Chapter-6-BINOMIAL THEOREM-(E)-01-Theory

6.1.1 Binomial Expression. An algebraic expression consisting of two terms with +ve or – ve sign between them is called a binomial expression. For example : etc. 6.1.2 Binomial Theorem for Positive Integral Index . The rule by which any power of binomial can be expanded is called the binomial theorem. If n is a positive integer and x, then i.e., .....(i) Here are called binomial coefficients and for . Important Tips  The number of terms in the expansion of 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 equidistant from beginning and end are equal i.e., .  Sum of odd terms + sum of even terms. 6.1.3 Some Important Expansions . (1) Replacing y by – y in (i), we get, i.e., .....(ii) The terms in the expansion of 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, i.e., This is expansion of in ascending power of x. (3) Replacing x by 1 and y by – x in (i) we get, i.e., (4) and (5) The coefficient of term in the expansion of is . (6) The coefficient of in the expansion of is . Note :  If n is odd, then and , both have the same number of terms equal to  If n is even, then has terms and has terms. Example: 1 (a) (b) (c) (d) Solution: (c) Conversely = = . Example: 2 The value of will be (a) – 198 (b) 198 (c) 98 (d) – 99 Solution: (b) We know that, = Example: 3 The larger of and is (a) (b) Both are equal (c) (d) None of these Solution: (c) We have, .....(i) and ..... (ii) Subtracting, . Hence . Example: 4 Sum of odd terms is A and sum of even terms is B in the expansion of , then (a) (b) (c) (d) None of these Solution: (c) .....(i) Similarly, .....(ii) From (i) and (ii), we get Trick: Put in . Then, . Comparing both sides . Option (c) L.H.S. , R.H.S. . i.e., L.H.S. = R.H.S 6.1.4 General Term . The first term = The second term = . The third term = and so on The term is the term from beginning in the expansion of . Let denote the (r + 1)th term 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 In the binomial expansion of In the binomial expansion of Note :  In the binomial expansion of , the pth term from the end is term from beginning. Important Tips  In the expansion of  The coefficient of in the expansion of  The coefficient of in the expansion of Example: 5 If the 4th term in the expansion of is 2.5 for all then (a) (b) (c) (d) None of these Solution: (b) We have   .......(i) Clearly, R.H.S. of the above equality is independent of x , Putting in (i) we get . Hence . Example: 6 If the second, third and fourth term in the expansion of are 240, 720 and 1080 respectively, then the value of n is (a) 15 (b) 20 (c) 10 (d) 5 Solution: (d) It is given that Now,  .....(i) and  .....(ii)  .....(iii) To eliminate x,  . Now . Putting and 2 in above expression, we get  Example: 7 The 5th term from the end in the expansion of is (a) (b) (c) (d) None of these Solution: (b) 5th term from the end = term from the beginning in the expansion of =  . Example: 8 If in the expansion of and in the expansion of are equal, then (a) 3 (b) 4 (c) 5 (d) 6 Solution: (c) and (given) ;   6.1.5 Independent Term or Constant Term. Independent term or constant term of a binomial expansion is the term in which exponent of the variable is zero. Condition : [Power of x] + r . [Power of y] = 0, in the expansion of . Example: 9 The term independent of x in the expansion of will be (a) (b) (c) (d) None of these Solution: (b) Example: 10 The term independent of x in the expansion of is (a) (b) (c) (d) None of these Solution: (c) We know that, Obviously, the term independent of x will be Trick : Put in the expansion of .....(i) We want coefficient of . Comparing to equation (i). Then, we get 2 i.e., independent of x. Option (c) : ; Put ; Then . Example: 11 The coefficient of in the expansion of will be (a) (b) (c) (d) Solution: (b) For coefficient of , ; Example: 12 If the coefficients of second, third and fourth term in the expansion of are in A.P., then is equal to (a) –1 (b) 0 (c) 1 (d) 3/2 Solution: (b) , are in A.P. then, On solving, Example: 13 The coefficient of in the expansion of is (a) 30 (b) 60 (c) 40 (d) None of these Solution: (b) We have = So coefficient of in = Example: 14 If A and B are the coefficient of in the expansions of and respectively, then (a) (b) (c) (d) None of these Solution: (b) coefficient of in = = .....(i) coefficient of in .....(ii) By (i) and (ii) we get, Example: 15 The coefficient of in the expansion of is (a) (b) (c) (d) Solution: (b) Coefficient of in = Coefficient of in coefficient of = Coefficient of in = = . 6.1.6 Number of Terms in the Expansion of (a + b + c)n and (a + b + c + d)n . can be expanded as : Total number of terms = . Similarly, Number of terms in the expansion of . Example: 16 If the number of terms in the expansion of is 45, then (a) 7 (b) 8 (c) 9 (d) None of these Solution: (b) Given, total number of terms =   . Example: 17 The number of terms in the expansion of is (a) 14 (b) 28 (c) 32 (d) 56 Solution: (b) We have = ; Number of terms = 6.1.7 Middle Term. The middle term depends upon the value of n. (1) When n is even, then total number of terms in the expansion of is (odd). So there is only one middle term i.e., term is the middle term. (2) When n is odd, then total number of terms in the expansion of is (even). So, there are two middle terms i.e., and are two middle terms. and Note :  When there are two middle terms in the expansion then their binomial coefficients are equal.  Binomial coefficient of middle term is the greatest binomial coefficient. Example: 18 The middle term in the expansion of is (a) (b) (c) (d) Solution: (b) ∵ n is even so middle term  Example: 19 The middle term in the expansion of is (a) (b) (c) (d) Solution: (d) Since 2n is even, so middle term =  = . 6.1.8 To Determine a Particular Term in the Expansion. In the expansion of , if occurs in , then r is given by  Thus in above expansion if constant term which is independent of x, occurs in then r is determined by  Example: 20 The term independent of x in the expansion of is (a) 7/12 (b) 7/18 (c) – 7/12 (d) – 7/16 Solution: (b) n =9, , . Then . Hence, . Example: 21 If the coefficient of in is equal to the coefficient of in then ab = (a) 1 (b) 1/2 (c) 2 (d) 3 Solution: (a) For coefficient of in ; n = 11, , Coefficient of in ......(i) and for coefficient of in ; , ; Coefficient of in .....(ii) From equation (i) and (ii), we get 6.1.9 Greatest Term and Greatest Coefficient. (1) Greatest term : If and be the rth and terms in the expansion of , then Let numerically, be the greatest term in the above expansion. Then or or ......(i) Now substituting values of n and x in (i), we get or where m is a positive integer and f is a fraction such that . When n is even is the greatest term, when n is odd and are the greatest terms and both are equal. Short cut method : To find the greatest term (numerically) in the expansion of . (i) Calculate m = (ii) If m is integer, then and are equal and both are greatest term. (iii) If m is not integer, there is the greatest term, where [.] denotes the greatest integral part. (2) Greatest coefficient (i) If n is even, then greatest coefficient is (ii) If n is odd, then greatest coefficient are and Important Tips  For finding the greatest term in the expansion of . we rewrite the expansion in this form . Greatest term in (x + y)n . Greatest term in Example: 22 The largest term in the expansion of , where is (a) 5th (b) 8th (c) 7th (d) 6th Solution: (c,d) , Now greatest term in (an integer)  and and and are numerically greatest terms Example: 23 The greatest coefficient in the expansion of is (a) (b) (c) (d) Solution: (b) n is even so greatest coefficient in is = Example: 24 The interval in which x must lie so that the greatest term in the expansion of has the greatest coefficient is (a) (b) (c) (d) None of these Solution: (b) Here the greatest coefficient is  and . Hence the result is (b) 6.1.10 Properties of Binomial Coefficients. In the binomial expansion of . where are the coefficients of various powers of x and called binomial coefficients, and they are written as . Hence, .....(i) (1) The sum of binomial coefficients in the expansion of is . Putting in (i), we get .....(ii) (2) Sum of binomial coefficients with alternate signs : Putting in (i) We get, …..(iii) (3) Sum of the coefficients of the odd terms in the expansion of is equal to sum of the coefficients of even terms and each is equal to . From (iii), we have ......(iv) i.e., sum of coefficients of even and odd terms are equal. From (ii) and (iv), ......(v) (4) and so on. (5) Sum of product of coefficients : Replacing x by in (i) we get (vi) Multiplying (i) by (vi), we get Now comparing coefficient of on both sides. We get, .....(vii) (6) Sum of squares of coefficients : Putting in (vii), we get (7) Example: 25 The value of is equal to (a) (b) (c) (d) Solution: (a) We have = = = Trick: For , = Which is given by option (a) . Example: 26 The value of is equal to (a) (b) (c) (d) None of these Solution: (c) We have = = = = = = . Trick: Put in given expansion . Which is given by option (c) . Example: 27 If and . Then is equal to (a) (b) (c) (d) Solution: (d) Take , then, = = = Example: 28 If . Then = (a) (b) (c) (d) Solution: (a) Putting we get ; .....(i) Again putting , we get ......(ii) Adding (i) and (ii), we get, Example: 29 If , then (a) (b) (c) (d) Solution: (c) We have = = Trick : Put Which is given by option (c) , ; For , Example: 30 In the expansion of , the sum of the coefficient of the terms is (a) 80 (b) 16 (c) 32 (d) 64 Solution: (c) Putting in , the required sum of coefficient = Example: 31 If the sum of coefficient in the expansion of vanishes, then the value of is (a) 2 (b) –1 (c) 1 (d) – 2 Solution: (c) The sum of coefficient of polynomial is obtained by putting in . Therefore by hypothesis = 0 Example: 32 If stands for , the sum of given series where n is an even positive integer, is (a) 0 (b) (c) (d) Solution: (d) We have = = = Therefore the value of given expression Example: 33 If , then the value of will be (a) (b) (c) (d) Solution: (a) Trick: Put the expansion is equivalent to . Which is given by option (a) = = (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., Then replace x by in the given series and multiplying both sides of expansion by . (ii) After process (i), differentiate both sides, w.r.t. x and put or 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. Example: 34 (a) (b) (c) (d) Solution: (c) We know that, ......(i) Differentiating both sides w.r.t. x, we get = 0 + Putting we get, . Example: 35 If n is an integer greater than 1, then (a) a (b) 0 (c) (d) Solution: (b) We have = We know that ; Put , Then differentiating both sides w.r.t. to x , we get Put , = . (2) Use of Integration : This method is applied only when the numericals occur as the denominator of the binomial coefficients. Solution process : If , then we integrate both sides between the suitable limits which gives the required series. (i) If the sum contains 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., 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. Example: 36 (a) (b) (c) (d) None of these Solution: (c) Consider the expansion .....(i) Integrating both sides of (i) within limits 0 to 1 we get, ; . Example: 37 (a) (b) (c) (d) Solution: (a) It is clear that it is a expansion of Integrating w.r.t. x both sides between the limit 0 to 2.  . Example: 38 The sum to terms of the following series is (a) (b) (c) (d) None of these Solution: (d)   (i) The integral on L.H.S. of (i) = by putting ,  Whereas the integral on the R.H.S. of (i) = = to terms = Trick : Put in given series = . Which is given by option (d). 6.1.11 An Important Theorem. If = where I and n are positive integers, n being odd and then where = and . Note :  If n is even integer then Hence L.H.S. and I are integers. is also integer;  ; Hence = = = . Example: 39 Let and where [.] denotes the greatest integer function. The value of R.f is (a) (b) (c) (d) Solution: (a) Since , , where [R] is integer Now let = = = = Even integer Hence = even integer ¬– [R], but . Therefore, Hence R.f = . 6.1.12 Multinomial Theorem (For positive integral index). If n is positive integer and then Where are all non-negative integers subject to the condition, . (1) The coefficient of in the expansion of is (2) The greatest coefficient in the expansion of is Where q is the quotient and r is the remainder when n is divided by m. (3) If n is +ve integer and then coefficient of in the expansion of is Where are all non-negative integers subject to the condition: and . (4) The number of distinct or dissimilar terms in the multinomial expansion is . Example: 40 The coefficient of in the expansion of is (a) – 83 (b) – 82 (c) – 81 (d) 0 Solution: (c) Coefficient of in the expansion of is . where and . The possible value of and are shown in margin 1 3 1 2 1 2 0 5 0 The coefficient of = + = Example: 41 Find the coefficient of in the expansion of (a) 0 (b) 60 (c) – 60 (d) None of these Solution: (b) In this case, z + x = 3, ; ; . Then Therefore the coefficient of in the expansion of = . 6.1.13 Binomial Theorem for any Index. Statement : When n is a negative integer or a fraction, where , otherwise expansion will not be possible. If , 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 . Important Tips  Expansion is valid only when .  can not be used because it is defined only for natural number, so will be written as  The number of terms in the series is infinite.  If first term is not 1, then make first term unity in the following way, , if . General term : Some important expansions: (i) (ii) (iii) (iv) (a) Replace n by 1 in (iii) : , General term, (b) Replace n by 1 in (iv) : , General term, . (c) Replace n by 2 in (iii) : , General term, . (d) Replace n by 2 in (iv) : General term, . (e) Replace n by 3 in (iii) : General term, (f) Replace n by 3 in (iv) : General term, Example: 42 To expand as an infinite series, the range of x should be (a) (b) (c) (d) (– 2, 2) Solution: (b) can be expanded if i.e., if i.e., if i.e., if . Example: 43 If the value of x is so small that and higher power can be neglected, then is equal to (a) (b) (c) (d) Solution: (b) Given expression can be written as = = = = , when are neglected. Example: 44 If then the value of a and n is (a) (b) 2, 3 (c) 3, 6 (d) 1, 2 Solution: (a) We know that  Comparing coefficients of both sides we get, and on solving, , b = 4. Example: 45 Coefficient of in the expansion of (a) (b) (c) (d) Solution: (b) Coefficient of Example: 46 The coefficient of in is (a) 25 (b) – 25 (c) 1 (d) – 1 Solution: (c) Coefficient of in = Coefficient of in = Coefficient of in = Coefficient of in [ ] = = Coefficient of in – Coefficient of in = . Example: 47 (a) (b) (c) (d) None of these Solution: (c) We know that Here   . Example: 48 If x is so small that its two and higher power can be neglected and = then k = (a) 1 (b) – 2 (c) 10 (d) 11 Solution: (d) Higher power can be neglected. Then ; ; Example: 49 The cube root of is (a) (b) (c) (d) None of these Solution: (c) We have =  Example: 50 The coefficient of in the expansion of is (a) (b) (c) (d) None of these Solution: (a) = = Coefficient of = . Trick: Put and find the coefficients of and comparing with the given option as : Coefficient of is = = ; Which is given by option (a) . 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. (2) If consecutive terms are given : In this case divide consecutive terms pair wise i.e. if four consecutive terms be then find  (say) then divide by and by and solve. Example: 51 If are the coefficients of any four consecutive terms in the expansion of , then (a) (b) (c) (d) Solution: (c) Let be respectively the coefficients of terms in the expansion of . Then . Now, = = = = 6.1.15 Some Important Points. (1) Pascal's Triangle : 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 Pascal's triangle gives the direct binomial coefficients. Example : (2) Method for finding terms free from radical or rational terms in the expansion of prime numbers : Find the general term Putting the values of , when indices of a and b are integers. Note :  Number of irrational terms = Total terms – Number of rational terms. Example: 52 The number of integral terms in the expansion of is (a) 32 (b) 33 (c) 34 (d) 35 Solution: (b) First term = and after eight terms, i.e., 9th term = Continuing like this, we get an A.P., ; Example: 53 The number of irrational terms in the expansion of is (a) 97 (b) 98 (c) 96 (d) 99 Solution: (a) As 2 and 5 are co-prime. will be rational if is multiple of 8 and r is multiple of 6 also ; ......(i) But is to be multiple of 8. So, = 0, 8, 16, 24,......96 .....(ii) Common terms in (i) and (ii) are 16, 40, 64, 88. r = 84, 60, 36, 12 give rational terms The number of irrational terms = 101 – 4 = 97.