficiet of a i e poduct ( + 2a) (2 a) 5 usig biomial eoem. Solutio We fist epad each of e factos of e give poduct usig Biomial Theoem. We have ( + 2a) C 0 + C (2a) + C 2 (2a) 2 + C 3 (2a) 3 + C (2a) + (2a) + 6(a 2 ) + (8a 3 ) + 6a. + 8a + 2a a 3 + 6a ad (2 a) 5 5 C 0 (2) 5 5 C (2) (a) + 5 C 2 (2) 3 (a) 2 5 C 3 (2) 2 (a) 3 Thus ( + 2a) (2 a) C (2) (a) 5 C 5 (a) a + 80a 2 0a 3 + 0a a 5 ( + 8a + 2a a 3 + 6a ) (32 80a + 80a 2 0a 3 + 0a a 5 ) The complete multiplicatio of e two backets eed ot be caied out. We wite oly ose tems which ivolve a. This ca be doe if we ote at a. a a. The tems cotaiig a ae (0a ) + (8a) ( 0a 3 ) + (2a 2 ) (80a 2 ) + (32a 3 ) ( 80a) + (6a ) (32) 38a

15 MATHEMATICS Thus, e coefficiet of a i e give poduct is 38. Eample Fid e tem fom e ed i e epasio of ( + a). Solutio Thee ae ( + ) tems i e epasio of ( + a). Obsevig e tems we ca say at e fist tem fom e ed is e last tem, i.e., ( + ) tem of e epasio ad + ( + ) ( ). The secod tem fom e ed is e tem of e epasio, ad ( + ) (2 ). The id tem fom e ed is e ( ) tem of e epasio ad ( + ) (3 ) ad so o. Thus tem fom e ed will be tem umbe ( + ) ( ) ( + 2) of e epasio. Ad e ( + 2) tem is C + a +. 3 Eample 5 Fid e tem idepedet of i e epasio of 3 2, > 0. Solutio We have T + 8 C C C 2 Sice we have to fid a tem idepedet of, i.e., tem ot havig, so take We get 9. The equied tem is 8 C Eample 6 The sum of e coefficiets of e fist ee tems i e epasio of 3 m 2, 0, m beig a atual umbe, is 559. Fid e tem of e epasio cotaiig 3. 3 m Solutio The coefficiets of e fist ee tems of 2 ad 9 m C 2. Theefoe, by e give coditio, we have ae m C 0, ( 3) m C m C 0 3 m C + 9 m C 2 559, i.e., 3m + 9 m ( m ) 559 2

16 BINOMIAL THEOREM 5 which gives m 2 (m beig a atual umbe). 3 Now T + 2 C C ( 3). 2 3 Sice we eed e tem cotaiig 3, so put i.e., 3. Thus, e equied tem is 2 C 3 ( 3) 3 3, i.e., Eample If e coefficiets of ( 5) ad (2 ) tems i e epasio of ( + ) 3 ae equal, fid. Solutio The coefficiets of ( 5) ad (2 ) tems of e epasio ( + ) 3 ae 3 C 6 ad 3 C 2 2, espectively. Sice ey ae equal so 3 C 6 3 C 2 2 Theefoe, eie o 6 3 (2 2) [Usig e fact at if C C p, e eie p o p] So, we get o. beig a atual umbe, is ot possible. So,. Miscellaeous Eecise o Chapte 8. Fid a, b ad i e epasio of (a + b) if e fist ee tems of e epasio ae 29, 290 ad 3035, espectively. 2. Fid a if e coefficiets of 2 ad 3 i e epasio of (3 + a) 9 ae equal. 3. Fid e coefficiet of 5 i e poduct ( + 2) 6 ( ) usig biomial eoem.. If a ad b ae distict iteges, pove at a b is a facto of a b, wheeve is a positive itege. [Hit wite a (a b + b) ad epad] Evaluate Fid e value of a a a a.. Fid a appoimatio of (0.99) 5 usig e fist ee tems of its epasio. 8. Fid, if e atio of e fif tem fom e begiig to e fif tem fom e ed i e epasio of 2 3 is : 6.

17 6 MATHEMATICS 2 9. Epad usig Biomial Theoem, Fid e epasio of (3 2 2a + 3a 2 ) 3 usig biomial eoem. Summay The epasio of a biomial fo ay positive itegal is give by Biomial Theoem, which is (a + b) C 0 a + C a b + C 2 a 2 b C a.b + C b. The coefficiets of e epasios ae aaged i a aay. This aay is called Pascal s tiagle. The geeal tem of a epasio (a + b) is T + C a. b. I e epasio (a + b), if is eve, e e middle tem is e 2 tem.if is odd, e e middle tems ae 2 ad 2 tems. Histoical Note The aciet Idia maematicias kew about e coefficiets i e epasios of ( + y), 0. The aagemet of ese coefficiets was i e fom of a diagam called Meu-Pastaa, povided by Pigla i his book Chhada shasta (200B.C.). This tiagula aagemet is also foud i e wok of Chiese maematicia Chu-shi-kie i 303. The tem biomial coefficiets was fist itoduced by e Gema maematicia, Michael Stipel (86-56) i appoimately 5. Bombelli (52) also gave e coefficiets i e epasio of (a + b), fo,2..., ad Oughted (63) gave em fo, 2,..., 0. The aimetic tiagle, populaly kow as Pascal s tiagle ad simila to e Meu- Pastaa of Pigla was costucted by e Fech maematicia Blaise Pascal ( ) i 665. The peset fom of e biomial eoem fo itegal values of appeaed i Tate du tiage aimetic, witte by Pascal ad published posumously i 665.