Binomial Expansion

8 Key excerpts on "Binomial Expansion"

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  Maths: A Student's Survival Guide

  A Self-Help Workbook for Science and Engineering Students

  • Jenny Olive(Author)
  • 2003(Publication Date)
  • Cambridge University Press

    (Publisher)

  Binomial Expansions are what we get when we raise these brackets to different powers and then multiply the brackets together to find the result. In this first section all these powers will be positive whole numbers. Here are some examples. ( a + b ) 1 is just a + b ( a + b ) 2 = ( a + b )( a + b ) = a 2 + 2 ab + b 2 . The 2 ab comes from the two middle terms of ab which add together because it doesn’t matter what order we multiply a and b in. 7.A Binomial series: positive whole numbers 261 Next comes ( a + b ) 3 = ( a + b )( a + b )( a + b ) = a 3 + 3 a 2 b + 3 ab 2 + b 3 . We find the answer by picking one letter from each bracket in every possible way and then multiplying these choices together. There is only one way of getting a 3 and b 3 . The a 2 b term comes in three ways, as we can choose the b from any of the three brackets, and then multiply it with the a terms in the other two brackets. Similarly, ab 2 can be made in three possible ways. What happens with ( a + b ) 4 = ( a + b )( a + b )( a + b )( a + b )? There will be just one a 4 and just one b 4 . There will also be some numbers of terms for each of a 3 b , a 2 b 2 and ab 3 . Because the a and the b are symmetrically placed in the brackets, there must be the same number of terms in a 3 b as there are in ab 3 . There will be four of each since we can pick either a single b or a single a in four different ways from the four brackets. The six possibilities for a 2 b 2 are given by aabb , abba , abab , baab , baba and bbaa . We see that by multiplying the four brackets together, we get ( a + b ) 4 = a 4 + 4 a 3 b + 6 a 2 b 2 + 4 ab 3 + b 4 .

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  Intermediate Algebra

  A Guided Approach

  • Rosemary Karr, Marilyn Massey, R. Gustafson, , Rosemary Karr, Marilyn Massey, R. Gustafson(Authors)
  • 2014(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Proper planning can help you avoid taking classes that might not be required for your degree, thus saving you time and money. Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 800 CHAPTER 11 Miscellaneous Topics SECTION 11.1 The Binomial Theorem 11 Review ` DEFINITIONS AND CONCEPTS EXAMPLES The symbol n ! (factorial) is defined as n ! 5 n 1 n 2 1 21 n 2 2 2 ? ? ? ? ? 1 3 21 2 21 1 2 where n is a natural number. 0! 5 1 n 1 n 2 1 2 ! 5 n ! ( n is a natural number) 7! 5 7 ? 6 ? 5 ? 4 ? 3 ? 2 ? 1 5 5,040 5! 2!3! 5 5 ? 4 ? 3! 2 ? 1 ? 3! 5 20 2 5 1 0 7 ? 6 ! 5 7 ? 6 ? 5 ? 4 ? 3 ? 2 ? 1 5 7! The binomial theorem: If n is any positive integer, then 1 a 1 b 2 n 5 a n 1 n ! 1! 1 n 2 1 2 ! a n 2 1 b 1 n ! 2! 1 n 2 2 2 ! a n 2 2 b 2 1 c 1 n ! r ! 1 n 2 r 2 ! a n 2 r b r 1 c 1 b n 1 x 2 2 y 2 4 5 3 x 1 1 2 2 y 24 4 5 x 4 1 4! 1!3! x 3 1 2 2 y 2 1 4! 2!2! x 2 1 2 2 y 2 2 1 4! 3!1! x 1 2 2 y 2 3 1 1 2 2 y 2 4 5 x 4 1 4 x 3 1 2 2 y 2 1 6 x 2 1 4 y 2 2 1 4 x 1 2 8 y 3 2 1 16 y 4 5 x 4 2 8 x 3 y 1 24 x 2 y 2 2 32 xy 3 1 16 y 4 Specific term of a Binomial Expansion: The binomial theorem can be used to find a specific term of a Binomial Expansion. The coefficient of the variables can be found using the formula n ! r ! 1 n 2 r 2 ! where n is the power of the expansion and r is the exponent of the second variable. To find the third term of 1 x 1 y 2 5 , note that the exponent on y will be 2, because the exponent on y is 1 less than the number of the term. The power on x will be 3 because the sum of the powers must be 5.

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  Sources in the Development of Mathematics

  Series and Products from the Fifteenth to the Twenty-first Century

  • Ranjan Roy(Author)
  • 2011(Publication Date)
  • Cambridge University Press

    (Publisher)

  4 The Binomial Theorem 4.1 Preliminary Remarks The discovery of the binomial theorem for general exponents exerted a tremendous impact on the development of analysis, especially the theory of power series. It also led to an understanding that an exponential function was defined by the property f (a + b) = f (a)f (b). The binomial theorem was pivotal not only in the initial dis- covery of series for other important functions but also in the eventual consolidation of the foundations of analysis as a whole. The development of the theorem is particularly fascinating because it was independently found by both Newton and Gregory; because of the various approaches to its proof, including one by Euler; and because the vali- dation of these proofs elicited the efforts of the best mathematicians of the nineteenth century. Islamic mathematicians were the original discoverers of the binomial theorem for positive integral exponents, although they did not have the notation to write the expan- sion for arbitrary integers. But they knew how to find the coefficients for any given integral exponent. The two important rules for binomial coefficients appear in the work of al-Kashi of around 1427, and it is likely that earlier Islamic mathematicians such as al-Tusi, Omar Khayyam, and al-Karji were also aware of them: Let C n,k denote the coefficient of x k in the expansion of (1 + x) n . Then C n,k = C n−1,k + C n−1,k−1 and C n,k = n(n − 1) ··· (n − k + 1) k! . The first formula, the additive rule for binomial coefficients, leads to the expansion of (1 + x) n , by means of the expansion of (1 + x) n−1 ; the second formula, the multiplicative rule, immediately yields the expansion of (1 + x) n . Henry Briggs (1561–1630) appears to be the first European to explicitly state both formulas, though Cardano may have known the results around 1570. In 1654, Pascal gave a proof by complete induction of the second formula.

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  Series and Products in the Development of Mathematics: Volume 1

  • Ranjan Roy(Author)
  • 2021(Publication Date)
  • Cambridge University Press

    (Publisher)

  4 The Binomial Theorem 4.1 Preliminary Remarks The discovery of the binomial theorem for general exponents exerted a tremendous impact on the development of analysis, especially the theory of power series. It also led to an understanding that an exponential function was defined by the property f (a + b) = f (a)f (b). The binomial theorem was pivotal not only in the initial discovery of series for other important functions but also in the eventual consolidation of the foundations of analysis as a whole. The development of the theorem is particularly fascinating because it was independently found by both Newton and Gregory; because of the various approaches to its proof, including one by Euler; and because the validation of these proofs elicited the efforts of the best mathematicians of the nineteenth century. The binomial theorem for a positive integer exponent n states that (a + b) n = a n + A n 1 a n−1 b + A n 2 a n−2 b 2 + · · · + A n n−1 ab n−1 + b n , (4.1) where the coefficients A n k satisfy the additive rule A n k = A n−1 k−1 + A n−1 k , (4.2) and the multiplicative rule A n k = n(n − 1) · · · (n − k + 1) 1 · 2 · · · k , (4.3) where it is understood that A n 0 = 1. We here use a notation unusual today, because the notation  n k  or C n k , or C n,k , may be suggestive of recent developments, whereas we wish to understand how these coefficients developed over time. Now we note that the additive rule (4.2) is not difficult to obtain. In terms of the notation used in (4.1), we can write 77 78 The Binomial Theorem (a + b) n−1 = a n−1 + A n−1 1 a n−2 b + · · · + A n−1 k−1 a n−k b k−1 + A n−1 k a n−k−1 b k + · · · + b n−1 . Multiplying both sides by a + b, we have (a + b) ( a n−1 + · · · + A n−1 k−1 a n−k b k−1 + A n−1 k a n−k−1 b k + · · · + b n−1 ) = a n + · · · + A n k a n−k b k + · · · + b n . Equating the coefficients of a n−k b k on each side, we obtain (4.2). The multiplicative rule is somewhat more difficult to obtain.

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  Precalculus

  Building Concepts and Connections 2E

  • Revathi Narasimhan(Author)
  • 2016(Publication Date)
  • XYZ Textbooks

    (Publisher)

  Example 5 Exercises 10.6 863 Skills In Exercises 1−4, write down the variable parts of the terms in the expansion of the binomial. 1. ( a + b ) 5 2. ( a + b ) 6 3. ( x + y ) 7 4. ( x + y ) 8 In Exercises 5−16, evaluate each expression. 5. 4! 6. 6! 7. 3! __ 2! 8. 4! __ 3! 9. ( 6 2 ) 10. ( 5 3 ) 11. ( 7 5 ) 12. ( 7 4 ) 13. ( 10 10 ) 14. ( 10 0 ) 15. ( 100 0 ) 16. ( 100 100 ) In Exercises 17−28, use the binomial theorem to expand the expression. 17. ( x + 2) 4 18. ( x − 3) 3 19. (2 x − l) 3 20. (2 x + 3) 4 21. (3 + y ) 5 22. (4 − z ) 4 23. ( x − 3 z ) 4 24. (2 z + y ) 3 25. ( x 2 + l) 3 26. ( x 2 − 2) 3 27. ( y − 2 x ) 4 28. ( z + 4 x ) 5 In Exercises 29−42, use the Binomial Theorem to find the indicated term or coefficient. 29. The coefficient of x 3 when expanding ( x + 4) 5 30. The coefficient of y 2 when expanding ( y − 3) 5 31. The coefficient of x 5 when expanding (3 x + 2) 6 32. The coefficient of y 4 when expanding (2 y + 1) 7 33. The coefficient of x 6 when expanding ( x + 1 ) 8 34. The coefficient of y 7 when expanding ( y − 3) 10 35. The third term in the expansion of ( x − 4) 6 36. The fourth term in the expansion of ( x + 3) 6 37. The sixth term in the expansion of ( x + 4 y ) 5 38. The seventh term in the expansion of ( a + 2 b ) 6 39. The fifth term in the expansion of (3 x − 2) 6 40. The fifth term in the expansion of (3 x + l) 8 41. The fourth term in the expansion of (4 x − 2) 6 42. The fourth term in the expansion of (3 x − l) 8 Concepts 43. Show that ( n r ) = ( n n − r ) , where 0 ≤ r ≤ n , with n and r integers. 44. Show that ( n 0 ) = 1. 45. Evaluate the following. ( 4 0 ) ( 1 __ 3 ) 4 + ( 4 1 ) ( 1 __ 3 ) 3 ( 2 __ 3 ) + ( 4 2 ) ( 1 __ 3 ) 2 ( 2 __ 3 ) 2 + ( 4 3 ) ( 1 __ 3 )( 2 __ 3 ) 3 + ( 4 4 ) ( 2 __ 3 ) 4 10.7 10.7 Mathematical Induction 865 Mathematical Induction Many mathematical facts are established by first observing a pattern, then making a conjecture about the general nature of the pattern, and finally proving the conjecture.

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  Combinatorics

  • Nicholas Loehr(Author)
  • 2017(Publication Date)
  • Chapman and Hall/CRC

    (Publisher)

  To see why, rewrite the given identity as (p - q) (r) = 0, where p - q is a polynomial in the formal variable r. If p - q were nonzero, say of degree d, then p - q would have at most d real roots. Since we are assuming the equation holds for infinitely many values of r, we conclude that p - q is the zero polynomial. Thus, p (r) = q (r) must hold for every r. A similar result holds for polynomial identities involving multiple parameters. 2.3 The Binomial Theorem The Binomial Theorem is a famous formula for expanding the n th power of a sum of two terms. Binomial coefficients are so named because of their appearance in this theorem. 2.8. The Binomial Theorem. For all x, y ∈ R and all n ∈ Z ≥ 0, (x + y) n = ∑ k = 0 n n k x k y n - k. For example, (x + y) 4 = 1 y 4 + 4 x y 3 + 6 x 2 y 2 + 4 x 3 y + 1 x 4. We now give a combinatorial proof of the Binomial Theorem under the additional assumption that x and y are positive integers. (Since both sides of the formula are polynomials in x and y, we can deduce the general case using the result mentioned in Remark 2.7.) Let A = { V 1, …, V x, C 1, …, C y } be an alphabet consisting of x + y letters, where A consists of x vowels V 1, …, V x and y consonants C 1, …, C y. Let S be the set of all n -letter words using the alphabet A. By the Word Rule, | S | = (x + y) n. On the other hand, we can classify words in S based on how many vowels they contain. For 0 ≤ k ≤ n, let S k be the set of words in S that contain exactly k vowels (and hence n - k consonants). To build a word in S k, first choose a set of k positions out of n where the vowels will appear (n k ways, by the Subset Rule); then fill these positions with a sequence of vowels (x k ways, by the Word Rule); then fill the remaining positions with a sequence of consonants (y n - k ways, by the Word Rule). By the Product Rule, | S k | = n k x k y n - k

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  Principles and Techniques in Combinatorics

  • Chen Chuan-Chong, Koh Khee-Meng(Authors)
  • 1992(Publication Date)
  • WSPC

    (Publisher)

  B|, as claimed.

  We thus conclude that the coefficient of (2.8.6) in the expansion is given by

  Combining this with identity (2.8.5), we arrive at the following generalization of the binomial theorem, that was first formulated by G.W. Leibniz (1646-1716) and later on proved by Johann Bernoulli (1667-1748).

  Theorem 2.8.1 (The Multinomial Theorem). For n, m ∈ N,

  where the sum is taken over all m-ary sequences (n1 , n2 , …, n

  m

  ) of nonnegative integers with , and

  Example 2.8.1. For n = 4 and m = 3, we have by Theorem 2.8.1 ,

  Because of Theorem 2.8.1 , the numbers of the form (2.8.2) are usually called the multinomial coefficients. Since multinomial coefficients are generalizations of binomial coefficients, it is natural to ask whether some results about binomial coefficients can be generalized to multinomial coefficients. We end this chapter with a short discussion on this.

  1° The identity for binomial coefficients may be written as (here of course n1 + n2 = n). By identity (2.8.5), it is easy to see in general that

  where {α(l), α(2), …, α(m)} = {1, 2 m}.

  2° The identity for binomial coefficients may be written:

  In general, we have:

  3° For binomial coefficients, we have the identity By letting x1 = x2 = ··· = x

  m

  = 1 in the multinomial theorem, we have

  where the sum is taken over all m-ary sequences (n1 , n2 , …, n

  m

  ) of nonnegative integers with .

  Identity (2.8.9) simply says that the sum of the coefficients in the expansion of (x1 + x2 +…+ x

  m

  )

  n

  is given by m

  n

  . Thus, in Example 2.8.1 , the sum of the coefficients in the expansion of (x1 + x2 + x3 )4 is 81, which is 34 .

  4° In the Binomial Expansion the number of distinct terms is n + 1. How many distinct terms are there in the expansion of (x1 + x2 +…+ x

  m

  )

  n

  ? To answer this question, let us first look at Example 2.8.1 . The distinct terms obtained in the expansion of (x1 + x2 + x3 )4

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  Mathematical Foundations of Computer Science

  • Bhavanari Satyanarayana, T.V. Pradeep Kumar, Shaik Mohiddin Shaw(Authors)
  • 2019(Publication Date)
  • CRC Press

    (Publisher)

  C HAPTER - 14 Binomial Theorem LEARNING OBJECTIVES ♦ to understand the basic concepts of Binomial and Multinomial Coefficients ♦ to know some Generating functions of permutations and combinations ♦ to develop problem solving skills related to the Principle of Inclusion & Exclusion If x and a are real numbers then for all n ∈ N we have. that (x+a) n = n C 0 x n a 0 + n C 1 x n − 1 a 1 + n C 2 x n − 2 a 2 + … + n C r x n − r a r + … + n C n − 1 x 1 a n − 1 + n C n x 0 a n That is, (x +. a) n = ∑ r = 0 n C n r x n − r a r and it is called as Binomial Theorem. 14.1  Binomial and Multinomial Coefficients 14.1.1  Pascal’s Triangle We know that (x + a) 0 =. 1 (x+a) 1 = x + a = 1. x + 1. a (x+a) 2 = x 2 + 2 ax+a 2 = 1. x 2 + 2. ax+1.a 2 (x+a) 3 = x 3 + 3 x 2 a + 3 xa 2 + a 3 = 1. x 3 + 3. x 2 a+3.xa 2 + 1. a 3 We observe that the coefficients in the above expansions follow a particular pattern as shown in the Table 14.1. Table 14.1 We observe that each row is bounded by 1 on both sides. Any entry, except the first and last, in a row is the sum of two entries in the preceding row, one on the immediate left and the other on immediate right. The above pattern is known as Pascal’s triangle. It has been checked that the above pattern also holds good for the coefficients in the expansions of the binomial expressions having index (exponent) greater than 3 as shown in the Table 14.2. Table 14.2 Index of the binomial Coefficients of various terms 0 1 1 1 1 2 1  2  1 3 1  3  3  1 4 1   4  6  4   1 5 1  5  10  10  5  1 Using the above Pascal’s triangle, we may express (x + a) n for n = 1, 2, 3,. … (x + a) 1 = x + a (x+a) 2 = x 2 + 2 ax + a 2 (x + a) 3 = x 3 + 3 ax 2 + 3 a 2 x + a 3 (x + a) 4 = x 4 + 4 ax 3 + 6 a 2 x 2 + 4 a 3 x + a 4 (x+a) 5 = x 5 + 5 ax 4 + 10 a 2 x 3 + 10 a 3 x[-. -=PLGO-SEPARATOR=--]2 + 5a 4 x + a 5 and so on. 14.1.2  Properties of Binomial Coefficients or Combinatorial Identities An identity that results from some counting process is called a combinatorial identity

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