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THE BINOMIAL THEOREM Robert Yen Hurlstone Agricultural High School

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

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THE BINOMIAL THEOREM Robert Yen Hurlstone Agricultural High School

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  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 + 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

  4. 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

  5. nCk, A FORMULA FOR PASCAL’S TRIANGLE

  6. CALCULATING Mentally

  7. CALCULATING Mentally Formula

  8. CALCULATING Mentally Formula because ...

  9. 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

  10. PROPERTIES OF nCk 1. nC0 = nCn = 1 1st and last 2. nC1 = nCn-1 = n 2nd and 2nd-last 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

  11. 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

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

  13. 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 =

  14. 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

  15. 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 = ?

  16. 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 (28)(34)x8y4 Coefficient is 12C4 (28)(34) or 10 264 320.  It’s OK to leave the coefficient unevaluated, especially if the question asks for ‘an expression’.

  17. 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 referring to ‘the kth term’: too confusing HSC questions ask for ‘the term that contains x8’ rather than ‘the 9th term’

  18. Example 3 (2005 HSC, Question 2(b), 3 marks) General term Tk = 12Ck(2x)12-k = 12Ck212-k x12-k (-x-2)k = 12Ck212-k x12-k (-1)k x-2k = 12Ck212-k x12-3k (-1)k For term independent of x:

  19. Example 3 (2005 HSC, Question 2(b), 3 marks) General term Tk = 12Ck(2x)12-k = 12Ck212-k x12-k (-x-2)k = 12Ck212-k x12-k (-1)k x-2k = 12Ck212-k x12-3k (-1)k For term independent of x: 12 – 3k = 0 k = 4  T4 = 12C4 28x0 (-1)4 = 126 720 

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

  21. 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

  22. 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) Leave out xk 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 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

  26. Example 4 (c) Greatest coefficient tk+1 = t5 = 8C5 25 = 56  32 = 1792 tk = 8Ck 2k

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

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

  29. Example 5 (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn 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

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

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

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

  33. Example 6  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

  34. Example 7 (2006 HSC, Question 2(b), 2 marks) (i) (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn Differentiating both sides: n(1 + x)n-1 = 0 + nC1 + 2 nC2 x + ... + rnCr xr-1 + ... + nnCn xn-1 = nC1 + 2 nC2 x + ... + rnCr xr-1 + ... + nnCn xn-1 (ii) Substitute x = ? to prove result: [Aiming to prove: n 3n-1 = nC1 + ... + rnCr 2r-1 + ... + nnCn 2n-1] The general term

  35. Example 7 (2006 HSC, Question 2(b), 2 marks) (i) (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn Differentiating both sides: n(1 + x)n-1 = 0 + nC1 + 2 nC2 x + ... + rnCr xr-1 + ... + nnCn xn-1 = nC1 + 2 nC2 x + ... + rnCr xr-1 + ... + nnCn xn-1 (ii) Substitute x = 2 to prove result: n(1 + 2)n-1 = nC1 + 2 nC2 2 + ... + rnCr 2r-1 + ... + nnCn 2n-1 n 3n-1 = nC1 + 4 nC2 + ... + rnCr 2r-1 + ... + nnCn 2n-1

  36. Example 8 (2008 HSC, Question 6(c), 5 marks) (i) (1 + x)p+q = p+qC0 + p+qC1 x + p+qC2 x2 + ... + p+qCp+q xp+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+qC0 + p+qC1 x + p+qC2 x2 + ... + p+qCp+q xp+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+qC0 + p+qC1 x + p+qC2 x2 + ... + p+qCp+q xp+q  The term of independent of x will be  (ii) (1 + x)p = pC0 + pC1 x + pC2 x2 + ... + pCp xp 

  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 = pC0qC0 + pC1 xqC1 + pC2 x2qC2 + ... + pCp xp qCp = 1 + pC1qC1 + pC2 qC2 + ... + pCp qCp  p ≤ q

  41. Example 8 (2008 HSC, Question 6(c), 5 marks) (ii) Terms independent of x = pC0qC0 + pC1 xqC1 + pC2 x2qC2 + ... + pCp xp qCp = 1 + pC1qC1 + pC2 qC2 + ... + pCp qCp   By equating the terms independent of x on both sides of p+qCq = 1 + pC1qC1 + pC2 qC2 + ... + pCp qCp 1 + pC1qC1 + pC2 qC2 + ... + pCp qCp = p+qCq 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 = nC0 + nC1 x + nC2 x2 + ... + nCn xn = c0 + c1 x + c2 x2 + ... + cn xn [1] Identity to be proved involves (n + 2) 2n-1so try ... [Aiming to prove: c0 + 2c1 + 3c2 + ... + (n + 1)cn = (n + 2) 2n-1] To save time, this question asks us to abbreviate nCkto ck

  44. Example 9 (2002 HSC, Question 7(b), 6 marks) (i) (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn = c0 + c1 x + c2 x2 + ... + cn xn [1] Identity to be proved involves (n + 2) 2n-1 so try ... Differentiating both sides: n(1 + x)n-1 = c1 + 2c2 x + ... + ncn xn-1 Substituting x = ? to give 2n-1 on the LHS: [Aiming to prove: c0 + 2c1 + 3c2 + ... + (n + 1)cn = (n + 2) 2n-1]

  45. Example 9 (2002 HSC, Question 7(b), 6 marks) (i) (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn = c0 + c1 x + c2 x2 + ... + cn xn [1] Identity to be proved involves (n + 2) 2n-1 so try ... Differentiating both sides: n(1 + x)n-1 = c1 + 2c2 x + ... + ncn xn-1 Substituting x = 1 to give 2n-1 on the LHS: n(1 + 1)n-1 = c1 + 2c2 1 + ... + ncn 1n-1 n 2n-1 = c1 + 2c2 + ... + ncn [2] How do we make the LHS say (n + 2) 2n-1? [Aiming to prove: c0 + 2c1 + 3c2 + ... + (n + 1)cn = (n + 2) 2n-1]

  46. Example 9 (2002 HSC, Question 7(b), 6 marks) (i) (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn = c0 + c1 x + c2 x2 + ... + cn xn [1] Differentiating both sides: n(1 + x)n-1 = c1 + 2c2 x + ... + ncn xn-1 Substituting x = 1: n(1 + 1)n-1 = c1 + 2c2 1 + ... + ncn 1n-1 n 2n-1 = c1 + 2c2 + ... + ncn [2] Add 2 (2n-1) to both sides. But that’s 2n. [Aiming to prove: c0 + 2c1 + 3c2 + ... + (n + 1)cn = (n + 2) 2n-1]

  47. Example 9 (2002 HSC, Question 7(b), 6 marks) (i) (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn = c0 + c1 x + c2 x2 + ... + cn xn [1] Differentiating both sides: n(1 + x)n-1 = c1 + 2c2 x + ... + ncn xn-1 Substituting x = 1: n(1 + 1)n-1 = c1 + 2c2 1 + ... + ncn 1n-1 n 2n-1 = c1 + 2c2 + ... + ncn [2] To prove the 2nidentity, sub x = 1into [1]above: (1 + 1)n = c0 + c1 1 + c2 12 + ... + cn 1n 2n = c0 + c1 + c2 + ... + cn [3]  [Aiming to prove: c0 + 2c1 + 3c2 + ... + (n + 1)cn = (n + 2) 2n-1]

  48. Example 9 (2002 HSC, Question 7(b), 6 marks) (i)n 2n-1 = c1 + 2c2 + ... + ncn [2] 2n = c0 + c1 + c2 + ... + cn [3] [2] + [3]: n 2n-1+ 2n = c1 + 2c2 + ... + ncn+ c0 + c1 + c2 + ... + cn 2n-1 (n + 2) = c0 + 2c1 + 3c2 + ... + (n + 1)cn c0 + 2c1 + 3c2 + ... + (n + 1)cn = (n + 2) 2n-1 [Aiming to prove: c0 + 2c1 + 3c2 + ... + (n + 1)cn = (n + 2) 2n-1] Each coefficient ckincreases 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)nfrom (i) twice and substituting x = -1. (1 + x)n = c0 + c1 x + c2 x2 + ... + cn xn [1] [Aiming to find ]

  50. Example 9 (2002 HSC, Question 7(b), 6 marks) (ii)(1 + x)n = c0 + c1 x + c2 x2 + ... + cn xn [1] Integrate both sides: To find k, sub x = 0: [Aiming to find ] Don’t forget the constant of integration 

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