THE BINOMIAL THEOREM Robert Yen

# THE BINOMIAL THEOREM Robert Yen

## THE BINOMIAL THEOREM Robert Yen

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1. Please take a handout This PowerPoint can be downloaded from HSC Online THE BINOMIAL THEOREM Robert Yen

2. TO AVOID ‘DEATH BY POWERPOINT’ ... Don’t copy everything down, as this presentation is on the HSC Online website Instead, make notes and copy the key steps, then fill in the gaps when you get home Aim to understand the theory TODAY, not later The handout contains all the important facts, while this PowerPoint contain the solutions to the examples If you already know this topic well, then work ahead and check your answers against mine

3. INTRODUCTION • Binomial theorem = expanding (a + x)n • Difficult topic at end of Ext 1 course: high-level algebra • Targeted at better Extension 1 students aiming for Band E4 • Master this topic to get ahead in the exam • No shortcuts for this topic: must learn the theory

4. BINOMIAL EXPANSIONS AND PASCAL’S TRIANGLE Coefficients (a + x)1 = a + x1 1 (a + x)2 = a2 + 2ax + x21 21 (a + x)3 = a3 + 3a2x + 3ax2 + x31 3 31 (a + x)4 = a4 + 4a3x + 6a2x2 + 4ax3 + x41 4 6 41 (a + x)5 = a5 + 5a4x + 10a3x2 + 10a2x3 + 5ax4 + x51 5 10 10 51 For (a + x)n, the powers of a are ↓ and the powers of x are↑ The sum of the powers in each term is always n

5. nCk, A FORMULA FOR PASCAL’S TRIANGLE 1 0C0 1 1 1C0 1C1 1 21 2C0 2C1 2C2 1 3 31 3C0 3C1 3C2 3C3 1 4 6 41 4C0 4C1 4C2 4C3 4C4 1 5 10 10 51 5C0 5C1 5C2 5C3 5C4 5C5 1 6 15 20 15 6 1 6C0 6C1 6C2 6C3 6C4 6C5 6C6 1 7 21 35 35 21 71 7C0 7C1 7C2 7C3 7C4 7C5 7C6 7C7 1 8 28 56 70 56 28 8 1 8C0 8C1 8C2 8C3 8C4 8C5 8C6 8C78C8 nCkgivesthe value of row n, term k, if we start numbering the rows and terms from 0

6. nCk, A FORMULA FOR PASCAL’S TRIANGLE C stands for coefficient as well as combination

7. CALCULATING Mentally

8. CALCULATING Mentally Formula

9. CALCULATING Mentally Formula because ...

10. THE BINOMIAL THEOREM (a + x)n = nC0an + nC1an-1x + nC2an-2x2 + nC3an-3x3 + nC4an-4x4 + ... + nCnxn = 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

11. PROPERTIES OF nCk 1. nC0 = nCn = 1 1st and last 2. nC1 = nCn-1 = n 2nd and 2nd-last 1 0C0 1 1 1C0 1C1 1 21 2C0 2C1 2C2 1 3 31 3C0 3C1 3C2 3C3 1 4 6 41 4C0 4C1 4C2 4C3 4C4 1 5 10 10 51 5C0 5C1 5C2 5C3 5C4 5C5 1 6 15 20 15 6 1 6C0 6C1 6C2 6C3 6C4 6C5 6C6 1 7 21 35 35 21 71 7C0 7C1 7C2 7C3 7C4 7C5 7C6 7C7 1 8 28 56 70 56 28 8 1 8C0 8C1 8C2 8C3 8C4 8C5 8C6 8C78C8

12. PROPERTIES OF nCk 3. nCk = nCn-k Symmetry 4. n+1Ck = nCk-1 + nCkPascal’s triangle result: each coefficient is the sum of the two coefficients in the row above it 1 0C0 1 1 1C0 1C1 1 21 2C0 2C1 2C2 1 3 31 3C0 3C1 3C2 3C3 1 4 6 41 4C0 4C1 4C2 4C3 4C4 1 5 10 10 51 5C0 5C1 5C2 5C3 5C4 5C5 1 6 15 20 15 6 1 6C0 6C1 6C2 6C3 6C4 6C5 6C6 1 7 21 35 35 21 71 7C0 7C1 7C2 7C3 7C4 7C5 7C6 7C7 1 8 28 56 70 56 28 8 1 8C0 8C1 8C2 8C3 8C4 8C5 8C6 8C78C8

13. n+1Ck = nCk-1 + nCkPascal’s triangle result 1 0C0 1 1 1C0 1C1 1 21 2C0 2C1 2C2 1 3 31 3C0 3C1 3C2 3C3 1 4 6 41 4C0 4C1 4C2 4C3 4C4 1 5 10 10 51 5C0 5C1 5C2 5C3 5C45C5 1 6 15 20 15 6 1 6C0 6C1 6C2 6C3 6C46C5 6C6 1 7 21 35 35 21 71 7C0 7C1 7C2 7C3 7C4 7C5 7C6 7C7 1 8 28 56 70 56 28 8 1 8C0 8C1 8C2 8C3 8C4 8C5 8C6 8C78C8 15 = 10 + 5 6C4 = 5C3 + 5C4

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

15. Example 1 (a) (a + 3)5 = 5C0a5 + 5C1a431 + 5C2a332 + 5C3a233 + 5C4a134 + 5C535 = a5 + 5a4.3 + 10a3.9 + 10a2.27 + 5a.81 + 243 = a5 + 15a4 + 90a3 + 270a2 + 405a + 243 (b) (2x – y)4 =

16. Example 1 (a) (a + 3)5 = 5C0 a5 + 5C1 a4 31 + 5C2 a3 32 + 5C3 a2 33 + 5C4 a1 34 + 5C5 35 = a5 + 5a4.3 + 10a3.9 + 10a2.27 + 5a.81 + 243 = a5 + 15a4 + 90a3 + 270a2 + 405a + 243 (b) (2x – y)4 = 4C0 (2x)4 + 4C1 (2x)3(-y)1 + 4C2 (2x)2(-y)2 + 4C3 (2x)1(-y)3 + 4C4 (-y)4 = 16x4 + 4(8x3)(-y) + 6(4x2)(y2) + 4(2x)(-y3) + y4 = 16x4 –32x3y+ 24x2y2 – 8xy3 + y4

17. Example 2 (2008 HSC, Question 1(d), 2 marks) Find an expression for the coefficient of x8y4 in the expansion of (2x + 3y)12.

18. Example 2 (2008 HSC, Question 1(d), 2 marks) (2x + 3y)12 = 12C0 (2x)12 + 12C1 (2x)11(3y)1 + 12C2 (2x)10(3y)2 + ... General term Tk = 12Ck(2x)12-k(3y)k = 1 mark For coefficient of x8y4, substitute k = ?

19. Example 2 (2008 HSC, Question 1(d), 2 marks) (2x + 3y)12 = 12C0 (2x)12 + 12C1 (2x)11(3y)1 + 12C2 (2x)10(3y)2 + ... General term Tk = 12Ck(2x)12-k(3y)k = 1 mark For coefficient of x8y4, substitute k = 4: T4 = 12C4 (2x)8(3y)4 = 12C4 2834 x8 y4 Coefficient is 12C4 2834 or 10 264 320.  Find an expression for the coefficient of x8y4 in the expansion of (2x + 3y)12. It’s OK to leave the coefficient unevaluated, especially if the question asks for ‘an expression’.

20. Tk is not the kth term Tk is the term that contains xk Simpler to write out the first few terms rather than memorise the  notation Better to avoid calling it ‘the kth term’: too confusing HSC questions ask for ‘the term that contains x8’ rather than ‘the 9th term’

21. Example 3 (2011 HSC, Question 2(c), 2 marks) Find an expression for the coefficient of x2 in the expansion of

22. Example 3 (2011 HSC, Question 2(c), 2 marks) General term Tk = 8Ck(3x)8-k = 8Ck38-k x8-k (-4)kx-k = 8Ck38-k x8-2k (-4)k For the coefficient of x2: 8 – 2k = 2 -2k = -6 k = 3 T3 = 8C3 38-3 x8-2×3(-4)3 = 8C3 35 (-4)3x2 = -870 192 x2 Coefficient is -870 192 or 8C3 35 (-4)3

23. FINDING THE GREATEST COEFFICIENT (1 + 2x)8 = 1 + 16x + 112x2 + 448x3 + 1120x4 + 1792x5 + 1792x6 + 1024x7 + 256x8

24. (1 + 2x)8 = 1 + 16x + 112x2 + 448x3 + 1120x4 + 1792x5 + 1792x6 + 1024x7 + 256x8 Example 4 Suppose (1 + 2x)8 = (a) Find an expression for tk, the coefficient of xk. (b) Show that (c) Show that the greatest coefficient is 1792.

25. Example 4 (a) (1 + 2x)8 = 8C0 18 + 8C1 17 (2x)1 + 8C2 16 (2x)2 + 8C3 15 (2x)3 +... General term Tk = 8Ck18-k (2x)k = 8Ck1 (2k) xk = 8Ck2kxk tk = 8Ck2k Leave out xk as we are only interested in the coefficient

26. Example 4 (a) (1 + 2x)8 = 8C0 18 + 8C1 17 (2x)1 + 8C2 16 (2x)2 + 8C3 15 (2x)3 +... General term Tk = 8Ck18-k (2x)k = 8Ck1 (2k) xk = 8Ck2kxk tk = 8Ck2k (b) Show that Leave out xk as we are only interested in the coefficient

27. Example 4 (a) (1 + 2x)8 = 8C0 18 + 8C1 17 (2x)1 + 8C2 16 (2x)2 + 8C3 15 (2x)3 +... General term Tk = 8Ck18-k (2x)k = 8Ck1 (2k) xk = 8Ck2kxk tk = 8Ck2k (b) Show that Leave out xk as we are only interested in the coefficient

28. Example 4 (b) (b) Show that Ratio of consecutive factorials

29. Example 4 (b) (b) Show that Ratio of consecutive factorials

30. Example 4 (c) Show that the greatest coefficient is 1792. For the greatest coefficient tk+1, we want: tk+1 > tk 16 – 2k > k + 1 -3k > -15 k < 5 k = 4 for the largest possible integer value of k k must be a whole number

31. Example 4 (c) Show that the greatest coefficient is 1792. Greatest coefficient tk+1 = t5 = 8C5 25 = 56  32 = 1792 tk = 8Ck 2k

32. THE BINOMIAL THEOREM FOR (1 + x)n (1 + x)n = nC0+ nC1x + nC2x2+ nC3x3 + nC4x4 + ... + nCnxn =

33. Example 5 (similar to 2010 HSC, Question 7(b)(i), 1 mark) Expand (1 + x)n and substitute an appropriate value of x to prove that

34. Example 5 (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCnxn Sub x = ? [Aiming to prove: ]

35. Example 5 (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCnxn Sub x = 1: (1 + 1)n = nC0 + nC1 (1) + nC2 (12) + ... + nCn(1n) 2n = nC0 + nC1 + nC2 + ... + nCn This will make the x’s disappear and make the LHS become 2n

36. Example 6 By considering that (1 + x)2n = (1 + x)n(1 + x)n and examining the coefficient of xn on each side, prove that

37. Example 6 (1 + x)2n = (1 + x)n(1 + x)n (1 + x)2n = 2nC0 + 2nC1 x + 2nC2 x2 + ... + 2nC2n x2n Term with xn = 2nCnxn Coefficient of xn = 2nCn [Aiming to prove ]

38. Example 6 (1 + x)2n = (1 + x)n(1 + x)n (1 + x)2n = 2nC0 + 2nC1 x + 2nC2 x2 + ... + 2nC2n x2n Term with xn = 2nCnxn Coefficient of xn = 2nCn (1 + x)n.(1 + x)n = (nC0 + nC1 x + nC2 x2 + ... + nCnxn)  (nC0 + nC1 x + nC2 x2 + ... + nCnxn) If we expanded the RHS, there would be many terms [Aiming to prove ]

39. Example 6 (1 + x)2n = (1 + x)n(1 + x)n (1 + x)n.(1 + x)n = (nC0 + nC1 x + nC2 x2 + ... + nCnxn)  (nC0 + nC1 x + nC2 x2 + ... +nCnxn) Terms with xn = nC0(nCnxn) + nC1 x (nCn-1xn-1) + nC2 x2 (nCn-2xn-2) + ... + nCnxn(nC0) Coefficient of xn = nC0 nCn+ nC1 nCn-1+ nC2 nCn-2+ ... + nCnnC0 = (nC0)2 + (nC1)2 + (nC2)2 + ... + (nCn)2 by symmetry of Pascal’s triangle [Aiming to prove ] nCk= nCn-k

40. Example 6 (1 + x)2n = (1 + x)n(1 + x)n  By equating coefficients of xn on both sides of (1 + x)2n = (1 + x)n.(1 + x)n 2nCn = (nC0)2 + (nC1)2 + (nC2)2 + ... + (nCn)2

41. Example 7 (2006 HSC, Question 2(b), 2 marks) (i) By applying the binomial theorem to (1 + x)n and differentiating, show that (ii) Hence deduce that

42. Example 7 (2006 HSC, Question 2(b), 2 marks) [Need to prove ] (i) (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCnxn Differentiating both sides:

43. Example 7 (2006 HSC, Question 2(b), 2 marks) [Need to prove ] (i) (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCnxn Differentiating both sides: n(1 + x)n-1 = 0 + nC1 + 2 nC2 x + ... + rnCrxr-1 + ... + nnCnxn-1 = nC1 + 2 nC2 x + ... + rnCrxr-1 + ... + nnCnxn-1 The general term

44. Example 7 (2006 HSC, Question 2(b), 2 marks) [Need to prove ] (i) (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCnxn Differentiating both sides: n(1 + x)n-1 = 0 + nC1 + 2 nC2 x + ... + rnCrxr-1 + ... + nnCnxn-1 = nC1 + 2 nC2 x + ... + rnCrxr-1 + ... + nnCnxn-1 (ii) Substitute x = ? to prove

45. Example 7 (2006 HSC, Question 2(b), 2 marks) (i) (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCnxn Differentiating both sides: n(1 + x)n-1 = 0 + nC1 + 2 nC2 x + ... + rnCrxr-1 + ... + nnCnxn-1 = nC1 + 2 nC2 x + ... + rnCrxr-1 + ... + nnCnxn-1 (ii) Substitute x = 2 to prove n(1 + 2)n-1 = nC1 + 2 nC2 2 + ... + rnCr2r-1 + ... + nnCn2n-1 n 3n-1 = nC1 + 4 nC2 + ... + rnCr2r-1 + ... + nnCn2n-1

46. Example 8 (2008 HSC, Question 6(c), 5 marks) Let p and q be positive integers with p ≤ q. (i) Use the binomial theorem to expand (1 + x)p+q , and hence write down the term of which is independent of x. (ii) Given that apply the binomial theorem and the result of part (i) to find a simpler expression for

47. Example 8 (2008 HSC, Question 6(c), 5 marks) (i) (1 + x)p+q = p+qC0 + p+qC1 x + p+qC2 x2 + ... + p+qCp+qxp+q  The term of independent of x will be

48. Example 8 (2008 HSC, Question 6(c), 5 marks) (i) (1 + x)p+q = p+qC0 + p+qC1 x + p+qC2 x2 + ... + p+qCp+qxp+q  The term of independent of x will be 

49. Example 8 (2008 HSC, Question 6(c), 5 marks) (i) (1 + x)p+q = p+qC0 + p+qC1 x + p+qC2 x2 + ... + p+qCp+qxp+q  The term of independent of x will be  (ii) Given that apply the binomial theorem and the result of part (i) to find a simpler expression for

50. Example 8 (2008 HSC, Question 6(c), 5 marks) (ii) (1 + x)p = pC0 + pC1 x + pC2 x2 + ... + pCpxp 