BINOMIAL THEOREM hapte 8 8. Oveview: 8.. A epessio cosistig of two tems, coected by + o sig is called a biomial epessio. Fo eample, + a, y,,7 4 5y, etc., ae all biomial epessios. 8.. Biomial theoem If a ad b ae eal umbes ad is a positive itege, the (a + b) 0 a + a b + a b +...... + a b +... + b, whee fo 0 The geeal tem o ( + ) th tem i the epasio is give by T + a b 8.. Some impotat obsevatios. The total umbe of tems i the biomial epasio of (a + b) is +, i.e. oe moe tha the epoet.. I the epasio, the fist tem is aised to the powe of the biomial ad i each subsequet tems the powe of a educes by oe with simultaeous icease i the powe of b by oe, till powe of b becomes equal to the powe of biomial, i.e., the powe of a is i the fist tem, ( ) i the secod tem ad so o edig with zeo i the last tem. At the same time powe of b is 0 i the fist tem, i the secod tem ad i the thid tem ad so o, edig with i the last tem.. I ay tem the sum of the idices (epoets) of a ad b is equal to (i.e., the powe of the biomial). 4. The coefficiets i the epasio follow a cetai patte kow as pascal s tiagle.
0 EXEMPLAR PROBLEMS MATHEMATIS Ide of Biomial oefficiet of vaious tems 0 4 4 6 4 5 5 0 0 5 Each coefficiet of ay ow is obtaied by addig two coefficiets i the pecedig ow, oe o the immediate left ad the othe o the immediate ight ad each ow is bouded by o both sides. The ( + ) th tem o geeal tem is give by T + a b 8..4 Some paticula cases If is a positive itege, the (a + b) 0 a b 0 + a b + a b +... + a b +... + a 0 b... () I paticula. Replacig b by b i (i), we get (a b) 0 a b 0 a b + a b +... + ( ) a b +... + ( ) a 0 b... (). Addig () ad (), we get (a + b) + (a b) [ 0 a b 0 + a b + 4 a 4 b 4 +... ] [tems at odd places]. Subtactig () fom (), we get (a + b) (a b) [ a b + a b +... ] [sum of tems at eve places] 4. Replacig a by ad b by i (), we get ( + ) 0 0 + + +... + +... + + i.e. ( + ) 0
BINOMIAL THEOREM 5. Replacig a by ad b by i... (), we get ( ) 0 0 +... + ( ) - + ( ) i.e., ( ) ( ) 0 8..5 The p th tem fom the ed The p th tem fom the ed i the epasio of (a + b) is ( p + ) th tem fom the begiig. 8..6 Middle tems The middle tem depeds upo the value of. (a) If is eve: the the total umbe of tems i the epasio of (a + b) is + (odd). Hece, thee is oly oe middle tem, i.e., + tem. th tem is the middle (b) If is odd: the the total umbe of tems i the epasio of (a + b) is + (eve). So thee ae two middle tems i.e., middle tems. 8..7 Biomial coefficiet I the Biomial epessio, we have th + ad + th ae two (a + b) 0 a + a b + a b +... + b... () The coefficiets 0,,,..., ae kow as biomial o combiatoial coefficiets. Puttig a b i (), we get 0 + + +... + Thus the sum of all the biomial coefficiets is equal to. Agai, puttig a ad b i (i), we get 0 + + 4 +... + + 5 +... Thus, the sum of all the odd biomial coefficiets is equal to the sum of all the eve biomial coefficiets ad each is equal to. 0 + + 4 +... + + 5 +...
EXEMPLAR PROBLEMS MATHEMATIS 8. Solved Eamples Shot Aswe Type Eample Fid the th tem i the epasio of +. Solutio We have T () + + + + + Eample Epad the followig ( + ) 4 Solutio Put y. The ( + ) 4 (y + ) 4 4 0 y 4 ( ) 0 + 4 y ( ) + 4 y ( ) + 4 y ( ) + 4 4 ( ) 4 y 4 + 4y + 6y 4 + 4y 6 + 8 ( ) 4 + 4 ( ) + 6 4 ( ) + 4 6 ( ) + 8 4 + 0 6 + 4 6 5 + 0 6 4 7 + 8 Eample Fid the 4 th tem fom the ed i the epasio of Solutio Sice th tem fom the ed i the epasio of (a + b) is ( + ) th tem fom the begiig. Theefoe 4 th tem fom the ed is 4 +, i.e., 7 th tem fom the begiig, which is give by T 7 6 6 Eample 4 Evaluate: ( ) + ( + ) 64 8 7 64 67 8 4 4
BINOMIAL THEOREM Solutio Puttig y, we get The give epessio ( y) 4 + ( + y) 4 [ 8 + 4 4 y + 4 4 y 4 ] 8 4 4 + ( ) + ( ) [ 8 + 6 4 ( ) + ( + 4 ] 8 6 + 4 4 4 + Eample 5 Fid the coefficiet of i the epasio of Solutio Let the geeal tem, i.e., ( + ) th cotai. We have T + ( ) 6 ( ) ( ) 6 5 Now fo this to cotai, we obseve that Thus, the coefficiet of is 6 5, i.e., 5 5 ( ) 5 5 0 8 5 4 544 Eample 6 Detemie whethe the epasio of cotaiig 0? Solutio Let T + cotai 0. The 8 will cotai a tem T + 8 8 ( ) 8 6 ( ). ( ) 8 6 Thus, 6 0, i.e., 6
4 EXEMPLAR PROBLEMS MATHEMATIS Sice is a factio, the give epasio caot have a tem cotaiig 0. Eample 7 Fid the tem idepedet of i the epasio of Solutio Let ( + ) th tem be idepedet of which is give by + 0. T + 0 0 0 0 0 0 0 Sice the tem is idepedet of, we have 0 0 Hece d tem is idepedet of ad its value is give by T 0 0 5 4 b Eample 8 Fid the middle tem i the epasio of a. Solutio Sice the powe of biomial is eve, it has oe middle tem which is the + th tem ad it is give by T 7 b 6 ( 6 a) 6 6 6 6 6 a ( b) 6 6 6 6 6 6 ab 56ab 6 6 6
BINOMIAL THEOREM 5 p Eample Fid the middle tem (tems) i the epasio of + p. Solutio Sice the powe of biomial is odd. Theefoe, we have two middle tems which ae 5 th ad 6 th tems. These ae give by 5 4 5 4 4 p p 6 p T p ad T 6 4 5 5 5 p 6 p p p Eample 0 Show that 4 + 4 5 6, whee N is divisible by 5. Solutio We have 4 + 4 5 6 4 ( + ) 5 6 6 + 5 6 ( + 5) + 5 6 + 0 5 0 + + 5 + + 5 + + 5 +... + + + (5) + 5 6 + ( + ) 5 + + 5 + + 5 +... + + + (5) + 5 6 + 5 + 5 + + 5 + + 5 +... + + + (5) + 5 6 5 [ + + + 5 +... so o] Thus, 4 + 4 5 6 is divisible by 5. Log Aswe Type Eample Fid umeically the geatest tem i the epasio of ( + ), whee. Solutio We have ( + ) +
6 EXEMPLAR PROBLEMS MATHEMATIS Now, T + T 0 0 0 Sice 4 Theefoe, T T + 0 4 0 4 0 (Why?) 6 Thus the maimum value of is 6. Theefoe, the geatest tem is T + T 7. Hece, T 7 6 6, whee 6 4 6 8 7 7 Eample If is a positive itege, fid the coefficiet of i the epasio of ( + ) +. Solutio We have ( + ) + ( + ) + ( + )
BINOMIAL THEOREM 7 Now to fid the coefficiet of i ( + ) +, it is equivalet to fidig ( + ) coefficiet of i which i tu is equal to the coefficiet of i the epasio of ( + ). Sice ( + ) 0 0 + + +... + +... + Thus the coefficiet of is + + Eample Which of the followig is lage? 50 + 00 50 o 0 50 We have (0) 50 (00 + ) 50 Similaly 50 (00 ) 50 00 50 + 50 (00) 4 + 50.4. (00)48 + 00 50 50. 00 4 + 50.4. (00)48 Subtactig () fom (), we get 0 50 50 0 50 50 00 50 + 0 50 50 > 00 50 Hece 0 50 > 50 + 00 50 4 50 4 48 47 50 (00) + 00 +... 50 4 48 47 00 +... 50.4.48 (00) 47 +... ().. 50.4.48 (00) 47 +... ().. Eample 4 Fid the coefficiet of 50 afte simplifyig ad collectig the like tems i the epasio of ( + ) 000 + ( + ) + ( + ) 8 +... + 000. Solutio Sice the above seies is a geometic seies with the commo atio +, its sum is
8 EXEMPLAR PROBLEMS MATHEMATIS 000 ( + ) + + 00 00 000 ( + ) + + + ( + ) 00 00 Hece, coefficiet of 50 is give by 00 50 00 50 5 Eample 5 If a, a, a ad a 4 ae the coefficiet of ay fou cosecutive tems i the epasio of ( + ), pove that a a a + a + a a + a a + a 4 Solutio Let a, a, a ad a 4 be the coefficiet of fou cosecutive tems T +, T, T, ad T + espectively. The + + 4 a coefficiet of T + a coefficiet of T + + a coefficiet of T + + ad a 4 coefficiet of T + 4 + Thus a a + a + + + + (Q + + + + )
BINOMIAL THEOREM + + + + Similaly, a a + a 4 + + + + + + + + + Hece, L.H.S. ad R.H.S. a a + + + 4 + + a + a a + a + + + 4 a a ( + ) ( + ) + + a + + + + + ( + ) + + + Objective Type Questios (M..Q) Eample 6 The total umbe of tems i the epasio of ( + a) 5 ( a) 5 afte simplificatio is (a) 0 (b) 5 (c) 6 (d) Noe of these Solutio is the coect choice sice the total umbe of tems ae 5 of which 6 tems get cacelled. Eample 7 If the coefficiets of 7 ad 8 i + ae equal, the is (a) 56 (b) 55 (c) 45 (d) 5 Solutio B is the coect choice. Sice T + a i epasio of (a + ), Theefoe, T 8 7 7 7 7 7() 7 7
40 EXEMPLAR PROBLEMS MATHEMATIS ad T 8 () 8 8 8 8 8 8 Theefoe, 7 7 7 8 8 8 8 8 7 7 8 7 (sice it is give that coefficiet of 7 coefficiet 8 ) 6 8 7 8 7 55 Eample 8 If ( + ) a 0 + a + a +... + a, the a 0 + a + a 4 +... + a equals. (A) + (B) () (D) + Solutio A is the coect choice. Puttig ad i ( + ) a 0 + a + a +... + a we get a 0 + a + a + a +... + a... () ad a 0 a + a a +... + a... () Addig () ad (), we get + (a 0 + a + a 4 +... + a ) + Theefoe a 0 + a + a 4 +... + a Eample The coefficiet of p ad q (p ad q ae positive iteges) i the epasio of ( + ) p + q ae (A) equal (B) equal with opposite sigs () ecipocal of each othe (D) oe of these Solutio A is the coect choice. oefficiet of p ad q i the epasio of ( + ) p + q ae p + q p ad p + q q ad p + q p p + q q p + q p q Hece (a) is the coect aswe.
BINOMIAL THEOREM 4 Eample 0 The umbe of tems i the epasio of (a + b + c), whee N is ( + )( + ) (A) (B) + () + (D) ( + ) Solutio A is the coect choice. We have (a + b + c) [a + (b + c)] a + a (b + c) + a (b + c) +... + (b + c) Futhe, epadig each tem of R.H.S., we ote that Fist tem cosist of tem. Secod tem o simplificatio gives tems. Thid tem o epasio gives tems. Similaly, fouth tem o epasio gives 4 tems ad so o. The total umbe of tems + + +... + ( + ) ( + )( + ) Eample The atio of the coefficiet of 5 to the tem idepedet of i + 5 is (A) : (B) : () : (D) : Solutio (B) is the coect choice. Let T + be the geeal tem of + 5, so, T + 5 ( ) 5 5 () 0... () Now, fo the coefficiet of tem cotaiig 5, 0 5, i.e., 5 Theefoe, 5 5 () 5 is the coefficiet of 5 (fom ()) To fid the tem idepedet of, put 0 0
4 EXEMPLAR PROBLEMS MATHEMATIS Thus 5 0 0 is the tem idepedet of (fom ()) Now the atio is 5 5 5 5 0 5 0 Eample If z 5 5 i i + +, the (A) Re (z) 0 (B) I m (z) 0 () Re (z) > 0, I m (z) > 0 (D) Re (z) > 0, I m (z) < 0 Solutio B is the coect choice. O simplificatio, we get z 4 5 5 i 5 i 0 + + 4 Sice i ad i 4, z will ot cotai ay i ad hece I m (z) 0. 8. EXERISE Shot Aswe Type. Fid the tem idepedet of, 0, i the epasio of 5.. If the tem fee fom i the epasio of of k. k. Fid the coefficiet of i the epasio of ( + 7 ) ( ) 6. 0 is 405, fid the value 4. Fid the tem idepedet of i the epasio of, 5. Fid the middle tem (tems) i the epasio of 5. (i) a a 0 (ii) 6
BINOMIAL THEOREM 4 6. Fid the coefficiet of 5 i the epasio of ( ) 0. 7. Fid the coefficiet of 7 i the epasio of 8. Fid the sith tem of the epasio ( y ) 4 +, if the biomial coefficiet of the thid tem fom the ed is 45. [Hit: Biomial coefficiet of thid tem fom the ed Biomial coefficiet of thid tem fom begiig.]. Fid the value of, if the coefficiets of ( + 4) th ad ( ) th tems i the epasio of ( + ) 8 ae equal. 0. If the coefficiet of secod, thid ad fouth tems i the epasio of ( + ) ae i A.P. Show that + 7 0.. Fid the coefficiet of 4 i the epasio of ( + + + ). Log Aswe Type 5.. If p is a eal umbe ad if the middle tem i the epasio of 0, fid p. 8 p + is. Show that the middle tem i the epasio of 5... ( ) ( ). is 4. Fid i the biomial + if the atio of 7 th tem fom the begiig to the 7 th tem fom the ed is 6. 5. I the epasio of ( + a) if the sum of odd tems is deoted by O ad the sum of
44 EXEMPLAR PROBLEMS MATHEMATIS eve tem by E. The pove that (i) O E ( a ) (ii) 4OE ( + a) ( a) 6. If p occus i the epasio of +, pove that its coefficiet is 4 p + p. 7. Fid the tem idepedet of i the epasio of ( + + ). Objective Type Questios hoose the coect aswe fom the give optios i each of the Eecises 8 to 4 (M..Q.). 8. The total umbe of tems i the epasio of ( + a) 00 + ( a) 00 afte simplificatio is (A) 50 (B) 0 () 5 (D) oe of these. Give the iteges >, >, ad coefficiets of () th ad ( + ) d tems i the biomial epasio of ( + ) ae equal, the (A) (B) () + (D) oe of these 0. The two successive tems i the epasio of ( + ) 4 whose coefficiets ae i the atio :4 ae (A) d ad 4 th (B) 4 th ad 5 th () 5 th ad 6 th (D) 6 th ad 7 th [Hit: 4 4 + 4 4 4 + 4 + 4 4 4 4 ]. The coefficiet of i the epasio of ( + ) ad ( + ) ae i the atio. (A) : (B) : () : (D) : [Hit : :. If the coefficiets of d, d ad the 4 th tems i the epasio of ( + ) ae i A.P., the value of is (A) (B) 7 (c) (D) 4 [Hit: + + 4 0 o 7
BINOMIAL THEOREM 45. If A ad B ae coefficiet of i the epasios of ( + ) ad ( + ) espectively, the A B equals (A) (B) () (D) [Hit: A ] B 4. If the middle tem of + si (A) π + 6 π (B) π + 6 π 0 is equal to 7 7 8, the value of is π π () π + ( ) (D) π + ( ) 6 [Hit: T 6 0 5 5 6 5 si 5 8 si 5 si 5 π + ( ) 6 π ] Fill i the blaks i Eecises 5 to. 5. The lagest coefficiet i the epasio of ( + ) 0 is. 6. The umbe of tems i the epasio of ( + y + z). [Hit: ( + y + z) [ + (y + z)] ] 7. I the epasio of 6, the value of costat tem is. 8. If the seveth tems fom the begiig ad the ed i the epasio of + ae equal, the equals. [Hit : T 7 T ( ) 7 + 6 6 6 ( ) 6 6 6
46 EXEMPLAR PROBLEMS MATHEMATIS ( ). The coefficiet of a 6 b 4 i the epasio of [Hit : T oly poblem whe 0 ]. b 4 0 b 0 6 4 5 4 a b a 7 b a ] 0 is. 0. Middle tem i the epasio of (a + ba) 8 is.. The atio of the coefficiets of p ad q i the epasio of ( + ) p + q is [Hit: p + q p p + q q ]. The positio of the tem idepedet of i the epasio of.. If 5 5 is divided by, the emide is. State which of the statemet i Eecises 4 to 40 is Tue o False. + 0 is 4. The sum of the seies 0 0 0 is 0 0 5. The epessio 7 + 7 is divisible by 64. Hit: 7 + 7 ( + 8) 7 ( 8) 6. The umbe of tems i the epasio of [( + y ) 4 ] 7 is 8 7. The sum of coefficiets of the two middle tems i the epasio of ( + ) is equal to. 8. The last two digits of the umbes 400 ae 0.. If the epasio of multiple of. + cotais a tem idepedet of, the is a 40. Numbe of tems i the epasio of (a + b) whee N is oe less tha the powe.