1. THE BINOMIAL THEOREM Robert Yen Hurlstone Agricultural High School This presentation is accompanied by workshop notes

2. INTRODUCTION   Expanding (a + x) n   Difficult topic: high-level algebra   Targeted at better Extension 1 students   Master this topic to get ahead in the HSC exam   No shortcuts for this topic

3. BINOMIAL EXPANSIONS AND PASCAL’S TRIANGLE Coefficients (a + x) 1 = a + x 1 1 (a + x) 2 = a 2 + 2ax + x 2 1 2 1 (a + x) 3 = a 3 + 3a 2 x + 3ax 2 + x 3 1 3 3 1 (a + x) 4 = a 4 + 4a 3 x + 6a 2 x 2 + 4ax 3 + x 4 1 4 6 4 1 (a + x) 5 = a 5 + 5a 4 x + 10a 3 x 2 + 10a 2 x 3 + 5ax 4 + x 5 1 5 10 10 5 1

4. n C k, A FORMULA FOR PASCAL’S TRIANGLE 1 0 C 0 1 1 1 C 0 1 C 1 1 2 1 2 C 0 2 C 1 2 C 2 1 3 3 1 3 C 0 3 C 1 3 C 2 3 C 3 1 4 6 4 1 4 C 0 4 C 1 4 C 2 4 C 3 4 C 4 1 5 10 10 5 1 5 C 0 5 C 1 5 C 2 5 C 3 5 C 4 5 C 5 1 6 15 20 15 6 1 6 C 0 6 C 1 6 C 2 6 C 3 6 C 4 6 C 5 6 C 6 1 7 21 35 35 21 7 1 7 C 0 7 C 1 7 C 2 7 C 3 7 C 4 7 C 5 7 C 6 7 C 7 1 8 28 56 70 56 28 8 1 8 C 0 8 C 1 8 C 2 8 C 3 8 C 4 8 C 5 8 C 6 8 C 7 8 C 8 n C k gives the value of row n, term k, if we start numbering the rows and terms from 0

5. n C k, A FORMULA FOR PASCAL’S TRIANGLE

6. CALCULATINGMentally

7. CALCULATINGMentallyFormula

8. CALCULATINGMentallyFormula because...

9. THE BINOMIAL THEOREM (a + x) n = n C 0 a n + n C 1 a n-1 x + n C 2 a n-2 x 2 + n C 3 a n-3 x 3 + n C 4 a n-4 x 4 +... + n C n x n = Don’t worry too much about writing in  notation: just have a good idea of the general term The sum of terms from k = 0 to n

10. PROPERTIES OF n C k 1. n C 0 = n C n = 11 st and last 2. n C 1 = n C n-1 = n2 nd and 2 nd -last 3. n C k = n C n-k Symmetry 4. n+1 C k = n C k-1 + n C k Pascal’s triangle result: each coefficient is the sum of the two coefficients in the row above it

11. n+1 C k = n C k-1 + n C k Pascal’s triangle result 1 0 C 0 1 1 1 C 0 1 C 1 1 2 1 2 C 0 2 C 1 2 C 2 1 3 3 1 3 C 0 3 C 1 3 C 2 3 C 3 1 4 6 4 1 4 C 0 4 C 1 4 C 2 4 C 3 4 C 4 1 5 10 10 5 1 5 C 0 5 C 1 5 C 2 5 C 3 5 C 4 5 C 5 1 6 15 20 15 6 1 6 C 0 6 C 1 6 C 2 6 C 3 6 C 4 6 C 5 6 C 6 1 7 21 35 35 21 7 1 7 C 0 7 C 1 7 C 2 7 C 3 7 C 4 7 C 5 7 C 6 7 C 7 1 8 28 56 70 56 28 8 1 8 C 0 8 C 1 8 C 2 8 C 3 8 C 4 8 C 5 8 C 6 8 C 7 8 C 8 15 = 10 + 5 6 C 4 = 5 C 3 + 5 C 4

12. Example 1 (a) (a + 3) 5 =

13. Example 1 (a) (a + 3) 5 = 5 C 0 a 5 + 5 C 1 a 4 3 1 + 5 C 2 a 3 3 2 + 5 C 3 a 2 3 3 + 5 C 4 a 1 3 4 + 5 C 5 3 5 = a 5 + 5a 4.3 + 10a 3.9 + 10a 2.27 + 5a.81 + 243 = a 5 + 15a 4 + 90a 3 + 270a 2 + 405a + 243 (b) (2x – y) 4 =

14. Example 1 (a) (a + 3) 5 = 5 C 0 a 5 + 5 C 1 a 4 3 1 + 5 C 2 a 3 3 2 + 5 C 3 a 2 3 3 + 5 C 4 a 1 3 4 + 5 C 5 3 5 = a 5 + 5a 4.3 + 10a 3.9 + 10a 2.27 + 5a.81 + 243 = a 5 + 15a 4 + 90a 3 + 270a 2 + 405a + 243 (b) (2x – y) 4 = 4 C 0 (2x) 4 + 4 C 1 (2x) 3 (-y) 1 + 4 C 2 (2x) 2 (-y) 2 + 4 C 3 (2x) 1 (-y) 3 + 4 C 4 (-y) 4 = 16x 4 + 4(8x 3 )(-y) + 6(4x 2 )(y 2 ) + 4(2x)(-y 3 ) + y 4 = 16x 4 – 32x 3 y + 24x 2 y 2 – 8xy 3 + y 4

15. Example 2 (2008 HSC, Question 1(d), 2 marks) (2x + 3y) 12 = 12 C 0 (2x) 12 + 12 C 1 (2x) 11 (3y) 1 + 12 C 2 (2x) 10 (3y) 2 +... General term T k = 12 C k (2x) 12-k (3y) k = 1 mark For coefficient of x 8 y 4, substitute k = ?

16. Example 2 (2008 HSC, Question 1(d), 2 marks) (2x + 3y) 12 = 12 C 0 (2x) 12 + 12 C 1 (2x) 11 (3y) 1 + 12 C 2 (2x) 10 (3y) 2 +... General term T k = 12 C k (2x) 12-k (3y) k = 1 mark For coefficient of x 8 y 4, substitute k = 4: T 4 = 12 C 4 (2x) 8 (3y) 4 = 12 C 4 (2 8 )(3 4 ) x 8 y 4  Coefficient is 12 C 4 (2 8 )(3 4 ) or 10 264 320. It’s OK to leave the coefficient unevaluated, especially if the question asks for ‘an expression’.

17. T k is not the k th term   T k is the term that contains x k   Simpler to write out the first few terms rather than memorise the  notation   Better to avoid referring to ‘the k th term’: too confusing   HSC questions ask for ‘the term that contains x 8 ’ rather than ‘the 9th term’

18. Example 3 (2005 HSC, Question 2(b), 3 marks) General term T k = 12 C k (2x) 12-k = 12 C k 2 12-k x 12-k (-x -2 ) k = 12 C k 2 12-k x 12-k (-1) k x -2k = 12 C k 2 12-k x 12-3k (-1) k For term independent of x:

19. Example 3 (2005 HSC, Question 2(b), 3 marks) General term T k = 12 C k (2x) 12-k = 12 C k 2 12-k x 12-k (-x -2 ) k = 12 C k 2 12-k x 12-k (-1) k x -2k = 12 C k 2 12-k x 12-3k (-1) k For term independent of x: 12 – 3k = 0 k = 4 T 4 = 12 C 4 2 8 x 0 (-1) 4 = 126 720

20. FINDING THE GREATEST COEFFICIENT (1 + 2x) 8 = 1 + 16x + 112x 2 + 448x 3 + 1120x 4 + 1792x 5 + 1792x 6 + 1024x 7 + 256x 8

21. Example 4 (a) (1 + 2x) 8 = 8 C 0 1 8 + 8 C 1 1 7 (2x) 1 + 8 C 2 1 6 (2x) 2 + 8 C 3 1 5 (2x) 3 +... General term T k = 8 C k 1 8-k (2x) k = 8 C k 1 (2 k ) x k = 8 C k 2 k x k  t k = 8 C k 2 k Leave out x k as we are only interested in the coefficient

22. Example 4 (a) (1 + 2x) 8 = 8 C 0 1 8 + 8 C 1 1 7 (2x) 1 + 8 C 2 1 6 (2x) 2 + 8 C 3 1 5 (2x) 3 +... General term T k = 8 C k 1 8-k (2x) k = 8 C k 1 (2 k ) x k = 8 C k 2 k x k  t k = 8 C k 2 k (b) Leave out x k as we are only interested in the coefficient

23. Example 4 (b) Ratio of consecutive factorials

24. Example 4 (b) Ratio of consecutive factorials

25. Example 4 (c) For the greatest coefficient t k+1, we want: t k+1 > t k 16 – 2k > k + 1 -3k > -15 k < 5 k = 4 k must be a whole number for the largest possible integer value of k

26. Example 4 (c) Greatest coefficient t k+1 = t 5 = 8 C 5 2 5 = 56  32 = 1792 t k = 8 C k 2 k

27. THE BINOMIAL THEOREM FOR (1 + x) n (1 + x) n = n C 0 + n C 1 x + n C 2 x 2 + n C 3 x 3 + n C 4 x 4 +... + n C n x n =

28. Example 5 (1 + x) n = n C 0 + n C 1 x + n C 2 x 2 +... + n C n x n Sub x = ? [Aiming to prove: ]

29. Example 5 (1 + x) n = n C 0 + n C 1 x + n C 2 x 2 +... + n C n x n Sub x = 1: (1 + 1) n = n C 0 + n C 1 (1) + n C 2 (1 2 ) +... + n C n (1 n ) 2 n = n C 0 + n C 1 + n C 2 +... + n C n This will make the x’s disappear and make the LHS become 2 n

30. Example 6 (1 + x) 2n = 2n C 0 + 2n C 1 x + 2n C 2 x 2 +... + 2n C 2n x 2n Term with x n = 2n C n x n Coefficient of x n = 2n C n [Aiming to prove ]

31. Example 6 (1 + x) 2n = 2n C 0 + 2n C 1 x + 2n C 2 x 2 +... + 2n C 2n x 2n Term with x n = 2n C n x n Coefficient of x n = 2n C n (1 + x) n = n C 0 + n C 1 x + n C 2 x 2 +... + n C n x n  (1 + x) n.(1 + x) n = ( n C 0 + n C 1 x + n C 2 x 2 +... + n C n x n )  ( n C 0 + n C 1 x + n C 2 x 2 +... + n C n x n ) If we expanded the RHS, there would be many terms [Aiming to prove ]

32. Example 6  (1 + x) n.(1 + x) n = ( n C 0 + n C 1 x + n C 2 x 2 +... + n C n x n )  ( n C 0 + n C 1 x + n C 2 x 2 +... + n C n x n ) Terms with x n = n C 0 ( n C n x n ) + n C 1 x ( n C n-1 x n-1 ) + n C 2 x 2 ( n C n-2 x n-2 ) +... + n C n x n ( n C 0 ) Coefficient of x n = n C 0 n C n + n C 1 n C n-1 + n C 2 n C n-2 +... + n C n n C 0 = ( n C 0 ) 2 + ( n C 1 ) 2 + ( n C 2 ) 2 +... + ( n C n ) 2 by symmetry of Pascal’s triangle [Aiming to prove ] n C k = n C n-k

33. Example 6  By equating coefficients of x n on both sides of (1 + x) 2n = (1 + x) n.(1 + x) n 2n C n = ( n C 0 ) 2 + ( n C 1 ) 2 + ( n C 2 ) 2 +... + ( n C n ) 2

34. Example 7 (2006 HSC, Question 2(b), 2 marks) (i) (1 + x) n = n C 0 + n C 1 x + n C 2 x 2 +... + n C n x n Differentiating both sides: n(1 + x) n-1 = 0 + n C 1 + 2 n C 2 x +... + r n C r x r-1 +... + n n C n x n-1 = n C 1 + 2 n C 2 x +... + r n C r x r-1 +... + n n C n x n-1 (ii) Substitute x = ? to prove result: [Aiming to prove: n 3 n-1 = n C 1 +... + r n C r 2 r-1 +... + n n C n 2 n-1 ] The general term

35. Example 7 (2006 HSC, Question 2(b), 2 marks) (i) (1 + x) n = n C 0 + n C 1 x + n C 2 x 2 +... + n C n x n Differentiating both sides: n(1 + x) n-1 = 0 + n C 1 + 2 n C 2 x +... + r n C r x r-1 +... + n n C n x n-1 = n C 1 + 2 n C 2 x +... + r n C r x r-1 +... + n n C n x n-1 (ii) Substitute x = 2 to prove result: n(1 + 2) n-1 = n C 1 + 2 n C 2 2 +... + r n C r 2 r-1 +... + n n C n 2 n-1 n 3 n-1 = n C 1 + 4 n C 2 +... + r n C r 2 r-1 +... + n n C n 2 n-1

36. Example 8 (2008 HSC, Question 6(c), 5 marks) (i) (1 + x) p+q = p+q C 0 + p+q C 1 x + p+q C 2 x 2 +... + p+q C p+q x p+q  The term of independent of x will be

37. Example 8 (2008 HSC, Question 6(c), 5 marks) (i) (1 + x) p+q = p+q C 0 + p+q C 1 x + p+q C 2 x 2 +... + p+q C p+q x p+q  The term of independent of x will be

38. Example 8 (2008 HSC, Question 6(c), 5 marks) (i) (1 + x) p+q = p+q C 0 + p+q C 1 x + p+q C 2 x 2 +... + p+q C p+q x p+q  The term of independent of x will be (ii) (1 + x) p = p C 0 + p C 1 x + p C 2 x 2 +... + p C p x p

39. Example 8 (2008 HSC, Question 6(c), 5 marks) (ii) If we expanded the RHS, there would be many terms Terms independent of x =

40. Example 8 (2008 HSC, Question 6(c), 5 marks) (ii) If we expanded the RHS, there would be many terms Terms independent of x = p C 0 q C 0 + p C 1 x q C 1 + p C 2 x 2 q C 2 +... + p C p x p q C p = 1 + p C 1 q C 1 + p C 2 q C 2 +... + p C p q C p p ≤ q

41. Example 8 (2008 HSC, Question 6(c), 5 marks) (ii) Terms independent of x = p C 0 q C 0 + p C 1 x q C 1 + p C 2 x 2 q C 2 +... + p C p x p q C p = 1 + p C 1 q C 1 + p C 2 q C 2 +... + p C p q C p  By equating the terms independent of x on both sides of p+q C q = 1 + p C 1 q C 1 + p C 2 q C 2 +... + p C p q C p 1 + p C 1 q C 1 + p C 2 q C 2 +... + p C p q C p = p+q C q From (i)

42. And now... A fairly hard identity to prove: Example 9 from 2002 HSC Question 7(b), 6 marks

43. Example 9 (2002 HSC, Question 7(b), 6 marks) (i) (1 + x) n = n C 0 + n C 1 x + n C 2 x 2 +... + n C n x n = c 0 + c 1 x + c 2 x 2 +... + c n x n [1] Identity to be proved involves (n + 2) 2 n-1 so try... [Aiming to prove: c 0 + 2c 1 + 3c 2 +... + (n + 1)c n = (n + 2) 2 n-1 ] To save time, this question asks us to abbreviate n C k to c k

44. Example 9 (2002 HSC, Question 7(b), 6 marks) (i) (1 + x) n = n C 0 + n C 1 x + n C 2 x 2 +... + n C n x n = c 0 + c 1 x + c 2 x 2 +... + c n x n [1] Identity to be proved involves (n + 2) 2 n-1 so try... Differentiating both sides: n(1 + x) n-1 = c 1 + 2c 2 x +... + nc n x n-1 Substituting x = ? to give 2 n-1 on the LHS: [Aiming to prove: c 0 + 2c 1 + 3c 2 +... + (n + 1)c n = (n + 2) 2 n-1 ]

45. Example 9 (2002 HSC, Question 7(b), 6 marks) (i) (1 + x) n = n C 0 + n C 1 x + n C 2 x 2 +... + n C n x n = c 0 + c 1 x + c 2 x 2 +... + c n x n [1] Identity to be proved involves (n + 2) 2 n-1 so try... Differentiating both sides: n(1 + x) n-1 = c 1 + 2c 2 x +... + nc n x n-1 Substituting x = 1 to give 2 n-1 on the LHS: n(1 + 1) n-1 = c 1 + 2c 2 1 +... + nc n 1 n-1 n 2 n-1 = c 1 + 2c 2 +... + nc n [2] How do we make the LHS say (n + 2) 2 n-1 ? [Aiming to prove: c 0 + 2c 1 + 3c 2 +... + (n + 1)c n = (n + 2) 2 n-1 ]

46. Example 9 (2002 HSC, Question 7(b), 6 marks) (i) (1 + x) n = n C 0 + n C 1 x + n C 2 x 2 +... + n C n x n = c 0 + c 1 x + c 2 x 2 +... + c n x n [1] Differentiating both sides: n(1 + x) n-1 = c 1 + 2c 2 x +... + nc n x n-1 Substituting x = 1: n(1 + 1) n-1 = c 1 + 2c 2 1 +... + nc n 1 n-1 n 2 n-1 = c 1 + 2c 2 +... + nc n [2] Add 2 (2 n-1 ) to both sides. But that’s 2 n. [Aiming to prove: c 0 + 2c 1 + 3c 2 +... + (n + 1)c n = (n + 2) 2 n-1 ]

47. Example 9 (2002 HSC, Question 7(b), 6 marks) (i) (1 + x) n = n C 0 + n C 1 x + n C 2 x 2 +... + n C n x n = c 0 + c 1 x + c 2 x 2 +... + c n x n [1] Differentiating both sides: n(1 + x) n-1 = c 1 + 2c 2 x +... + nc n x n-1 Substituting x = 1: n(1 + 1) n-1 = c 1 + 2c 2 1 +... + nc n 1 n-1 n 2 n-1 = c 1 + 2c 2 +... + nc n [2] To prove the 2 n identity, sub x = 1 into [1] above: (1 + 1) n = c 0 + c 1 1 + c 2 1 2 +... + c n 1 n 2 n = c 0 + c 1 + c 2 +... + c n [3] [Aiming to prove: c 0 + 2c 1 + 3c 2 +... + (n + 1)c n = (n + 2) 2 n-1 ]

48. Example 9 (2002 HSC, Question 7(b), 6 marks) (i)n 2 n-1 = c 1 + 2c 2 +... + nc n [2] 2 n = c 0 + c 1 + c 2 +... + c n [3] [2] + [3]: n 2 n-1 + 2 n = c 1 + 2c 2 +... + nc n + c 0 + c 1 + c 2 +... + c n 2 n-1 (n + 2) = c 0 + 2c 1 + 3c 2 +... + (n + 1)c n c 0 + 2c 1 + 3c 2 +... + (n + 1)c n = (n + 2) 2 n-1 [Aiming to prove: c 0 + 2c 1 + 3c 2 +... + (n + 1)c n = (n + 2) 2 n-1 ] Each coefficient c k increases by 1 as required

49. Example 9 (2002 HSC, Question 7(b), 6 marks) (ii)Answer involves dividing by (k + 1)(k + 2) and alternating –/+ pattern so try integrating (1 + x) n from (i) twice and substituting x = -1. (1 + x) n = c 0 + c 1 x + c 2 x 2 +... + c n x n [1] [Aiming to find ]

50. Example 9 (2002 HSC, Question 7(b), 6 marks) (ii)(1 + x) n = c 0 + c 1 x + c 2 x 2 +... + c n x n [1] Integrate both sides: To find k, sub x = 0: [Aiming to find ] Don’t forget the constant of integration

51. Example 9 (2002 HSC, Question 7(b), 6 marks) (ii) Integrate again to get 1.2, 2.3, 3.4 denominators: To find d, sub x = 0: [Aiming to find ]

52. Example 9 (2002 HSC, Question 7(b), 6 marks) (ii) Sub x = ? for –/+ pattern: [Aiming to find ]

53. Example 9 (2002 HSC, Question 7(b), 6 marks) (ii) Sub x = -1 for –/+ pattern: [Aiming to find ]

54. Example 9 (2002 HSC, Question 7(b), 6 marks) (ii) [Aiming to find ]

55. Example 10 (2007 HSC, Question 4(a), 6 marks) (i) P(both green) = 0.1  0.1 = 0.01 (ii) p = 0.1, q = 1 – 0.1 = 0.9, n = 20 P(X = 2) =

56. Example 10 (2007 HSC, Question 4(a), 6 marks) (i) P(both green) = 0.1  0.1 = 0.01 (ii) p = 0.1, q = 1 – 0.1 = 0.9, n = 20 P(X = 2) = 20 C 2 0.1 2 0.9 18 = 0.28517...  0.285

57. Example 10 (2007 HSC, Question 4(a), 6 marks) (i) P(both green) = 0.1  0.1 = 0.01 (ii) p = 0.1, q = 1 – 0.1 = 0.9, n = 20 P(X = 2) = 20 C 2 0.1 2 0.9 18 = 0.28517...  0.285 (iii) P(X > 2) = 1 – P(X = 0) – P(X = 1) – P(X = 2) = 1 – 20 C 0 0.1 0 0.9 20 – 20 C 1 0.1 1 0.9 19 – 0.285 from (ii) = 1 – 0.9 20 – 20(0.1)0.9 19 – 0.285 = 0.32325...  0.32

58. HOW TO STUDY FOR MATHS (P-R-A-C) 1. Practise your maths 2. Rewrite your maths 3. Attack your maths 4. Check your maths WORK HARD AND BEST OF LUCK FOR YOUR HSC EXAMS!