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44 CHAPTER 5 BINOMIAL THEOREM 1.0 Introduction This theorem is simply a method for expanding expressions such as (ax + b)n where n is a positive integer. For example (3x + 4)7 can be expanded as follows, 2187x7 + 20412x6 + 81648x5 + 181440x4 +241920x3 + 193536x2 + 86016x + 16384. The binomial theorem was introduced by Newton and it finds a use in many parts of mathematics. It can also be expanded to include negative and fractional indicies. 2.0 The Pattern To see the pattern to be used when expanding the binomial expressions we expand the simple expressions algebraically. (a + b)2 = a2 + 2ab + b2 (a + b)3 = a3 + 3a2b + 3ab2 + b3 (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4 In each case the powers of a start at n and run down to 0 while the powers of b start at 0 and run up to n. In each term the powers of a and b add to n. The coefficients of each term can be found in several ways as shown in the next section. 3.0 Pascal’s Triangle. This is a triangle of numbers whose properties were first studied by the French mathematician Blaise Pascal. 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 Each row is calculated from the row above by adding the two terms either side together. Each row starts and ends with a 1. It should be noticed that the second, third and fourth rows of this table are giving the coefficients of the terms in the expansions of (a + b)2; (a + b)3 and (a + b)4 respectively (see above). 45 To complete the expansion of the binomial expression we select the coefficients from the appropriate row and remember that the powers of a (the first term) run down from n and the powers of b (the second term) run up from 0. In expanding (a + b)5 the coefficients will be 1 5 10 10 5 1. Following the pattern we get, (a + b)5 = 1a5b0 +5a4b1 + 10a3b2 + 10a2b3 + 5a1b4 + 1a0b5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 There is another way of calculating the coefficients. Pascal’s triangle becomes large when we have high powers hence using the graphics calculator becomes an attractive option. On the casio 9800 from the RUN menu select OPTN F6 F3(prob) . The expression nCr should appear on the screen. This calculates the binomial coefficient for a given n and value r. To find the value in the 5th row of Pascal’s triangle, third power of x you enter 5F33 (you see 5C3) on your screen) then EXE gives the answer 10. The value of n is easy to find but r sometimes requires a calculation. More details on finding the value of r to use in your calculations follows in the examples below. See appendix for more on how to use the calculator. 4.0 Notation The symbol nCr , which appears on the calculator screen, is written in different ways by different authors. cn and cn,r are just two of the various ways in which the symbol is r written. A more common symbol that is used, especially when expanding binomial n 7 terms, is the symbol Thus has the same meaning as 7 C2 . It is easier to use the . r 2 former symbol rather than the later is this work. Exercise: 1 Use the calculator to find the value of the following. 8 6 16 (a) 7 C3 (b) 12 C4 (c) (d) 11 C11 (e) (f) 2 4 4 12 11 11 15 (g) (h) 10 C7 (i) 13 C10 (j) (k) (l) 6 0 6 9 Further Examples In the following use Pascal’s triangle to find the expansion of the following. 46 (x + 3)4 = 1(x)4(3)0 + 4(x)3(3)1 + 6(x)2(3)2 + 4(x)1(3)3 + 1(x)0(3)4 = x4 +12x3 + 54x2 + 106x + 81 Notes: The coefficients are from row 4 of the triangle. The powers of x, the first term, run down. The powers of 3, the second term, run up. It is worthwhile putting the terms in brackets on the first expansion to avoid mistakes. (x – 4)3 = 1(x)3(–4)0 + 3(x)2(–4)1 + 3(x)1(–4)2 + 1(x)0(–4)3 = x3 – 12x2 + 48x – 64 Note: It is –4 which is raise to the power, not just 4. (2x – 1)5 = 1(2x)5(–1)0 + 5(2x)4(–1)1 + 10(2x)3(–1)2 + 10(2x)2(–1)3 + 5(2x)1(–1)4 + 1(2x)0(–1)5 = 32x5 – 80x4 +80x3 – 40x2 +10x – 1 (on simplifying) Exercise: 2 Use Pascal’s triangle to expand the following binomial expressions. Follow the method outlined in the example. (a) (x + 3)4 (b) (2x + 1)3 (c) (x – 2)4 (d) (1 – 3x)4 (e) (2x + 3)5 5.0 The General Expansion of the Binomial Expression Because there is a pattern to the expansion of the binomial expression it is possible to write out a general term for the expansion. If we consider the expansion of (a + b)5 we get the following a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 Ignoring the coefficients for a moment, we can see that each term is of the form a5–r br where r has the values 0,1,2,3,4,5 Now checking the coefficients we find that they come from position (r – 1) (remembering to start at 0) in 5th row of Pascal’s triangle. The coefficient of the second term is 5, which comes from the 2 position in the 5th row, hence r is 2 – 1 = 1. From our previous discussions on the use of the calculator we recall that the symbol nCr will give the coefficient from the 5th row position (r – 1). So the final general expansion of the binomial expression (a + b)n is given by. n (a + b)n = anr br where r has the values 0,1,2,3,4,5 . r 47 Once the value of r is known the calculator can be used to obtain the coefficient. Example: 1 Find the general expansion of 2x 38 8 8 r 3 r 2x 3 = 2x 8 r 8 = 28 r x8 - r 3r r 8 = 28 r 3r x8 r for r = 0,1,2,3,4,5,6,7,8 r Exercise: 3 Find the general term of the following binomial expressions. Remember to simplify as far as possible. (a) (x – 2)10 (b) (4 – x)12 remember –x = –1 x (c) (3x – 1)8 (d) (2x + 1)6 (e) (½x – 3)7 (f) (2x + 3y)8 (g) (x2 + 2)4 (h) (x – 1/x)6 6.0 Finding the value of r Now that we have the general expansion of the binomial theorem it is easy to select the value of r to be used in the calculating of the coefficient. Follow the example below. Example: 2 Find the coefficient of x3 in the following expansion (3 – x)12. 12 The general expansion is 312 r x r r 12 = 312 r 1r xr r now since we want the coefficient of x3 it is easy to see that r must equal 3 so substituting r = 3 we get the coefficient to be 12 9 3 1 = –220 3 3 9 3 Exercise: 4 In the following exercises write out the general expansion in full, determine the value of r required, then find the coefficient of the term. 48 (a) Find the coefficient of the x3 term in the expansion of (x – 2)9. (b) Find the coefficient of x5 in the expansion of (2 – x)10. (c) Find the coefficient of the constant term in the expansion of (2x – 3)6. 6 (d) Find the coefficient of the constant term in the expansion of x 2 2 . 1 x (e) Using the general term of question (d) find the coefficient of x–4. 7 (f) Find the coefficient of x3 in the expansion of x . 2 x (g) Find the coefficient of x6y2 in the expansion of (x – y)8. (h) Expand (3x – 2)3. 3 5 (i) Find the coefficient of the term p in the expansion of 2 p p (j) If (2x – 3)4 = 16x4 + Ax3 + 216x2 + Bx + 81 find the values of A and B. 3 (k) Write 3 2 in the form a + b3. 8 2 (l) Find the constant term in the expansion of 3x 2 2 . x 1 4 (m) Expand x 2 1 . (n) Find the coefficient of a4b2 in the expansion of (a + b)6. 49 Chapter 5 Review 1. Find the value of the following 2 8 7 8 7 10 (a) 7 C 3 (b) 8 C2 8 C 4 (c) 17 C13 (d) (e) (f) (g) 0 4 3 2 5 2. Expand the expression (x – 3)5. 3. Find the coefficient of x3 in the expansion of (x + 2)5. 4. Expand the expression (2x + 3)4. 5. Expand and simplify the expression (2x + 1)4. 6. If x4 + 8x3 + 24x2 + 32x + 16 is the expansion of (x + a)n, find a and n. 10 2 7. Find the term that is independent of x in the expansion of x . x 6 8. Write 1 3 in the form a + b 6 where a and b . 9. Find the coefficient of x6 in the expansion of (2x – 3)10. 10. (a) Expand (a + b)4 (b) By making a suitable substitution find the value of 1.014. 6 1 11. Find the coefficient of the constant term in the expansion of x 2 . x 12. Find the coefficient of the term containing x4y2 in the expansion of 2 x 3 y . 6 13. Find the coefficient of x7 in the expansion of (2x – 3)10. 12 –2 1 14. Find the coefficient of x in the expansion of x . x 15. If (2x – 3)5 can be expanded as 32x5 + Bx4 + Cx3……………. – 243 find the values of B and C. 16. Expand the binomial (1 + x)4. Hence, find the value of (1 + 0.01)4 without using the calculator. 17. Using the binomial theorem write (1 + 2)3 in the form a + b2. 50 18. Find the coefficient of a4b2 in the expansion of (a + b)6. 19. Expand (1 – x)3 and hence find a value for 0.993 ( hint 0.99 = 1– 0.01). 20. If (1 + x)n can be expanded as 1 + ax + bx2 +…………….+ x7. Find n, a and b. 21. Expand (1 + x)5 and hence show that 115 = 161051. 22. (a) Given that a row of Pascal’s triangle is 1 5 10 10 5 1; write down the next two rows. (b) Find the coefficient of x4y3 in the expansion of (x + y)7 23. Write (3 – 1)4 in the form a3 + b. 24. Expand (x – 1/x2)5 25. Find the first three terms of the expansion ( ½x – 3)4 26. Simplify (1 + x)3 – (1 – x)3. 27. Find the coefficient of a5b2 in the expansion of (a + 2b)7. 28. (a) Expand (a + b)4 (b) By making a suitable substitution for a and b show that 24 = 1 + 4 + 6 +4 + 1 29. If (2x + 1)5 = Ax5 + Bx4 + Cx3 + ……… when expanded, find, using the binomial theorem, the values of A, B and C. 30. Given the first two terms of the binomial expansion, find the values of n and b. ( x b)n x7 64 x6 .....................

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