Mathematics-10.Unit-6-Binomial theorem

SYLLABUS Mathematical induction and its applications, Binomial theorem for a positive integral index, properties of binomial coefficients, 1. MATHEMATICAL INDUCTION AND ITS APPLICATIONS It is often used to prove a statement depending upon a natural number n. Type I: If P(n) is a statement depending upon n, then to prove it by induction, we proceed as follows: (i) Verify the validity of P(n) for n = 1. (ii) Assume that P(n) is true for some positive integer m and then using it establish the validity of P(n) for n = m + 1. Then, P(n) is true for each n N. Illustration 1: Prove that if sin   0, then , holds for each n n. Solution: If P(n) denotes the given statement, then for n = 1, P(1): , which is true because = cos  cos 2. Suppose that P(n) is true for some positive integer m, i.e. Using (1), we shall prove P(n) is true for n = m + 1 i.e. L.H.S. R.H.S. Hence, P(n) is true for each n. Type II: If P(n) is a statement depending upon n but beginning with some positive integer k, then to prove P(n), we proceed as follows: (i) Verify the validity of P(n) for n = k. (ii) Assume that the statement is true for n = m  k. Then, using it establish the validity of P(n) for n = m + 1. Then, P(n) is true for each n  k Illustration 2: Prove the inequality: , for n  2. Solution : Let P(n) : . For n = 2, P(2): or which is true. Suppose that P(m) is true for n = m  2 i.e. . . . (1) Using (1), we shall prove P(m + 1) i.e. L.H.S. [Using (1)] Hence, P(n) is true for n  2 Note: 1. Product of r consecutive integers is divisible by r !. 2. For x  y, xn – yn is divisible by (i) x + y if n is even (ii) x – y if n is even or odd. 2. BINOMIAL EXPRESSION An algebraic expression containing two terms is called a binomial expression. For example, (a + b), (2x – 3y), etc. are binomial expressions. 1.1 BINOMIAL THEOREM FOR POSITIVE INDEX Such formula by which any power of a binomial expression can be expanded in the form of a series is known as Binomial Theorem. For a positive integer n , the expansion is given by (a+x)n = nC0an + nC1an–1 x + nC2 an-2 x2 + . . . + nCr an–r xr + . . . + nCnxn = . where nC0 , nC1 , nC2 , . . . , nCn are called Binomial co-efficients. Similarly (a – x)n = nC0an – nC1an–1 x + nC2 an-2 x2 – . . . + (–1)r nCr an–r xr + . . . +(–1)n nCnxn i.e. (a – x)n = Replacing a = 1, we get (1 + x)n = nC0 +nC1x+nC2x2 + . . . + nCr xr + . . . + nCnxn and (1 – x)n = nC0 –nC1x+nC2x2 – . . . + (–1)r nCr xr + . . . +(–1)n nCnxn Observations:  There are (n+1) terms in the expansion of (a +x)n.  Sum of powers of x and a in each term in the expansion of (a +x)n is constant and equal to n.  The general term in the expansion of ( a+x)n is (r+1)th term given as Tr+1 = nCr an-r xr  The pth term from the end = ( n –p + 2)th term from the beginning .  Coefficient of xr in expansion of (a + x)n is nCr an - r xr.  nCx = nCy  x = y or x + y = n.  In the expansion of (a + x)n and (a –x)n, xr occurs in (r + 1)th term. Illustration 3: If the coefficients of the second, third and fourth terms in the expansion of (1 + x)n are in A.P., show that n = 7. Solution: According to the question nC1  nC2  nC3 are in A.P. n2 – 9n + 14 = 0  (n – 2)(n – 7) = 0  n = 2 or 7 Since the symbol nC3 demands that n should be  3 n cannot be 2,  n = 7 only. Illustration 4: Find the (i) last digit (ii) last two digit (iii) last three digit of 17256. Solution: 17256 = 289128 = (290 –1)128 = 128C0(290)128 –128C1(290)127 + ………..+ 128C126(290)2 –128C127(290)+1 = 1000m + 128C2(290)2 –128C1(290) + 1 = 1000m + (290)2 – + 1 = 1000m + 683527680 + 1 Hence the last digit is 1. Last two digits is 81. Last three digit is 681. Illustration 5: If the binomial coefficients of (2r + 4)th, (r –2)th term in the expansion of (a + bx)18 are equal find r. Solution: This is possible only when either 2r + 3 = r –3 …….(1) or 2r + 3 + r –3 = 18 ……..(2) from (1) r = –6 not possible but from (2) r = 6 Hence r = 6 is the only solution. Illustration 6: Find the coefficient of (i) x7 in , (ii) and x–7 in . Find the relation between a and b if these coefficients are equal. Solution : The general term in = If in this term power of x is 7, then 22 – 3r = 7  r = 5  coefficient of x7 = …(1) The general term in = If in this term power of x is –7, then 11 – 3r = –7  r = 6  coefficient of x–7 = (–1)6 If these two coefficient are equal, then  1.2 MIDDLE TERM There are two cases (a) When n is even Clearly in this case we have only one middle term namely Tn/2 + 1. Thus middle term in the expansion of (a + x)n will be nCn/2 an/2xn/2 term. (b) When n is odd Clearly in this case we have two middle terms namely . That means the middle terms in the expansion of (a +x)n are and . Illustration 7: Find the middle term in the expansion of . Solution: There will be two middle terms as n = 9 is an odd number. The middle terms will be and terms. t5 = 9C4(3x)5 t6 = 9C5(3x)4 . Illustration 8: Find the middle term in the expansion of . Solution : 7th term is the middle term T6+1 = 12C6 . . (b x)6 = 12C6 a6 b6 1.3 GREATEST BINOMIAL COEFFICIENT In the binomial expansion of (1 + x)n , when n is even, the greatest binomial coefficient is given by nCn/2. Similarly if n be odd, the greatest binomial coefficient will be 1.4 NUMERICALY GREATEST TERM If tr and tr + 1 be the rth and (r + 1)th term in the expansion of (1 + x)n, then x. Let numerically, tr + 1 be the greatest term in the above expansion. Then tr + 1  tr or  1  |x|  1  r  ……(2) Now shifting values of n and x in (2), we get r  m + f or r  m Where m is a positive integer, f is a fraction such that 0  f < 1. Now if f = 0 then tm + 1 and tm both the terms will be numerically equal and greatest while if f  0, then tm +1 is the greatest term of the binomial expansion. i.e. to find the greatest term (numerically) in the expansion of (1 + x)n. (i) Calculate m = . (ii) If m is integer, then tm and tm + 1 are equal and are greatest term. (iii) If m is not integer, then t[m] + 1 is the greatest term (where [.] denotes the greatest integer function). Illustration 9: Find the value of the greatest term in the expansion of . Solution: Since if only 21 – r  if only r  = 7.686 Hence t1 < t2 < t3 < t4 < t5 < t6 < t7 < t8 > t9 > t10 Hence t8 is the greatest term and its value is = Illustration 10: Find numerically the greatest term in the expansion of when x = . Solution : Since = Now in the expansion of , we have = = = so, the greatest terms are and . Greatest terms (where r = 2) = = = = and greatest term (where r = 3) = = = = From above we say that the values of both greatest terms are equal. Alternative Method (Short Cut Method) : Since Now, calculate = 3 The greatest terms in the expansion are and Greatest term (when r = 2) = = = and greatest term (when r = 3) = = = From above we say that the values of both greatest terms are equal. 1.5 PROPERTIES OF BINOMIAL COEFFICIENT For the sake of convenience the coefficients nC0 , nC1 , . . ., nCr , . . . ,nCn are usually denoted by C0, C1 , . . . , Cr , . . . ,Cn respectively • C0 + C1 + C2 +. . . . . + Cn = 2n • C0 - C1 + C2 -. . . . . + (–1)n Cn = 0 • C0 + C2 + C4 +. . . . . = C1 + C3 + C5 +. . . . . = 2n-1 •  r1 = r2 or r1 + r2 = n • nCr + nCr-1 = n+1Cr • r nCr =n n-1Cr-1 • . Illustration 11: Find the value of Solution: The given value is Illustration 12: If (1 + x)n = C0 + C1x + C2x2 + . . . . . + Cnxn, Show that (C0 + C1)(C1 + C2)(C2 + C3) . . . . . . (Cn-1 + Cn) = Solution: As we know tr = Cr – 1 + Cr = n + 1Cr = = = Hence C0 + C1 = C1 + C2 = …… …… Cn - 1 + Cn = Þ (C0 + C1)(C1 + C2) . . . . (Cn – 1 + Cn) = 1.6 PROBLEMS RELATED TO SERIES OF BINOMIAL COEFFICIENTS  Problems involving binomial coefficients with alternate sign: Illustration 13: Evaluate C0 - C1 + C2 - C3 +...+ (-1)nCn. Solution: Here alternately +ve and - ve sign occur This can be obtained by putting (-1) instead of 1 in place of x in (1 + x)n = C0 + C1x +...+ nCnxn, we get C0 - C1 +...+ (-1)nCn = 0 Now to obtain the sum C0 + C2 + C4 + ... we add (1 + 1)n and (1 - 1)n. Similarly, the cube roots of unity may be used to evaluate C0 + C3 + C6 + ... OR C1 + C4 +... OR C2 + C5 +... put x = 1, x = w, x = w2 in (1 + x)n = C0 + C1x +...+ Cnxn and add to get C0 + C3 + C6 +... the other two may be obtained by suitably multiplying (1 + w)n and (1 + w2)n by w and w2 respectively.  Problems Related to series of Binomial coefficients in which each term is a product of an integer and a binomial coefficient i.e. in the form k nCr Illustration 14: If (1+x)n = then prove that C1 + 2C2 + 3C3+. . .+ nCn= n2n-1. Solution: Method (i): By summation rth term of the given series, tr = r nCr  tr = n n-1Cr-1 Sum of the series = = = = n 2n-1 . Method (ii) By calculus We have ( 1+ x )n = C0 + C1x + C2 x2 + . . . + Cnxn . . .(1) Differentiating (1) with respect to x n(1 +x )n-1 = C1 +2C2x + 3C3 x2 + . . . + n Cnxn-1 . . . (2) Putting x = 1 in (2), n 2n-1 = C1 + 2C2 + . . . + n nCn  Problems related to series of binomial coefficients in which each term is a binomial coefficient divided by an integer i.e. in the form of . Illustration 15: Prove that Solution: Consider the expansion (1  x)n  C0  C1x  C2x2  C3x3  C4x4  …  Cnxn …(i) Integrating both sides of (i) within limits –1 to 1, we get   (By Prop. Of definite integral)(since second integral contains odd function) Hence Alternative Method. L.H.S.  C0   1    {n1C1n1C3n1C5…}  {sum of even binomial coefficients of (1  x)n1}    R. H. S. coefficient divided by an integer i.e. in the form of .  Problem related to series of binomial coefficients in which each term is a product of two binomial coefficients. Solution Process: (1) If difference of the lower suffixes of binomial coefficients in each term is same. i.e. C1C3  C2C4  C3C5  … Here 3 – 1  4 – 2  5 – 3  …  2 Case I: If each term of series is positive then (1  x)n  C0  C1x  C2x2  ….  Cnxn …(i) Interchanging 1 and x, (x  1)n  C0xn  C1xn–1  C2xn–2  … Cn …(ii) Then multiplying (i) and (ii) and equate the coefficient of suitable power of x on both sides Or Replacing x by in (i), then ….(iii) Then multiplying (i) and (ii) and equate the coefficient of suitable power of x on both sides. Case II: If terms of the series alternately positive and negative then (1–x)n  C0 – C1 x  C2x2 - …  (–1)nCnxn …(i) and (x 1)n  C0xnC1xn–1  C2xn–2  …  Cn …(ii) Then multiplying (i) and (ii) and equate the coefficient of suitable power of x on both sides. Or Replacing x by in (i), then …(iii) Then multiplying (i) and (iii) and equate the coefficient of suitable power of x on both sides. Illustration 16: If I is integral part of (2 + )n and f is fraction part of (2 + )n, then prove that (I + f) (1 –f) = 1. Also prove that I is an odd Integer. Solution: (2 + 3)n = I + f where I is an integer and 0  f < 1 Here note that (2 - 3)n (2 +3)n = (4 - 3)n = 1 Since (2 + 3)n (2 -3)n = 1 it is thus required to prove that (2 - 3)n = 1 - f but, (2 - 3)n + (2 +3)n = [2n - C1.2n - 1.3 + C22n - 2..(3)2 - ...] + [2n + C1.2n - 1.3 + C22n -2..(3)2 - ...] = 2[2n + C2.2n - 2.3+C42n - 4.32 + ...] = even integer Now 0 < (2 - 3) < 1 0 < (2 - 3)n < 1 if (2 - 3)n = f ', then I + f + f ' = Even Now O f < 1 and 0 < f ' < 1 ……(1) Also I + f + f ' = Even integer f + f ' = integer ……(2) (1) and (2) imply that f + f ' = 1 ( since 0 < f + f ' < 2)  I is odd and f ' = 1 - f  (I + f) (1 - f) = 1. 1.7 BINOMIAL THEOREM FOR ANY INDEX (1+x)n = 1+ nx + + . . . + Observations: • Expansion is valid only when –1 1 (C) for all n > m, m is some fixed positive integer (D) nothing can be said. Solution: Nothing can be said about the truth of P(n), for all n N because truth of P(1) is not given. Hence (D) is the correct answer. 42. If x > –1, then the statement P(n) : (1 + x)n > 1 + nx is true for (A) all n  N (B) all n > 1 (C) all n > a and x  0 (D) None of these. Solution: P(1) is not true. For n = 2, P(2) : (1 + x)2 > 1 + 2x is true if x  0 Let P(k) : (1 + x)k > 1 + kx be true  (1 + x)k+1 = (1 + x) (1 + x)k > (1 + x) (1 + kx) = 1 + (k + 1)x + kx2 > 1 + (k + 1)x ( kx2 > 0) Hence (C) is the correct answer. 43. The greatest positive integer, which divides (n + 16) (n + 17) (n + 18) (n + 19), for all n N, is (A) 2 (B) 4 (C) 24 (D) 120 Solution: Since product of any r consecutive integers is divisible by r ! and not by (r+1)!  The given product is divisible by 4 ! = 24. Hence (C) is the correct answer. 44. A student was asked to prove a statement by induction. He proved (i) P(5) is true and (ii) truth of P(n)  truth of P(n + 1), n N. On the basis of this, he could conclude that P(n) is true (A) for no n (B) for all n  5 (C) for all n (D) None of these. Solution: Obviously (B) is the answer. 45. The inequality n ! > 2n–1 is true (A) for all n > 1 (B) for all n > 2 (C) for all n N (D) for no n N Solution: It is not true for n = 1, 2 For n > 2, n ! > 1 . 2 . 3 ......... (n – 1)n > 2n–1 ( 2  2, 3 > 2, 4 > 2, ......., n > 2) Hence (B) is the correct answer. 46. The smallest positive integer for which the statement 3n+1 < 4n holds is (A) 1 (B) 2 (C) 3 (D) 4 Solution: The given statement is true for n  4. Hence (D) is the correct answer. 47. 23n – 7n – 1 is divisible by (A) 64 (B) 36 (C) 49 (D) 25 Solution: For n = 1, 23n – 7n – 1 has value 23 – 7 – 1 = 0 For n = 2, 23n – 7n – 1 has value 26 – 14 – 1 = 49. which is divisible by 49 and not by 36 or 64. Hence (C) is the correct answer. 48. For each n N, 23n – 1 is divisible by (A) 8 (B) 16 (C) 32 (D) None of these. Solution: For n  1, 23n – 1 = (23)n – 1 = 8n – 1 = (8 – 1) [8n–1 + 8n–2 + ....... + 1] = 7  positive integer Hence (D) is the correct answer. 49. If the ratio of the 7th term from the beginning to the 7th term from the end in the expansion of is , then x is (A) 9 (B) 6 (C) 12 (D) None of these. Solution: in 7th term from the end in = T7 in . Hence (A) is the correct answer. 50. If , then is equal to (A) (B) (C) (D) Solution: . Hence (D) is the correct answer. 51. The term independent of x in the expansion of is (A) (B) (C) (D) Solution: Let be independent of x.  6 – 2r = 0 or r = 3 Hence (C) is the correct answer. 52. The middle term in the expansion of is (A) (B) (C) (D) Solution: 2n is even.  Middle term Hence (D) is the correct answer. 53. If the binomial expansion of is , where a > 0, then (a, b) is (A) (2, 12) (B) (2, 8) (C) (–2, 12) (D) None of these. Solution: Also, . . . (1) and . . . (2) (1) and from (2) b = 12 Hence (A) is the correct answer. 54. can be expanded as a power series of x if (A) (B) (C) – 1 < x < 1 (D) None of these. Solution: . or or or or . Hence (B) is the correct answer. 55. In the binomial expansion of (a – b)n, n  5, the sum of the 5th and 6th terms is zero. The a/b is equal to (A) (B) (C) (D) Solution: . Hence (B) is the correct answer. 56. If the coefficient of mth, (m + 1)th and (m + 2)th terms in the expansion are in A.P., then (A) (B) (C) (D) Solution: We have in A.P. On simplification, we get . Hence (C) is the correct answer. 57. If , then is equal to (A) (B) (C) (D) Solution: We have Putting x = 1 and – 1, we get and Adding, we get . Hence (A) is the correct answer. 58. The positive value of a so that the coefficients of x5 and x15 are equal in the expansion of (A) (B) (C) 1 (D) Solution:  Coefficient of x5 = 120 a3 Also,  Coefficient of x15 = 10a or . Hence (A) is the correct answer. 59. The term independent of x in the expansion of is (A) (B) (C) (D) None of these. Solution: We have Term independent of x on the R.H.S. . Hence (C) is the correct answer. 60. The coefficient of x3 in is (A) 0 (B) 120 (C) 420 (D) 540 Solution: Let contains x3. or r = 3  Coefficient of x3 = 540. Hence (D) is the correct answer. 61. In the expansion of , let S be the sum of coefficients of odd power of x, then S is (A) 0 (B) 249 (C) 250 (D) 251 Solution:  Sum of coefficients of odd powers of x . Hence (B) is the correct answer. 62. The coefficient of x53 in is (A) (B) (C) (D) Solution:  Coefficient of . Hence (C) is the correct answer.