Binomial Theorem PDF
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This document provides a detailed explanation of the binomial theorem, including its definition, theorem for a positive integral index, expansions, general terms, and middle term problems. Examples and solutions are also included.
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240 60 240 Binomial Theorem 6.1.1 Binomial Expression. E3 An algebraic expression consisting of two terms with +ve or – ve sign between them is called a binomial expression. q 1 4 p For example : (a b), (2 x 3 y ), 2 4 , 3 etc. x x y x 6.1.2 Binomial Theorem for Positi...
240 60 240 Binomial Theorem 6.1.1 Binomial Expression. E3 An algebraic expression consisting of two terms with +ve or – ve sign between them is called a binomial expression. q 1 4 p For example : (a b), (2 x 3 y ), 2 4 , 3 etc. x x y x 6.1.2 Binomial Theorem for Positive Integral Index. ID The rule by which any power of binomial can be expanded is called the binomial theorem. If n is a positive integer and x, y C then (x y)n n C 0 x n 0 y 0 n C1 x n 1 y 1 n C 2 x n 2 y 2 ........ n C r x n r y r ...... n C n 1 xy n 1 n C n x 0 y n (x y )n n n C r. x n r.y r U i.e.,.....(i) r 0 D YG Here n C 0 , n C1 , n C 2 ,...... n C n are called binomial coefficients and n C r n! for 0 r n. r !(n r)! Important Tips The number of terms in the expansion of (x y)n are (n + 1). The expansion contains decreasing power of x and increasing power of y. The sum of the powers of x and y in each term is equal to n. The binomial coefficients n C0 , n C1 , n C 2........ equidistant from beginning and end are equal i.e., n Cr n Cn r. (x y)n Sum of odd terms + sum of even terms. U 6.1.3 Some Important Expansions. (1) Replacing (x y ) C 0 x n n 0 y.y C1 x ST n 0 n by n 1.y C 2 x 1 n n2 –. y.... (1) 2 y r n Cr x in n r (i),.y .... (1) r n n we 0 C n x.y get, n i.e., (x y )n (1)r n C r x n r.y r.....(ii) r 0 The terms in the expansion of (x y )n are alternatively positive and negative, the last term is positive or negative according as n is even or odd. (2) Replacing x by 1 and y by x in equation (i) we get, (1 x )n n C 0 x 0 n C1 x 1 n C 2 x 2 ...... n C r x r ...... n C n x n i.e., (1 x )n n r 0 This is expansion of (1 x) in ascending power of x. n n Cr x r Binomial Theorem 241 (3) Replacing x by 1 and y by – x in (i) (1 x )n n C 0 x 0 n C1 x 1 n C 2 x 2 ...... (1)r n C r x r .... (1)n n C n x n i.e., (1 x )n we get, n (1) r n Cr x r r 0 (4) (x y) (x y) 2 [ C 0 x y C 2 x n n n n 0 n n2 y C4 x 2 n n 4 y .......] and 4 60 (x y)n (x y)n 2 [n C1 x n 1 y 1 n C 3 x n 3 y 3 n C 5 x n 5 y 5 .......] (5) The coefficient of (r 1)th term in the expansion of (1 x)n is n Cr. (6) The coefficient of x r in the expansion of (1 x)n is n Cr. Note : If n is odd, then (x y)n (x y)n and (x y)n (x y)n , both have the same number of E3 n 1 . 2 terms equal to n n If n is even, then (x y)n (x y)n has 1 terms and (x y)n (x y)n has terms. 2 2 x 5 10 x 4 a 40 x 3 a 2 80 x 2 a 3 80 xa 4 32 a 5 (b) (3 x a)5 (a) (x a)5 Solution: (c) ID Example: 1 (c) (x 2a)5 (d) (x 2a)3 Conversely (x y )n n C0 n C1 x n 1 y 1 n C 2 x n 2 y 2 .... n Cn x 0 y n U (x 2a)5 = 5 C0 x 5 5 C1 x 4 (2a)1 5 C 2 x 3 (2a)2 5 C 3 x 2 (2a)3 5 C 4 x 1 (2a)4 5 C5 x 0 (2a)5 = x 5 10 x 4 a 40 x 3 a 2 80 x 2 a 3 80 xa 4 32 a5. Example: 2 The value of ( 2 1)6 ( 2 1)6 will be (b) 198 (c) 98 D YG (a) – 198 Solution: (b) (d) – 99 We know that, (x y )n (x y )n 2[ x n n C 2 x n 2 y 2 n C4 x n 4 y 4 .....] ( 2 1)6 ( 2 1)6 2[( 2 )6 6 C2 ( 2 )4 (1)2 6 C4 ( 2 )2 (1)4 6 C6 ( 2 )0 (1)6 ] = 2[8 15 4 30 1] 198 Example: 3 The larger of 99 50 100 50 and 101 50 is 100 50 We have, 101 50 (a) 99 Solution: (c) 50 [IIT 1980] 50 (b) Both are equal (100 1) 50 100 50 50.100 49 (c) 101 50.49 100 48 2.1 (d) None of these.....(i) 50. 49 100 48 ............ (ii) 2.1 50.49.48 2. 100 47 ..... 100 50. Hence 101 50 100 50 99 50. 3.2.1 U and 99 50 (100 1)50 100 50 50.100 49 Subtracting, 101 50 99 50 100 50 Sum of odd terms is A and sum of even terms is B in the expansion of (x a)n , then ST Example: 4 (a) AB 1 (x a)2 n (x a) 2 n 4 (b) 2 AB (x a)2n (x a)2n (c) 4 AB (x a)2n (x a)2n Solution: (c) (x a) C0 x C1 x n n n n n 1 1 a C2 x n (d) None of these n2 2 a ... Cn x n n n.a (x n C 2 x n 2 a 2 ..) (n C1 x n 1a1 n C3 x n 3 a 3 ....) A B.....(i) n n Similarly, (x a)n A B.....(ii) From (i) and (ii), we get 4 AB (x a) 2n (x a) 2n Trick: Put n 1 in (x a). Then, x a A B. Comparing both sides A x, B a. n Option (c) L.H.S. 4 AB 4 xa , R.H.S. (x a)2 (x a)2 4 ax. i.e., L.H.S. = R.H.S 6.1.4 General Term. (x y)n n C 0 x n y 0 n C1 x n 1 y 1 n C 2 x n 2 y 2 ..... n C r x n r y r .... n C n x 0 y n 242 Binomial Theorem The first term = n C 0 x n y 0 The third term = n C 2 x n 2 y 2 and so on The second term = n C1 x n 1 y 1. The term n C r x n r y r is the (r 1)th term from beginning in the expansion of (x y)n. 60 Let Tr 1 denote the (r + 1)th term Tr 1 n C r x n r y r This is called general term, because by giving different values to r, we can determine all terms of the expansion. In the binomial expansion of (x y)n , Tr 1 (1)r n C r x n r y r In the binomial expansion of (1 x )n , Tr 1 n Cr x r Note : In the binomial expansion of E3 In the binomial expansion of (1 x )n , Tr 1 (1)r nCr x r (x y)n , the pth term from the end is (n p 2)th term from beginning. Important Tips In the expansion of (x y )n , n N ID Tr 1 n r 1 y Tr r x n(n 1) 2 n(n 1) in the expansion of (x 1)(x 2).....( x n) 2 The coefficient of x n 1 in the expansion of (x 1)(x 2).......( x n) The coefficient of x n 1 D YG Example: 5 U If the 4th term in the expansion of ( px x 1 )m is 2.5 for all x R then (a) p 5 / 2, m 3 Solution: (b) We have T4 (b) p 5 5 T3 1 2 2 1 ,m 6 2 (c) p 1 ,m 6 2 (d) None of these 3 m 5 5 1 C3 ( px )m 3 m C3 p m 3 x m 6 2 2 x.......(i) U Clearly, R.H.S. of the above equality is independent of x m 6 0 , m 6 5 1 Putting m 6 in (i) we get 6 C 3 p 3 p . Hence p 1 / 2, m 6. 2 2 ST Example: 6 If the second, third and fourth term in the expansion of (x a)n are 240, 720 and 1080 respectively, then the value of n is Solution: (d) [Kurukshetra CEE 1991; DCE 1995, 2001] (a) 15 (b) 20 It is given that T2 240 , T3 720 , T4 1080 (c) 10 (d) 5 Now, T2 240 T2 n C1 x n 1 a1 240.....(i) and T3 720 T3 n C 2 x n 2 a 2 720 T4 1080 T4 C 3 x To eliminate x, Now Example: 7 n T2.T4 T32 n 3 3 a 1080 .....(ii).....(iii) T T 240. 1080 1 1 2. 4 . 720. 720 2 T3 T3 2 n Tr 1 C n2 2 1 n r 1. n r . Putting r 3 and 2 in above expression, we get n5 Tr r 3 n 1 2 C r 1 x3 2 3 The 5th term from the end in the expansion of x 2 9 is Binomial Theorem 243 (b) (a) 63 x 3 252 x3 672 x 18 (c) (d) None of these 9 Solution: (b) x3 2 5th term from the end = (9 5 2)th term from the beginning in the expansion of 3 = T6 2 x 4 60 Example: 8 x 3 2 5 1 252 T6 T5 1 9 C5 9 C4.2. 3 3. 3 x x 2 x T T If 2 in the expansion of (a b)n and 3 in the expansion of (a b)n 3 are equal, then n T4 T3 [Rajasthan PET 1987, 96] T2 T3 T3 T4 E3 (d) 6 2 b 3 b 2n 2 3n 3 n 5 n 1 a n 1 a (given) ; 6.1.5 Independent Term or Constant Term. ID Solution: (c) (a) 3 (b) 4 (c) 5 T 2 b 2 b T 3 3 b b. . and 3 2 T3 n 2 1 a n 1 a T4 n 3 3 1 a n 1 a Independent term or constant term of a binomial expansion is the term in which exponent of the variable is zero. Example: 9 U Condition : (n r) [Power of x] + r. [Power of y] = 0, in the expansion of [ x y]n. x 3 The term independent of x in the expansion of 3 2x 2 10 will be D YG [IIT 1965; BIT Ranchi 1993; Karnataka CET 2000; UPSEAT 2001] 3 (a) 2 5 (b) 4 8/2 5 2 (c) (d) None of these 2 Solution: (b) 1 1 (10 r) r(2) 0 r 2 T3 10 C 2 2 3 Example: 10 1 The term independent of x in the expansion of (1 x )n 1 x 5 3 4 2 (a) C02 2C12 ...... (n 1) Cn2 (b) (C0 C1 ...... Cn )2 We know that, (1 x ) C0 C1 x C 2 x .......... Cn x n n 1 n 2 is n [EAMCET 1989] (c) C02 C12 ...... Cn2 n U Solution: (c) n n n 1 1 1 1 1 n C0 n C1 1 n C 2 2 ..... n Cn n x x x x ST Obviously, the term independent of x will be nC0.nC0 nC1nC1 ..... nCn.nCn C02 C12 ....... Cn2 1 1 1 1 Trick : Put n 1 in the expansion of (1 x )1 1 1 x 1 2 x .....(i) x x x We want coefficient of x 0. Comparing to equation (i). Then, we get 2 i.e., independent of x. Option (c) : C 02 C12 ..... C n2 ; Put n 1 ; Then 1 C 02 1 C12 1 1 2. Example: 11 1 The coefficient of x 7 in the expansion of ax 2 bx 11 will be [IIT 1967; Rajasthan PET 1996] (a) 462 a 6 b5 (b) 462 a 5 b 6 (c) 462 a 5 b 6 (d) 462 a 6 b5 6 Solution: (b) 462 a5 1 For coefficient of x 7 , (11 r)(1) (2). r 7 11 r 2r 7 r 6 ; T7 11 C6 (a)5 b6 b (d) 244 Binomial Theorem Example: 12 If the coefficients of second, third and fourth term in the expansion of (1 x )2n are in A.P., then 2n 2 9n 7 is equal to [AMU 2001] (a) –1 (c) 1 (d) 3/2 T2 2 n C1 , T3 2 n C 2 , T4 2 n C 3 are in A.P. then, 2. 2 n C 2 2 n C1 2 n C 3 2. 2n(.2n 1) 2n 2n(2n 1)(2n 2) 2.1 1 3.2.1 60 Solution: (b) (b) 0 On solving, 2n 2 9n 7 0 Example: 13 The coefficient of x 5 in the expansion of (1 x 2 )5 (1 x )4 is (a) 30 Solution: (b) (b) 60 (c) 40 4 5 5 2 5 (b) A 2B 4 1 4 2 C 2. 4 C1 4 C 3. 5 C1 60 Cn (d) None of these.....(i) (2n 1)! n!(n 1)! U 2 n 1 ID A coefficient of x n in (1 x )2n By (i) and (ii) we get, A 2 B.....(ii) The coefficient of x n in the expansion of (1 x )(1 x )n is (a) (1)n 1 n (b) (1)n (1 n) D YG Solution: (b) 5 4 (c) 2 A B ( 2 n ) ! 2.( 2n 1)! = 2 n Cn = (n 1)!n! n!n! B coefficient of x n in (1 x ) 2n 1 Example: 15 4 If A and B are the coefficient of x n in the expansions of (1 x )2n and (1 x )2n1 respectively, then[MP PET 1999 (a) A B Solution: (b) (d) None of these We have (1 x ) (1 x ) = ( C 0 C1 x C 2 x .....) ( C 0 C1 x C 2 x .......) 2 5 So coefficient of x 5 in [(1 x 2 )5 (1 x )4 ] = Example: 14 [EAMCET 1996; E3 UPSEAT 2001; Pb. CET 2002] Coefficient of x n (c) (1)n 1 (n 1)2 in (1 x )(1 x ) = Coefficient of x n n (d) (n 1) in (1 x) coefficient of x n 1 in (1 x )n n = Coefficient of x n in [n Cn ( x )n x.n Cn 1 ( x )n 1 ] = (1)n n Cn (1)n 1.n C1 = (1)n (1)n.(n) (1)n [1 n]. 6.1.6 Number of Terms in the Expansion of (a + b + c) n and (a + b + c + d) n. (a b c)n can be expanded as : (a b c)n {(a b) c}n (a b)n n C1 (a b)n 1 (c)1 n C 2 (a b)n 2 (c)2 ..... n C n c n (n 1) term n term (n 1)term ... 1term (n 1)(n 2). 2 (n 1)(n 2)(n 3) Similarly, Number of terms in the expansion of (a b c d )n . 6 ST U Total number of terms = (n 1) (n) (n 1) ...... 1 Example: 16 If the number of terms in the expansion of (x 2y 3 z)n is 45, then n (a) 7 Solution: (b) Example: 17 (c) 9 (d) None of these (n 1) (n 2) 45 (n 1)(n 2) 90 n 8. Given, total number of terms = 2 The number of terms in the expansion of [(x 3 y) 2 (3 x y) 2 ]3 is (a) 14 Solution: (b) (b) 8 (b) 28 (d) 56 ( 6 1 ) ( 6 2 ) 28 We have [(x 3 y)(3 x y)]6 = [3 x 2 8 xy 3 y 2 ]6 ; Number of terms = 2 6.1.7 Middle Term. The middle term depends upon the value of n. (c) 32 [Rajasthan PET 1986] Binomial Theorem 245 (1) When n is even, then total number of terms in the expansion of (x y )n is n 1 (odd). So n there is only one middle term i.e., 1 2 th term is the middle term. T n 2 1 n Cn / 2 x n / 2 y n / 2 (2) When n is odd, then total number of terms in the expansion of (x y )n is n 1 (even). So, and T n 3 n C n 1 x 2 n3 and 2 th are two middle terms. T n 1 n C n 1 x 2 n 1 n 1 2 y 2 2 n 1 n 1 2 y 2 2 E3 Note th 60 n 1 there are two middle terms i.e., 2 : When there are two middle terms in the expansion then their binomial coefficients are equal. Example: 18 ID Binomial coefficient of middle term is the greatest binomial coefficient. 1 The middle term in the expansion of x x 1991] 10 C4 1 x (b) 10 is (c) C5 Solution: (b) ∵ n is even so middle term T 10 Example: 19 The middle term in the expansion of (1 x ) 2 n is Solution: (d) 10 C5 x T6 T6 T5 1 10 C 5 x 5. D YG 1 2 1.3.5......( 2n 1) 2 n 1 x (a) n! [BIT Ranchi 1991; Rajasthan PET 2002; Pb. CET U (a) 10 2.4.6...... 2n 2n 1 x (b) n! Since 2n is even, so middle term = T 2 n 2 1 1 (d) 10 C7 x 4 10 C 5 x5 [Pb. CET 1998] 1.3.5......( 2n 1) n x (c) n! Tn 1 Tn 1 2n Cn x n 1. 3.5......( 2n 1) n n x.2 (d) n! (2n)! n 1.3. 5........( 2n 1) n n x =.2 x. n! n!. n! 6.1.8 To Determine a Particular Term in the Expansion. n ST U 1 In the expansion of x , if x m occurs in Tr 1 , then r is given by n r( ) m x n m r Thus in above expansion if constant term which is independent of x, occurs in Tr 1 then r is determined by n n r( ) 0 r Example: 20 9 3x 2 1 The term independent of x in the expansion of is 2 3 x (a) 7/12 (b) 7/18 (c) – 7/12 3 (d) – 7/16 6 9 8 7 1 7 9(2) 3 1. 6. Hence, T7 9 C6 . 12 3 2 1 2 3.3 3 18 2 3 Solution: (b) n =9, 2 , 1. Then r Example: 21 1 If the coefficient of x 7 in ax 2 bx 11 11 1 is equal to the coefficient of x 7 in ax 2 bx then ab = [MP PET 1999; AMU 2001; Pb. CET 2002] 246 Binomial Theorem (a) 1 (b) 1/2 (c) 2 (d) 3 Solution: (a) 1 For coefficient of x 7 in ax 2 ; n = 11, 2, 1 , m 7 bx 11. 2 7 15 r 5 2 1 3 1 Coefficient of x 7 in T6 11 C 5 a 6. 5......(i) b 11 60 11 E3 1 11. 1 7 and for coefficient of x 7 in ax 6 ; n 11, 1, 2 , m 7 ; r 2 3 bx 1 Coefficient of x 7 in T7 11 C 6.a 5. 6.....(ii) b From equation (i) and (ii), we get ab 1 6.1.9 Greatest Term and Greatest Coefficient. (1) Greatest term : If Tr and Tr 1 be the rth and (r 1)th terms in the expansion of (1 x)n , then ID n Tr 1 C xr n r 1 n r r 1 x Tr r C r 1 x n r 1 | x| 1 r or r U Let numerically, Tr 1 be the greatest term in the above expansion. Then Tr 1 Tr or (n 1) | x| (1 | x |) Tr 1 1 Tr......(i) D YG Now substituting values of n and x in (i), we get r m f or r m where m is a positive integer and f is a fraction such that 0 f 1. When n is even Tm 1 is the greatest term, when n is odd Tm and Tm 1 are the greatest terms and both are equal. Short cut method : To find the greatest term (numerically) in the expansion of (1 x)n. x (n 1) x 1 U (i) Calculate m = (ii) If m is integer, then Tm and Tm 1 are equal and both are greatest term. ST (iii) If m is not integer, there T[m ]1 is the greatest term, where [.] denotes the greatest integral part. (2) Greatest coefficient (i) If n is even, then greatest coefficient is n Cn / 2 (ii) If n is odd, then greatest coefficient are n C n 1 and n C n 3 2 2 Important Tips For finding the greatest term in the expansion of n y (x y )n x n 1 . x (x y )n. we rewrite the expansion in this form Binomial Theorem 247 y Greatest term in (x + y) n x n. Greatest term in 1 x The largest term in the expansion of (3 2 x )50 , where x 2x Solution: (c,d) (3 2 x )50 3 50 1 3 (b) 8th 50 (c) 7th 2x , Now greatest term in 1 3 1 2. 2x 5 (51) (50 1) x (n 1) r 3 3 6 (an integer) 2x 2 1 x 1 1 3 15 (d) 6th 50 60 (a) 5th 1 is 5 E3 Example: 22 n Tr and T[r]1 T6 and T[6 ]1 T6 and T7 are numerically greatest terms The greatest coefficient in the expansion of (1 x ) 2n 2 is (2 n ) ! n!2 (a) (b) (2n 2)! [(n 1)!]2 (2n 2)! n!(n 1)! (c) (d) (2n)! n!. (n 1)! (2n 2)! [(n 1)!]2 ID Example: 23 Solution: (b) n is even so greatest coefficient in (1 x ) 2n 2 is = Example: 24 coefficient is The interval in which x must lie so that the greatest term in the expansion of (1 x )2n has the greatest Here the greatest coefficient is n n2 , (c) n 2 n 2n 2n C n x n 2 n C n 1 x n 1 x (d) None of these Cn D YG Solution: (b) n n 1 , (b) n 1 n Cn 1 U n n 1 , (a) n 1 n 2n 2 n and n 1 2n C n x n 2 n C n 1 x n 1 x n 1. Hence the result is (b) n 6.1.10 Properties of Binomial Coefficients. In the binomial expansion of (1 x )n , (1 x )n n C 0 n C1 x n C 2 x 2 ..... n C r x r .... n C n x n. where n C 0 , n C1 , n C 2 ,......, n C n are the coefficients of various powers of x and called binomial coefficients, and they are written as C 0 , C 1 , C 2 ,.....C n. U Hence, (1 x )n C 0 C1 x C 2 x 2 ..... C r x r ..... C n x n.....(i) (1) The sum of binomial coefficients in the expansion of (1 x)n is 2 n......(ii) ST Putting x 1 in (i), we get 2 n C 0 C1 C 2 ..... C n (2) Sum of binomial coefficients with alternate signs : Putting x 1 in (i) We get, 0 C 0 C1 C 2 C 3 ...... …..(iii) (3) Sum of the coefficients of the odd terms in the expansion of (1 x)n is equal to sum of the coefficients of even terms and each is equal to 2n 1. From (iii), we have C 0 C 2 C 4 ..... C1 C 3 C 5 ..............(iv) i.e., sum of coefficients of even and odd terms are equal. From (ii) and (iv), C0 C2 C4 ..... C1 C3 C5 ..... 2n 1......(v) (4) n Cr n r n 1 n n 1 n2 Cr 1 . Cr 2 and so on. r r 1 248 Binomial Theorem (5) Sum of product of coefficients : Replacing x 1 x by in (i) we get n 1 C C C 1 C0 1 22 ... nn .... (vi) x x x x (1 x )2 n C C (C0 C1 x C 2 x 2 ......) C0 1 22 ..... n x x x Now comparing coefficient of x r on both sides. 2n.....(vii) Cn r C0 Cr C1Cr 1 ......Cn r.Cn 60 Multiplying (i) by (vi), we get We (6) Sum of squares of coefficients : Putting r 0 in (vii), we get 2n 1 n 1 (a) We have = (b) n.2 n (d) 2n 1 n 1 1 (n 1)n (n 1)(n)(n 1)(n 2) 1 2n 1 ..... = [2 (n 1)1 1] n 1 n 1 n 1 2! 4! D YG C1 1 C1 1 2 2 2 Which is given by option (a) 2 n 1 21 1 1 . n 1 1 1 2 The value of C0 3C1 5C2 ..... (2n 1)Cn is equal to (b) 2 n n. 2 n 1 (a) 2 n (c) 2 n (n 1) n Solution: (c) 2n n (c) n n(n 1)(n 2) n(n 1)(n 2)(n 3)(n 4 ) C1 C 3 C 5 ..... ...... = 2.1 4.3.2.1 6.5.4.3.2.1 2 4 6 Trick: For n 1 , = Example: 26 [Karnataka CET 2000] U Solution: (a) C1 C 3 C 5 ..... is equal to 2 4 6 ID The value of Cn C02 C12 ......Cn2 E3 (7) nCr nCr 1 n 1Cr Example: 25 2n We have C0 3C1 5C2 ..... (2n 1) Cn = n U n r.. n 1Cr 1 r r 1 n n n Cr = 2n r 0 n 1 (2r 1) C r r 1 (2r 1) n C r = r 0 n Cr 1 (d) None of these n n r 0 = 2. n n 2r n Cr r 0 n Cr r 0 Cr = 2n [(1 1)n 1 ] [1 1]n = 2n. 2 n 1 2 n 2 n.[n 1]. r 0 ST Trick: Put n 1 in given expansion C 0 3.1 C1 1 3 4. 1 Which is given by option (c) 2 n.(n 1) 2 1 (1 1) 4. n Example: 27 If S n r 0 1 and t n n Cr n r 0 2n 1 (a) 2 Solution: (d) n r 0 r n Cr t r. Then n is equal to Sn Cr 1 n 1 2 (b) Take n 2m , then, S n tn n 2m r 0 r Cr 2m 1 C0 2m 1 C1 2m (c) n 1 1 ...... C1 2m 2 ...... C2 2m get, 2m 1 1 = 2 2m C2m C0 2m C2 m 2m (d) 1 ...... C1 2m 2m n 2 1 Cm 1 1 Cm 2m Binomial Theorem 249 1 2m 1 C1 2 m C 2 m 1 2m 1 = 2m 2 m C1 tn mS n Example: 28 1 ..... C2 2m 2 2m 2 C2 2m C2m 2 2m m 1 m 1 ...... 2m 2m C Cm 1 m 1 1 m 2 m = 2m 2 m Cm C0 2m 1 ... C1 2m 2m m Cm 2m 1 Cm 1 tn n m Sn 2 2m C2m 2m m m S n Cm 2m 1 Cm 2m m mSn Cm 2m 60 tn If (1 x x 2 )n a0 a1 x a2 x 2 ...... a2 n x 2 n. Then a0 a 2 a4 ...... a 2n = [MNR 1992; DCE 1996; AMU 1998; Rajasthan PET 1999; Karnataka CET 1999; UPSEAT 1999] 3n 1 (a) 2 (c) (1 x x 2 )n a 0 a1 x a 2 x 2 ...... a 2 n x 2 n 1 3n 2 (d) 3 n 1 2 E3 Solution: (a) 3n 1 2 (b) Putting x 1, we get (1 1 1)n a 0 a1 a 2 .... a 2 n ; 1 a0 a1 a 2 ..... a 2n Again putting x 1 , we get 3 a0 a1 a2 ...... a2 n n.....(i)......(ii) Adding (i) and (ii), we get, 3 1 2[a 0 a 2 a 4 ...... a 2 n ] 3n 1 a 0 a 2 a 4 ....... a 2n 2 If (1 x )n C x r r r 0 C C C , then 1 1 1 2 ...... 1 n C C C 0 1 n 1 n 1 (a) (b) (n 1)n 1 (n 1) ! (c) (n 1)n n! (d) (n 1)n1 n! n n(n 1) / 2 ! 1 C C C We have 1 1 1 2 ........ 1 n = 1 1 ...... 1 1 n n C C C 0 1 n 1 D YG Solution: (c) n (n 1) ! U n Example: 29 ID n 1n 1n 1n 1 n (n 1)n = ....... n! 1 2 3 n Trick : Put n 1, 2, 3......., S 1 1 2 2 C1 1 C1 1 C 2 9 2 , S 2 1 2 2 C0 C1 C1 2 1 Which is given by option (c) n 1 , In the expansion of (1 x )5 , the sum of the coefficient of the terms is [Rajasthan PET 1992, 97; Kurukshetra CEE U Example: 30 (2 1)2 (1 1)1 9 2 ; For n 2 , 1! 2! 2 (a) 80 (b) 16 (c) 32 (d) 64 Putting x 1 in (1 x ) , the required sum of coefficient = (1 1) 2 32 Example: 31 If the sum of coefficient in the expansion of ( 2 x 2 2 x 1)51 vanishes, then the value of is [IIT ST Solution: (c) 5 5 5 1991; Pb. CET 1988] (a) 2 Solution: (c) (b) –1 (c) 1 (d) – 2 The sum of coefficient of polynomial ( 2 x 2 2 x 1)51 is obtained by putting x 1 in ( 2 x 2 2 x 1)51. Therefore by hypothesis ( 2 2 1)51 = 0 1 Example: 32 If Cr stands for n C r , the sum of given series 2(n / 2)!(n / 2)! 2 [C0 2C12 3 C22 ...... (1)n (n 1)Cn2 ] where n is n! an even positive integer, is (a) 0 Solution: (d) (b) (1)n / 2 (n 1) (c) (1)n (n 2) (d) (1)n / 2 (n 2) We have C02 2C12 3 C 22 ........ (1)n (n 1)Cn2 = [C02 C12 C22 ....... (1)n Cn2 ] [C12 2C22 3 C32...... (1)n n.Cn2 ] n 1 = (1)n / 2.n Cn / 2 (1)n / 2 1. n.n Cn / 2 = (1)n / 2 1 n C n / 2 2 2 250 Binomial Theorem 2. Therefore the value of given expression Example: 33 n n ! ! 2 2 (1)n / 2. 1 n n! (1)n / 2 (n 2) n! 2 n n ! ! 2 2 If (1 x )n C0 C1 x C 2 x 2 ....... Cn x n , then the value of C0 2C1 3C2 ...... (n 1)Cn will be [MP PET 1996; Rajasthan PET 1997; DCE 1995; IIT 1971; AMU 1995; EAMCET 2001] Solution: (a) (c) (n 1) 2 n 1 (b) (n 1)2n (d) (n 2) 2 n 60 (a) (n 2) 2 n 1 Trick: Put n 1 the expansion is equivalent to 1C0 2.1 C1 1 2 3. Which is given by option (a) = (n 2)2n 1 = (1 2) 2 0 3 E3 (1) Use of Differentiation : This method applied only when the numericals occur as the product of binomial coefficients. Solution process : (i) If last term of the series leaving the plus or minus sign be m, then divide m by n if q be the quotient and r be the remainder. i.e., m nq r Example: 34 C1 2C 2 3 C 3 ..... n C n 2002] Solution: (c) [Rajasthan PET 1995; MP PET (c) n. 2 n 1 (b) n. 2 n D YG (a) 2 n U ID Then replace x by x q in the given series and multiplying both sides of expansion by x r. (ii) After process (i), differentiate both sides, w.r.t. x and put x 1 or 1 or i or – i etc. according to given series. (iii) If product of two numericals (or square of numericals) or three numericals (or cube of numerical) then differentiate twice or thrice. (d) n. 2 n 1 We know that, (1 x )n C0 C1 x C 2 x 2 ...... Cn x n......(i) n 1 Differentiating both sides w.r.t. x, we get n(1 x ) = 0 + C1 2. C 2 x 3 C3 x ....... nC n x n 1 2 Putting x 1, we get, n. 2n 1 C1 2C2 3 C3 ...... n Cn. Example: 35 If n is an integer greater than 1, then a n C1 (a 1) n C 2 (a 2) ........ (1)n (a n) Solution: (b) (a) a (b) 0 (c) a 2 (d) 2 n We have a[C0 C1 C 2......] [C1 2C 2 3C 3.....] = a[C0 C1 C 2......] [C1 2C 2 3C 3 ......] U We know that (1 x )n C0 C1 x C 2 x 2...... (1)n Cn x n ; Put x 1 , 0 C0 C1 C2 ....... Then differentiating both sides w.r.t. to x , we get n(1 x )n 1 0 C 1 2C 2 x 3 C 3 x 2 ..... Put x 1 , 0 C1 2C 2 3C 3 .... = a[0 ] [0 ] 0. ST (2) Use of Integration : This method is applied only when the numericals occur as the denominator of the binomial coefficients. Solution process : If (1 x )n C 0 C1 x C 2 x 2 ..... C n x n , then we integrate both sides between the suitable limits which gives the required series. (i) If the sum contains C 0 , C1 , C 2 ,....... C n with all positive signs, then integrate between limit 0 to 1. (ii) If the sum contains alternate signs (i.e. +, –) then integrate between limit – 1 to 0. (iii) If the sum contains odd coefficients i.e., (C0, C2, C4.....) then integrate between –1 to 1. (iv) If the sum contains even coefficients (i.e., C1 , C 3 , C 5.....) then subtracting (ii) from (i) and then dividing by 2. (v) If in denominator of binomial coefficients is product of two numericals then integrate two times, first taking limit between 0 to x and second time take suitable limits. Binomial Theorem 251 C0 C1 C 2 C ..... n 1 2 3 n 1 (a) Solution: (c) 2n n 1 [Rajasthan PET 1996] 2n 1 n 1 (b) 2 n 1 1 n 1 (c) (d) None of these Consider the expansion (1 x )n C0 C1 x C 2 x 2 .... Cn x n Integrating both sides (1 x) dx C C x C x 1 1 n 0 0 1 0 0 1 1 (i) within.....(i) limits 0 to 1 we get, 1 ....... Cn x ndx 2 2 0 1 of 60 Example: 36 0 1 1 (1 x )n 1 x2 x n 1 1 C0 [ x ]0 C1 ........ Cn n 1 0 2 0 n 1 0 Example: 37 2C0 E3 C C C 2n 1 1 1 1 2n 1 1 1 ; C0 1 2 .... n . C0 C1 C2 ....... Cn. n 1 n 1 n 1 n 1 2 3 2 3 n 1 22 23 211 C1 C 2 ...... C10 2 3 11 [MP 1992] Solution: (a) 3 11 1 11 (b) 211 1 11 (c) 11 3 1 11 (d) ID (a) PET 1999; EAMCET 11 2 1 11 It is clear that it is a expansion of (1 x )10 C0 C1 x C 2 x 2 ..... C10 x 10 Integrating w.r.t. x both sides between the limit 0 to 2. 2 2 2 2 The sum to (n 1) terms of the following series (a) Solution: (d) D YG Example: 38 U (1 x )11 x2 x3 x 11 311 1 22 23 211 2 2 C0 .C1 .C2 ...... C10. C0 [ x ]0 C1 C 2 ...... C10 11 2 2 11 11 0 2 0 3 0 11 0 1 n 1 (b) 1 n2 C 0 C1 C 2 C 3 ..... is 2 3 4 5 (c) 1 n(n 1) (d) None of these (1 x )n C 0 C1 x C 2 x 2 C 3 x 3 ....... x (1 x )n C 0 x C1 x 2 C 2 x 3 C 3 x 4 ..... 0 The integral on L.H.S. of (i) = (1 t) t n (dt ) by putting 1 x t , U 1 1 1 x2 x3 x4 x (1 x )n dx C 0 C1 C 2 ....... (i) 0 2 0 3 0 4 0 1 1 (t 1 n t n 1 ) dt 0 1 1 n 1 n 2 Whereas the integral on the R.H.S. of (i) ST 1 1 1 C C C 1 1 C = C0 C1 2 ....... = 0 1 2 ....... to (n 1) terms = n 1 n 2 (n 1)(n 2) 2 3 4 2 3 4 Trick : Put n 1 in given series = 1 C0 1 C1 1 . Which is given by option (d). 2 3 6 6.1.11 An Important Theorem. If ( A B)n = I f where I and n are positive integers, n being odd and 0 f 1 then (I f ). f K n where A B 2 = K 0 and Note : If n is even integer then ( A B 1. A B)n ( A B)n I f f Hence L.H.S. and I are integers. f f is also integer; f f 1 ; f (1 f ) 252 Binomial Theorem Hence (I f ) (1 f ) (I f ) f = ( A B)n ( A B)n = ( A B 2 )n = K n. Example: 39 Let R (5 5 11)2 n 1 and f R [R ] where [.] denotes the greatest integer function. The value of R.f is [IIT 19 (a) 4 2 n 1 Solution: (a) (d) 4 2 n (c) 4 2 n 1 (b) 4 2 n Since f R [R ] , R f [R ] Now let f ' [5 5 11]2n1 , 0 f ' 1 f [R] f ' [5 5 11]2 n 1 [5 5 11]2 n 1 = 2 2 n 1 60 [5 5 11]2 n 1 f [R] , where [R] is integer C1 (5 5 )2n (11)1 2n 1 C 3 (5 5 )2n 2 (11)3 ....... E3 = 2.(Integer ) = 2 K (K N ) = Even integer Hence f f ' = even integer – [R], but 1 f f ' 1. Therefore, f f ' 0 f f ' Hence R.f = R. f ' (5 5 11)2n 1 (5 5 11)2n 1 4 2n 1. 6.1.12 Multinomial Theorem (For positive integral index). is (a1 a 2 a 3 ... am )n positive integer n! a n1 a n2...amnm n1 !n 2 !n 3 !...nm ! 1 2 Where n1 , n 2 , n 3 ,.....n m n1 n2 n3 .....nm n. are and ID n all non-negative integers U If a1 , a 2 , a 3 ,....a n C subject to D YG (1) The coefficient of a1n1.a 2n 2.....a mnm in the expansion of (a1 a2 a3 ....am )n is (2) The greatest coefficient in the expansion of (a1 a 2 a3 ....am )n is m r the then condition, n! n1 ! n2 ! n3 !....nm ! n! [(q 1)!]r (q!) Where q is the quotient and r is the remainder when n is divided by m. (3) If n is +ve integer and a1 , a 2 ,.....a m C, a1n1. a 2n 2......... a mnm then coefficient of x r in the expansion of (a1 a2 x .....am x m 1 )n is n !n !n !.....n n! 1 2 3 m! U Where n1 , n 2.....n m are all non-negative integers subject to the condition: n1 n 2 .....n m n and n 2 2n 3 3n 4 .... (m 1)n m r. ST (4) The number of distinct (a1 a 2 a3 ....am )n is n m 1 C m 1. Example: 40 dissimilar terms in the multinomial expansion The coefficient of x 5 in the expansion of (x 2 x 2)5 is (a) – 83 Solution: (c) or (b) – 82 (c) – 81 Coefficient of x 5 in the expansion of (x 2 x 2)5 is (d) 0 5! (1)n1 (1)n 2 (2)n3. n1 !. n 2 ! n 3 ! where n1 n2 n3 5 and n2 2n3 5. The possible value of n1 , n2 and n3 are shown in margin n1 1 2 0 n2 3 1 5 n3 1 2 0 The coefficient of x 5 = 5! 5! 5! (1)1 (1)3 (2)1 (1)2 (1)1 (2)2 + (1)0 (1)5 (2)0 = 40 120 1 81 1! 3!1! 2!1!2! 0!5!0! Binomial Theorem 253 Find the coefficient of a 3 b 4 c 5 in the expansion of (bc ca ab)6 (a) 0 (b) 60 (c) – 60 (d) None of these In this case, a3b 4 c 5 (ab)x (bc )y (ca)z a x z.b x y.c y z z + x = 3, x y 4, y z 5 ; 2(x y z ) 12 ; x y z 6. Then x 1, y 3, z 2 Example: 41 Solution: (b) Therefore the coefficient of a 3 b 4 c 5 in the expansion of (bc ca ab)6 = 60 6.1.13 Binomial Theorem for any Index. 6! 60. 1! 3! 2! n(n 1)x 2 n(n 1) (n 2) 3 n(n 1)......(n r 1) r x .... x ... terms up to 2! 3! r! When n is a negative integer or a fraction, where 1 x 1 , otherwise expansion will not be possible. If x 1 , the terms of the above expansion go on decreasing and if x be very small a stage may be reached when we may neglect the terms containing higher power of x in the expansion, then (1 x )n 1 nx. Important Tips Expansion is valid only when 1 x 1. n ID E3 Statement : (1 x )n 1 nx C r can not be used because it is defined only for natural number, so n C r will be written as The number of terms in the series is infinite. y If first term is not 1, then make first term unity in the following way, (x y )n x n 1 , if x U n (n) (n 1)......( n r 1) r! y 1. x n(n 1)(n 2)......(n r 1) r x r! Some important expansions: n(n 1)(n 2)......(n r 1) r n(n 1) 2 (i) (1 x )n 1 nx x ....... x ...... 2! r! n(n 1) 2 n(n 1)(n 2).....(n r 1) (ii) (1 x )n 1 nx x ....... ( x )r ....... 2! r! n(n 1) 2 n(n 1) (n 2) 3 n(n 1)......(n r 1) r (iii) (1 x )n 1 nx x x ..... x ..... 2! 3! r! n(n 1) 2 n(n 1)(n 2) 3 n(n 1)......(n r 1) (iv) (1 x )n 1 nx x x ..... ( x )r ...... 2! 3! r! U D YG General term : Tr 1 ST (a) Replace n by 1 in (iii) : (1 x )1 1 x x 2 ..... x r ...... , General term, Tr 1 x r (b) Replace n by 1 in (iv) : (1 x )1 1 x x 2 x 3 ..... ( x )r ...... , General term, Tr 1 ( x ). r (c) Replace n by 2 in (iii) : (1 x )2 1 2 x 3 x 2 ..... (r 1)x r ..... , General term, Tr 1 (r 1)x. r (d) Replace n by 2 in (iv) : (1 x )2 1 2 x 3 x 2 4 x 3 ...... (r 1)( x )r ..... General term, Tr 1 (r 1) ( x )r. (e) Replace n by 3 in (iii) : (1 x ) 3 1 3 x 6 x 2 10 x 3 ..... General term, Tr 1 (r 1) (r 2) / 2!. x r (r 1) (r 2) r x .......... 2! 254 Binomial Theorem (f) Replace n by 3 in (iv) : (1 x ) 3 1 3 x 6 x 2 10 x 3 ..... General term, Tr 1 (r 1)(r 2) ( x )r 2! To expand (1 2 x ) 1 / 2 as an infinite series, the range of x should be 1 1 (a) , 2 2 1 1 (b) , 2 2 (c) [2, 2] [AMU 2002] 60 Example: 42 (r 1) (r 2) ( x )r ..... 2! (d) (– 2, 2) 1 1 1 1 1 i.e., if x i.e., if x , . 2 2 2 2 2 (1 2 x ) 1 / 2 can be expanded if | 2 x | 1 i.e., if | x | Example: 43 If the value of x is so small that x 2 and higher power can be neglected, then equal to 5 (b) 1 x 6 Solution: (b) Given expression can be written as is [Roorkee 1962] (1 x )1 / 2 (1 x ) 2 / 3 1 x (1 x )1 / 2 = D YG U 1 2 1 1 1 x x 2 ..... 1 x x 2 .... 2 3 9 8 1 1 1 x 1 x x 2 ..... 2 8 = 1 x 1 x 2 (d) 1 x 3 2 (c) 1 x 3 ID 5 (a) 1 x 6 1 x 3 (1 x ) 2 E3 Solution: (b) 1 1 x x 2 ..... 5 5 12 144 1 x ..... = 1 x , when x 2 , x 3.... are neglected. 3 1 2 6 6 1 x x ..... 4 16 1 = Example: 44 Solution: (a) If (1 ax )n 1 8 x 24 x 2 ..... then the value of a and n is (a) 2, 4 (b) 2, 3 We know that (1 x )n 1 nx n(n 1)x 2 ..... 1! 2! ST Comparing coefficients of both sides we get, na 8, and Example: 45 n(n 1)a 2 24 on solving, a 2 , b = 4. 2! Coefficient of x r in the expansion of (1 2 x )1 / 2 (a) Solution: (b) (d) 1, 2 n (ax ) n(n 1)(ax )2 n(ax ) n(n 1)(ax )2 .... ...... 1 8 x 24 x 2 ...... 1 1! 2! 1! 2! U (1 ax )n 1 (c) 3, 6 (2 r)! (b) (r! ) 2 (2r)! 2 r.(r!)2 [Kurukshetra CEE 2001] (c) (2r)! (r! )2.2 2r (d) (2r)! 2 r.(r 1)!(r 1)! Coefficient of 1 1 1 1 1 2 .... r 1 r r r 2 2 2 2 (2)r 1.3.5...( 2r 1). (1).(1).2 1.3.5...( 2r 1) (2r)! xr r! r! 2r r! r! r!2r Example: 46 The coefficient of x 25 in (1 x x 2 x 3 x 4 )1 is (a) 25 Solution: (c) Coefficient of x (b) – 25 25 (c) 1 4 1 in (1 x x x x ) 2 3 (d) – 1 Binomial Theorem 255 1 (1 x 5 ) = Coefficient of x 25 in 1 x 1 = Coefficient of x 25 in (1 x 5 )1. (1 x ) = Coefficient of x 25 in [ (1 x 5 )1 x (1 x 5 )1 ] = [1 (x 5 )1 (x 5 )2 ......] x[1 (x 5 )1 (x 5 )2 ] ......] = Coefficient of x 25 in [1 x 5 x 10 x 15 .....] – Coefficient of x 24 in [1 x 5 x 10 x 15 .....] = 1 0 1. (a) Solution: (c) 1 1 3 . ......... 8 8 16 2 5 (b) We know that (1 x )n 1 nx Here nx Example: 48 [EAMCET 1990] 2 5 (c) 2 (d) None of these 5 n(n 1) 2 x ...... 2! 1 1 1 3 1 n(n 1) 2 1 1 3 x ,n 1 . , x ...... 1 4 8 8 16 8 4 2 8.16 2 1 / 2 2. 5 If x is so small that its two and higher power can be neglected and (1 2 x )1 / 2 (1 4 x )5 / 2 = 1 kx then k (a) 1 [Rajasthan PET 1993] ID = Solution: (d) 60 1 E3 Example: 47 (b) – 2 (c) 10 (1 2 x )1 / 2 (1 4 x )5 / 2 1 kx (d) 11 U 5 / 2 4 x 5 / 2 7 / 2 (4 x )2 ...... 1 kx (1 / 2) 2 x 1 / 2 3 / 2 (2 x )2 ...... 1 1 1! 2! 1! 2! Example: 49 D YG x 10 x 1 kx ; 1 10 x x 1 kx ; k 11 Higher power can be neglected. Then 1 1 1! 1! The cube root of 1 3 x 6 x 2 10 x 3 ..... is (a) 1 x x 2 x 3 ..... Solution: (c) 1 (1 x ) (b) 1 x 3 x 6 x 9 ...... (c) 1 x x 2 x 3 .... We have (1 3 x 6 x 10 x .....) 2 3 1/3 3 1 / 3 = [(1 x ) ] 3 ; [ (1 x ) (d) None of these 1 3 x 6 x ... ] 2 1 x x ... 2 Example: 50 1 1 The coefficient of x n in the expansion of is 1 x 3 x 3 n 1 1 2.3 (b) n 1 3 n 1 1 x (1 x ) 1 (3 x ) 1 = 3 1 (1 x ) 1 1 (1 x ) (3 x ) 3 ST Solution: (a) 3 n 1 1 (c) 2 n 1 3 3 n 1 1 U (a) Coefficient of x n 1 3 n 1 1 = (d) None of these 1 x x2 x n 1 xn [1 x x 2 ..... x n ] 1 2 ..... n 1 n 3 3 3 3 3 1 [3 n 1 1] 3 n 1 1 1 1 n 1 .....( n 1) terms = n 1 . n 3 1 3 3 3 2. 3 n 1 Trick: Put n 1, 2, 3...... and find the coefficients of x , x 2 , x 3...... and comparing with the given option as : Coefficient 3 n 1 1 2. (3 n 1 ) 33 1 2.3 3 of x2 is = 1 3 3 1 3 2 1 3 1 = 1 [3 3 1] 13 ; 27 33 3 1 Which is given by option (a) 13. 27 6.1.14 Three / Four Consecutive terms or Coefficients. (1) If consecutive coefficients are given: In this case divide consecutive coefficients pair wise. We get equations and then solve them. 256 Binomial Theorem (2) If consecutive terms are given : In this case divide consecutive terms pair wise i.e. if four Tr Tr 1 Tr 2 consecutive terms be Tr , Tr 1 , Tr 2 , Tr 3 then find 1 , 2 , 3 (say) then divide 1 , , Tr 1 Tr 2 Tr 3 by 2 and 2 by 3 and solve. 60 If a1 , a 2 , a 3 , a 4 are the coefficients of any four consecutive terms in the expansion of (1 x)n , then Example: 51 a1 a3 a1 a 2 a 3 a 4 [IIT 1975] a2 a2 a3 (b) a2 1 2 a2 a3 2a 2 a2 a3 2a 3 a2 a3 E3 (a) (c) (d) Let a1 , a 2 , a 3 , a 4 be respectively the coefficients of (r 1) th , (r 2) th , (r 3) th , (r 4 ) th terms in the expansion Solution: (c) of (1 x)n. Then a1 n C r , a 2 n C r 1 , a 3 n C r 2 , a 4 n C r 3. = 6.1.15 Some Important Points. (1) Pascal's Triangle : 1 1 5 Cr n Cr2 C r2 n C r3 r 1 r 3 2(r 2) = 2. n 1 n 1 n 1 n 1 n C r 1 n 1 C r3 n = n Cr Cr2 n 1 n n 1 n Cr Cr2 r 1 r3 n Cr 1 Cr 1 2a2 2. n Cr 2 Cr 1 nCr 2 a2 a3 n 1 2 (x y )2 1 3 6 10 (x y )1 1 3 4 n = D YG 1 1 n Cr2 (x y)0 1 1 n ID n a1 a3 Cr n a1 a 2 a 3 a 4 C r n C r 1 U Now, (x y)3 1 4 10 (x y)4 1 5 1 (x y )5 U Pascal's triangle gives the direct binomial coefficients. Example : (x y)4 1 x 4 4 x 3 y 6 x 2 y 2 4 xy 3 y 4 ST (2) Method for finding terms free from radical or rational terms in the expansion of (a 1/ p b ) a, b prime numbers : Find the general term Tr 1 C r (a N 1/q N 1 / p N r ) (b ) Cr a 1/q r N N r p.b r q Putting the values of 0 r N , when indices of a and b are integers. Note : Example: 52 Number of irrational terms = Total terms – Number of rational terms. The number of integral terms in the expansion of ( 3 8 5 )256 is (a) 32 Solution: (b) Tr 1 256 C r (b) 33 256 r.3 2 First term = 256 [AIEEE 2003] (c) 34 (d) 35 r.5 8 C 0 3 128 5 0 integer and after eight terms, i.e., 9th term = 256 C 8 3 124.5 1 integer Continuing like this, we get an A.P., 1 st , 9 th....... 257 th ; Tn a (n 1) d 257 1 (n 1) 8 n 33 Binomial Theorem 257 Example: 53 The number of irrational terms in the expansion of (a) 97 Solution: (a) Tr 1 100 C r (b) 98 5 2 8 6 100 is (c) 96 (d) 99 100 r r 5 8.2 6 As 2 and 5 are co-prime. Tr 1 will be rational if 100 r is multiple of 8 and r is multiple of 6 also r 0, 6, 12....... 96 ; 100 r 4, 10, 16..... 100......(i) But 100 r is to be multiple of 8. So, 100 r = 0, 8, 16, 24,......96.....(ii) Common terms in (i) and (ii) are 16, 40, 64, 88. 60 0 r 100 ST U D YG U ID *** E3 r = 84, 60, 36, 12 give rational terms The number of irrational terms = 101 – 4 = 97. 276 60 276 Mathematical Induction 6.2.1 First Principle of Mathematical Induction. E3 The proof of proposition by mathematical induction consists of the following three steps : Step I : (Verification step) : Actual verification of the proposition for the starting value “i” Step II : (Induction step) : Assuming the proposition to be true for “k”, k i and proving that it is true for the value (k + 1) which is next higher integer. ID Step III : (Generalization step) : To combine the above two steps Let p(n) be a statement involving the natural number n such that (i) p(1) is true i.e. p(n) is true for n = 1. U (ii) p(m + 1) is true, whenever p(m) is true i.e. p(m) is true p(m + 1) is true. Then p(n) is true for all natural numbers n. D YG 6.2.2 Second Principle of Mathematical Induction. The proof of proposition by mathematical induction consists of following steps : Step I : (Verification step) : Actual verification of the proposition for the starting value i and (i + 1). Step II : (Induction step) : Assuming the proposition to be true for k – 1 and k and then proving that it is true for the value k + 1; k i + 1. Step III : (Generalization step) : Combining the above two steps. Let p(n) be a statement involving the natural number n such that U (i) p(1) is true i.e. p(n) is true for n = 1 and (ii) p(m + 1) is true, whenever p(n) is true for all n, where i n m ST Then p(n) is true for all natural numbers. For a b, The expression a n b n is divisible by (a) a + b if n is even. (b) a – b is n if odd or even. 6.2.3 Some Formulae based on Principle of Induction. For any natural number n n(n 1) 2 n(n 1)(2n 1) n 2 1 2 2 2 3 2 ....... n 2 6 (i) n 1 2 3 ....... n (iii) n 3 1 3 2 3 3 3 ...... n 3 n 2 (n 1) 2 4 (ii) n 2 Mathematical Induction 277 n Example: 1 n 1 The smallest positive integer n, for which n ! hold is 2 (a) 1 Solution: (b) (b) 2 n 1 Let P(n) : n ! 2 (c) 3 (d) 4 n 2 2 2.25 which is true. Therefore, P(2) is true. k 2 P(k 1) : (k 1) ! 2 k 2 k 1 (k 1)k 1 k 1 (k 1)k 1 k 1 k ! (k 1)k ! 2k 2 k …..(i) 2k 1 2 1 k 1 (k 1)k 1 k 2 2k 2 and k 1 k 1 2 1 (k 1) 2 U 1 1 1 k 1 C 2 ...... 2 k 1 Which is true, hence (ii) is true. …..(ii) 2 1 1 k 1 C 2 ........ 2 k 1 k 1 ID (k 1)! k 1 k E3 k 1 Step II : Assume that P(k) is true, then p(k) : k ! 2 Step III : For n = k + 1, 60 9 2 1 2 ! 2 4 2 Step I : For n = 2 k 1 k 1 (k 1)k 1 k 2 k 2 (k 1)! 2 2k 2 Hence P(k 1) is true. Hence by the principle of mathematical induction P(n) is true for all n N D YG From (i) and (ii), (k 1)! Trick : By check option 1 2 1 1 (a) For n = 1, 1 ! 1 1 which is wrong 2 9 3 (b) For n = 2, 2 ! 2 which is correct 4 2 3 3 1 (c) For n = 3, 3 ! 6 < 8 which is correct 2 4 4 U 4 1 5 (d) For n = 4, 4 ! 24 24 < 39.0625 which is correct. 2 2 But smallest positive integer n is 2. Example: 2 Let S (k ) 1 3 5 ....... (2k 1) 3 k 2. Then which of the following is true. [AIEEE 2004] ST (a) Principle of mathematical induction can be used to prove the formula (b) S (k ) S (k 1) (c) S (k ) S (k 1) (d) S(1) is correct Solution: (c) We have S (k ) 1 3 5 ...... (2k 1) 3 k 2 , S (1) 1 4 , Which is not true and S (2) 3 7 , Which is not true. Hence induction cannot be applied and S (k ) S (k 1) Example: 3 When P is a natural number, then P n 1 (P 1) 2n 1 is divisible by Solution: (c) (a) P (b) P P (c) P P 1 n 1 2 n 1 2 1 2 (P 1) P (P 1) P P 1 , For n =1, we get, P 2 [IIT 1994] (d) P 1 2 2 Which is divisible by P 2 P 1 , so result is true for n =1 Let us assume that the given result is true for n m N i.e. P m 1 (P 1) 2m 1 is divisible by P 2 P 1 i.e. P m 1 (P 1) 2m 1 k (P 2 P 1) k N Now, P (m 1)1 (P 1) 2(m 1)1 P m 2 (P 1) 2 m 1 P m 2 (P 1) (P 1) 2 2 m 1 …..(i) 278 Mathematical Induction P m 2 (P 1) 2 [k (P 2 P 1) P m 1 ] P m 2 by using (i) (P 1) k (P P 1) (P 1) (P) 2 2 2 m 1 P m 1 [P (P 1) 2 ] (P 1) 2 k (P 2 P 1) P m 1 [P P 2 2 P 1] (P 1)2 k (P 2 P 1) P m 1 [P 2 P 1] (P 1)2 k (P 2 P 1) (P 2 P 1)[k (P 1) 2 P m 1 ] Which is divisible by P 2 P 1. Given result is true for all n N Given Un 1 3 Un 2Un 1 and U0 2 , U1 3 , the value of Un for all n N is (a) 2n 1 Solution: (b) Un1 3 Un 2 Un1 (b) 2n 1 (c) 0 …..(i) Step I : Given U1 3 For n =1, U11 3 U1 2 U0 , U2 3.3 2.2 5 ID Option (b) Un 2n 1 (d) None of these E3 Example: 4 60 Which is divisible by p 2 p 1 , so the result is true for n m 1. Therefore, the given result is true for all n N by induction. Trick : For n = 2, we get, P n 1 (P 1) 2n 1 P 3 (P 1)3 P 3 P 3 1 3 P 2 3 P 2 P 3 3 P 2 3 P 1 For n = 1, U1 21 1 3 which is true. For n = 2, U2 2 2 1 5 which is true Therefore, the result is true for n = 1 and n = 2 Then Uk 2 k 1 U Step II : Assume it is true for n = k then it is also true for n = k – 1 …..(ii) and Uk 1 2k 1 1 …..(iii) Step III : Putting n = k in (i), we get D YG U k 1 3 U k 2 U k 1 3[2 k 1] 2[2 k 1 1] 3.2 k 3 2.2 k 1 2 3.2 k 1 2.2 k 1 3. 2 k 2 k 1 2. 2 k 1 2 k 1 1 U k 1 2 k 1 1 This shows that the result is true for n k 1 , by the principle of mathematical induction the result is true for all n N. 6.2.4 Divisibility Problems. To show that an expression is divisible by an integer U (i) If a, p, n, r are positive integers, then first of all we write a pn r a pn. ar (a p )n. ar. (ii) If we have to show that the given expression is divisible by c.S ST Then express, a p [1 (a p 1] , if some power of (a p 1) has c as a factor. a p [2 (a p 2)] , if some power of (a p 2) has c as a factor. a p [K (a p K)], if some power of (a p K) has c as a factor. Example: 5 (1 x )n nx 1 is divisible by (where n N ) (a) 2 x Solution: (b) (b) x 2 (c) 2 x 3 (d) All of these n (n 1) 2 n(n 1)(n 2) 3 n(n 1) n(n 1)(n 3) x ..... (1 x )n 1 nx x x ..... (1 x )n nx 1 x 2 2 ! 3 ! 2! 3! From above it is clear that (1 x )n nx 1 is divisible by x 2. Trick : (1 x )n nx 1. Put n 2 and x 3 ; Then 4 2 2.3 1 9 Is not divisible by 6, 54 but divisible by 9. Which is given by option (c) = x 2 9. Example: 6 The greatest integer which divides the number 101 100 1 is Mathematical Induction 279 (a) 100 Solution: (c) (1 100 )100 1 100. 100 (b) 1000 (c) 10000 (d) 100000 100.99 100.99 100.99. 98 100 ..... (100 ) 2 .... 101 100 1 100. 100 1 1.2 3.2.1 1.2 From above it is clear that, 101 100 1 is divisible by (100 )2 10000 ST U D YG U ID E3 60 ***
