EXAMPLES. Leader in CBSE Coaching. Solutions of BINOMIAL THEOREM A.V.T.E. by AVTE (avte.in) Class XI

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1 avtei EXAMPLES Solutios of AVTE by AVTE (avtei) lass XI Leade i BSE oachig 1

2 avtei SHORT ANSWER TYPE 1 Fid the th tem i the epasio of 1 We have T Epad the followig (1 + ) 4 Put 1 y The (1 + ) 4 (y + ) y 4 ( ) y ( ) y ( ) + 4 y ( ) ( ) 4 y 4 + 4y + y 4 + 4y + 8 (1 ) (1 ) + 4 (1 ) + 4 (1 ) AVTE Fid the 4 th tem fom the ed i the epasio of 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 +, ie, 7 th tem fom the begiig which is give by Leade i BSE oachig T Evaluate : Puttig 1 y, we get The give epessio ( y) 4 + ( + y) 4 [ y y 4 ] [ (1 ) + (1 + 4 )] Fid the coefficiet of 11 i the epasio of Let the geeal tem, ie, ( + 1) th cotai 11 1 We have T ( 1) 1 ( 1) 5 Now fo this to cotai 11, we obseve that 5 11, ie, 5

3 avtei Thus, the coefficiet of 11 is Detemie whethe the epasio of Let T + 1 cotai 10 The 18 will cotai a tem cotaiig 10? T ( 1) ( 1) 18 Thus, 0, ie, Sice is a factio, the give epasio caot have a tem cotaiig 10 7 Fid the tem idepedet of i the epasio of AVTE Let ( + 1) th tem be idepedet of which is give by T Leade i BSE oachig Sice the tem is idepedet of, we have 10 0 Hece d tem is idepedet of ad its value is give by T Fid the middle tem i the epasio of a b 1 Sice the powe of biomial is eve, it has oe middle tem which is the tem ad it is give by 1 th T 7 1 a b 1 a b 1 1 a b 51a b

4 avtei p Fid the middle tem (tems) i the epasio of p Sice the powe of biomial is odd Theefoe, we have two middle tems which ae 5 th ad th tems These ae give by T p p 1p p ad T p 1 p p p 10 Show that , whee N is divisible by 5 We have ( + 1) (1 + 15) (15) ( + 1) (15) (15) [ so o] Thus, is divisible by 5 LONG ANSWER TYPE AVTE 11 Fid umeically the geatest tem i the epasio of ( + ), whee Leade i BSE oachig We have ( + ) 1 Now, T T Sice 4 T 1 Theefoe, 1 0 T Thus the maimum value of is Theefoe, the geatest tem is T + 1 T 7 Hece, T 7 whee

5 avtei 1 If is positive itege, fid the coefficiet of 1 i the epasio of (1 + ) We have Now to fid the coefficiet of 1 i (1 + ) 1, it is equivalet to fidig coefficiet of 1 i 1 which i tu is equal to the coefficiet of 1 i the epasio of (1 + ) Sice (1 + ) Thus the coefficiet of 1 is Which of the followig is lage? o We have ( ) 50 AVTE Similaly 50 (100 1) (100) (i) (ii) 1 1 Leade i BSE oachig Subtactig (ii) fom (i), we get > Hece > Fid the coefficiet of 50 afte simplifyig ad collectig the like tems i the epasio of (1 + ) (1 + ) + (1 + ) Sice the above seies is a geometic seies with the commo atio, its sum is Hece, coefficiet of 50 is give by

6 avtei 15 If a 1, a, a ad a 4 ae the coefficiet of ay fou cosecutive tems i the epasio of (1 + ), pove that a1 a a a a a a a a 1 4 Let a 1, a, a ad a 4 be the coefficiet of fou cosecutive tems T + 1, T +, T +, ad T + 4 espectively The a 1 coefficiet of T + 1 a coefficiet of T a coefficiet of T + + ad a 4 coefficiet of T Thus Similaly, a1 a a AVTE a a a 4 Hece, LHS ad RHS 1 1 ( ) Leade i BSE oachig a1 a 1 4 a a a a a a a OBJETIVE TYPE QUESTIONS (MQ) 1 The total umbe of tems i the epasio of ( + a) 51 ( a) 51 afte simplificatio is (a) 10 (b) 5 (c) (d) Noe of these (c) is the coect choice sice the total umbe of tems ae 5 of which tems get cacelled 17 If the coefficiets of 7 ad 8 i ae equal, the is (a) 5 (b) 55 (c) 45 (d) 15 (b) is the coect choice Sice T + 1 a i epasio of (a + ), Theefoe, T

7 avtei ad T Theefoe (sice it is give that coefficiet of 7 coefficiet 8 ) If (1 + ) a 0 + a 1 + a + + a, the a 0 + a + a a equals (a) 1 (b) 1 (c) 1 1 (d) (a) is the coect choice Puttig 1 ad 1 i (1 + ) a 0 + a 1 + a + + a we get 1 a 0 + a 1 + a +a + + a (1) ad a 0 a 1 + a a + + a () Addig (1) ad () we get + 1 (a 0 + a + a a ) AVTE Leade i BSE oachig Theefoe a 0 + a + a a 1 1 The coefficiet of p ad q (p ad q ae positive iteges) i the epasio of (1 + ) p + q ae (a) equal (b) equal with opposite sigs (c) ecipocal of each othe (d) oe of these (a) is the coect choice oefficiet of p ad q i the epasio of (1 + ) p + q ae p + q p ad p + q q ad p q p q p q p q p q 0 The umbe of tems i the epasio of (a + b + c), whee N is 1 (a) (b) + 1 (c) + (d) ( + 1) (a) is the coect choice We have (a + b + c) [a + (b + c)] a + 1 a 1 (b c) 1 + a (b + c) + + (b + c) Futhe, epadig each tem of RHS, we ote that Fist tem cosist of 1 tem Secod tem o simplificatio gives tems Thid tem o epadig gives tems Similaly, fouth tem o epasio gives 4 tems ad so o The total umbe of tems ( + 1) 1 7

8 avtei 1 The atio of the coefficiet of 15 to tem idepedet of i (a) 1 : (b) 1 : (c) : 1 (d) : 1 15 is (b) is the coect choice Let T + 1 be the geeal tem of 15, so, T ( ) () 0 (1) Now, fo the coefficiet of tem cotaiig 15, 0 15, ie, 5 Theefoe, 15 5 () 5 is the coefficiet of 15 (fom (1)) To fid the tem idepedet of, put 0 0 Thus is the tem idepedet of (fom (1)) Now the atio is AVTE 5 5 i i If z, the (a) Re (z) 0 (b) I m (z) 0 (c) Re(z) 0, I m (z) > 0 (d) Re (z) > 0, I m (z) < 0 (b) is the coect choice O simplificatio, we get Leade i BSE oachig i 5 i z 0 4 Sice i 1 ad i 4 1, z will ot cotai ay i ad hece I m (z) 0 AVTE INDIA Pvt Ltd Head Office: 7, AVTE AMPUS, DDA ommecial omple, Kailash oloy Et, Zamudpu, New Delhi , INDIA (P) 855, (M) , avtei avte7@yahoocoi 8