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# 12 x1 t08 03 binomial theorem (12)

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### 12 x1 t08 03 binomial theorem (12)

1. 1. Calculating Coefficientswithout Pascal’s Triangle
2. 2. Calculating Coefficientswithout Pascal’s Triangle n! n Ck  k!n  k !
3. 3. Calculating Coefficients without Pascal’s Triangle n! n Ck  k!n  k !Proof
4. 4. Calculating Coefficients without Pascal’s Triangle n! n Ck  k!n  k !ProofStep 1: Prove true for k = 0
5. 5. Calculating Coefficients without Pascal’s Triangle n! n Ck  k!n  k !ProofStep 1: Prove true for k = 0 LHS  nC0 1
6. 6. Calculating Coefficients without Pascal’s Triangle n! n Ck  k!n  k !ProofStep 1: Prove true for k = 0 n! LHS  nC0 RHS  0!n! 1 n!  n! 1
7. 7. Calculating Coefficients without Pascal’s Triangle n! n Ck  k!n  k !ProofStep 1: Prove true for k = 0 n! LHS  nC0 RHS  0!n! 1 n!  n!  LHS  RHS 1
8. 8. Calculating Coefficients without Pascal’s Triangle n! n Ck  k!n  k !ProofStep 1: Prove true for k = 0 n! LHS  nC0 RHS  0!n! 1 n!  n!  LHS  RHS 1 Hence the result is true for k = 0
9. 9. Step 2 : Assume the result is true for k  r where r is an integer  0
10. 10. Step 2 : Assume the result is true for k  r where r is an integer  0 n! i.e.n Cr  r!n  r !
11. 11. Step 2 : Assume the result is true for k  r where r is an integer  0 n! i.e.n Cr  r!n  r !Step 3 : Prove the result is true for k  r  1
12. 12. Step 2 : Assume the result is true for k  r where r is an integer  0 n! i.e.n Cr  r!n  r !Step 3 : Prove the result is true for k  r  1 n! i.e.n Cr 1  r  1!n  r  1!
13. 13. Step 2 : Assume the result is true for k  r where r is an integer  0 n! i.e.n Cr  r!n  r !Step 3 : Prove the result is true for k  r  1 n! i.e.n Cr 1  r  1!n  r  1!Proof
14. 14. Step 2 : Assume the result is true for k  r where r is an integer  0 n! i.e.n Cr  r!n  r !Step 3 : Prove the result is true for k  r  1 n! i.e.n Cr 1  r  1!n  r  1!Proof n 1 Cr 1  nCr  nCr 1
15. 15. Step 2 : Assume the result is true for k  r where r is an integer  0 n! i.e.n Cr  r!n  r !Step 3 : Prove the result is true for k  r  1 n! i.e.n Cr 1  r  1!n  r  1!Proof n 1 Cr 1  nCr  nCr 1 n Cr 1  n1Cr 1  nCr
16. 16. Step 2 : Assume the result is true for k  r where r is an integer  0 n! i.e.n Cr  r!n  r !Step 3 : Prove the result is true for k  r  1 n! i.e.n Cr 1  r  1!n  r  1!Proof n 1 Cr 1  nCr  nCr 1 n Cr 1  n1Cr 1  nCr n  1! n!   r  1!n  r ! r!n  r!
17. 17. Step 2 : Assume the result is true for k  r where r is an integer  0 n! i.e.n Cr  r!n  r !Step 3 : Prove the result is true for k  r  1 n! i.e.n Cr 1  r  1!n  r  1!Proof n 1 Cr 1  nCr  nCr 1 n Cr 1  n1Cr 1  nCr n  1! n!   r  1!n  r ! r!n  r! n  1!n!r  1  r  1!n  r !
18. 18. Step 2 : Assume the result is true for k  r where r is an integer  0 n! i.e.n Cr  r!n  r !Step 3 : Prove the result is true for k  r  1 n! i.e.n Cr 1  r  1!n  r  1!Proof n 1 Cr 1  nCr  nCr 1 n Cr 1  n1Cr 1  nCr n  1! n!   r  1!n  r ! r!n  r! n  1!n!r  1  r  1!n  r ! n!n  1  r  1  r  1!n  r !
19. 19. n!n  1  r  1 r  1!n  r !
20. 20. n!n  1  r  1 r  1!n  r ! n!n  r  r  1!n  r !
21. 21. n!n  1  r  1 r  1!n  r ! n!n  r  r  1!n  r ! n! r  1!n  r  1!
22. 22. n!n  1  r  1 r  1!n  r ! n!n  r  r  1!n  r ! n! r  1!n  r  1!Hence the result is true for k = r + 1 if it is also true for k = r
23. 23. n!n  1  r  1  r  1!n  r ! n!n  r   r  1!n  r ! n!  r  1!n  r  1! Hence the result is true for k = r + 1 if it is also true for k = rStep 4: Hence the result is true for all positive integral values of n by induction
24. 24. The Binomial Theorem
25. 25. The Binomial Theorem 1  x n  nC0  nC1 x  nC2 x 2   nCk x k   nCn x n
26. 26. The Binomial Theorem 1  x n  nC0  nC1 x  nC2 x 2   nCk x k   nCn x n n   nCk x k k 0
27. 27. The Binomial Theorem 1  x n  nC0  nC1 x  nC2 x 2   nCk x k   nCn x n n   nCk x k k 0 n! where Ck  n and n is a positive integer k!n  k !
28. 28. The Binomial Theorem 1  x n  nC0  nC1 x  nC2 x 2   nCk x k   nCn x n n   nCk x k k 0 n! where Ck  n and n is a positive integer k!n  k !NOTE: there are (n + 1) terms
29. 29. The Binomial Theorem 1  x n  nC0  nC1 x  nC2 x 2   nCk x k   nCn x n n   nCk x k k 0 n! where Ck  n and n is a positive integer k!n  k ! NOTE: there are (n + 1) termsThis extends to; n a  b    n C k a n  k b k n k 0
30. 30. The Binomial Theorem 1  x n  nC0  nC1 x  nC2 x 2   nCk x k   nCn x n n   nCk x k k 0 n! where Ck  n and n is a positive integer k!n  k ! NOTE: there are (n + 1) termsThis extends to; n a  b    n C k a n  k b k n k 0e.g . Evaluate 11C4
31. 31. The Binomial Theorem 1  x n  nC0  nC1 x  nC2 x 2   nCk x k   nCn x n n   nCk x k k 0 n! where Ck  n and n is a positive integer k!n  k ! NOTE: there are (n + 1) termsThis extends to; n a  b    n C k a n  k b k n k 0 11! C4  11 11e.g . Evaluate C4 4!7!
32. 32. The Binomial Theorem 1  x n  nC0  nC1 x  nC2 x 2   nCk x k   nCn x n n   nCk x k k 0 n! where Ck  n and n is a positive integer k!n  k ! NOTE: there are (n + 1) termsThis extends to; n a  b    n C k a n  k b k n k 0 11! C4  11 11e.g . Evaluate C4 4!7! 1110  9  8  4  3  2 1
33. 33. The Binomial Theorem 1  x n  nC0  nC1 x  nC2 x 2   nCk x k   nCn x n n   nCk x k k 0 n! where Ck  n and n is a positive integer k!n  k ! NOTE: there are (n + 1) termsThis extends to; n a  b    n C k a n  k b k n k 0 11! C4  11 11e.g . Evaluate C4 4!7! 1110  9  8  4  3  2 1
34. 34. The Binomial Theorem 1  x n  nC0  nC1 x  nC2 x 2   nCk x k   nCn x n n   nCk x k k 0 n! where Ck  n and n is a positive integer k!n  k ! NOTE: there are (n + 1) termsThis extends to; n a  b    n C k a n  k b k n k 0 11! C4  11 11e.g . Evaluate C4 4!7! 3 1110  9  8  4  3  2 1
35. 35. The Binomial Theorem 1  x n  nC0  nC1 x  nC2 x 2   nCk x k   nCn x n n   nCk x k k 0 n! where Ck  n and n is a positive integer k!n  k ! NOTE: there are (n + 1) termsThis extends to; n a  b    n C k a n  k b k n k 0 11! C4  11 11e.g . Evaluate C4 4!7! 3 1110  9  8  4  3  2 1  330
36. 36. (ii) Find the value of n so that;
37. 37. (ii) Find the value of n so that; a) nC5  nC8
38. 38. (ii) Find the value of n so that; a) nC5  nC8 n Ck  nCnk
39. 39. (ii) Find the value of n so that; a) nC5  nC8 n Ck  nCnk 8  n  5 n  13
40. 40. (ii) Find the value of n so that; a) nC5  nC8 b) nC7  nC8  20C8 n Ck  nCnk 8  n  5 n  13
41. 41. (ii) Find the value of n so that; a) nC5  nC8 b) nC7  nC8  20C8 n 1 n Ck  Cnk n Ck  n1Ck 1  nCk 8  n  5 n  13
42. 42. (ii) Find the value of n so that; a) nC5  nC8 b) nC7  nC8  20C8 n 1 n Ck  Cnk n Ck  n1Ck 1  nCk 8  n  5 n  19 n  13
43. 43. (ii) Find the value of n so that; a) nC5  nC8 b) nC7  nC8  20C8 n 1 n Ck  Cnk n Ck  n1Ck 1  nCk 8  n  5 n  19 n  13 11  5a  3 iii  Find the 5th term in the expansion of    b
44. 44. (ii) Find the value of n so that; a) nC5  nC8 b) nC7  nC8  20C8 n 1 n Ck  Cnk n Ck  n1Ck 1  nCk 8  n  5 n  19 n  13 11  5a  3 iii  Find the 5th term in the expansion of    b k 11 k  3  Tk 1 11Ck 5a      b
45. 45. (ii) Find the value of n so that; a) nC5  nC8 b) nC7  nC8  20C8 n 1 n Ck  Cnk n Ck  n1Ck 1  nCk 8  n  5 n  19 n  13 11  5a  3 iii  Find the 5th term in the expansion of    b k 11 k  3  Tk 1 11Ck 5a      b 4 11  3 T5  C4 5a    7  b
46. 46. (ii) Find the value of n so that; a) nC5  nC8 b) nC7  nC8  20C8 n 1 n Ck  Cnk n Ck  n1Ck 1  nCk 8  n  5 n  19 n  13 11  5a  3 iii  Find the 5th term in the expansion of    b k 11 k  3  Tk 1 11Ck 5a      b 4 11  3 T5  C4 5a    7  b 11 C 4 5 7 34 a 7  b4
47. 47. (ii) Find the value of n so that; a) nC5  nC8 b) nC7  nC8  20C8 n 1 n Ck  Cnk n Ck  n1Ck 1  nCk 8  n  5 n  19 n  13 11  5a  3 iii  Find the 5th term in the expansion of    b k 11 k  3  Tk 1 11Ck 5a      b 4 11  3 T5  C4 5a    7  b 11 C 4 5 7 34 a 7  unsimplified b4
48. 48. (ii) Find the value of n so that; a) nC5  nC8 b) nC7  nC8  20C8 n 1 n Ck  Cnk n Ck  n1Ck 1  nCk 8  n  5 n  19 n  13 11  5a  3 iii  Find the 5th term in the expansion of    b k 11 k  3  Tk 1 11Ck 5a      b 4 11  3 T5  C4 5a    7 330  78125  81a 7  b  b4 11 C 4 5 7 34 a 7  unsimplified b4
49. 49. (ii) Find the value of n so that; a) nC5  nC8 b) nC7  nC8  20C8 n 1 n Ck  Cnk n Ck  n1Ck 1  nCk 8  n  5 n  19 n  13 11  5a  3 iii  Find the 5th term in the expansion of    b k 11 k  3  Tk 1 11Ck 5a      b 4 11  3 T5  C4 5a    7 330  78125  81a 7  b  b4 11 C 4 5 7 34 a 7 2088281250a 7  unsimplified  b4 b4
50. 50. 9  x2  1 iv  Obtain the term independent of x in  3   2x 
51. 51. 9iv  Obtain the term independent of x in  3 x2  1   k  2x  Tk 1  Ck 3 x     2 9 k  9 1   2x 
52. 52. 9iv  Obtain the term independent of x in  3 x2  1   k  2x  Tk 1  Ck 3 x     2 9 k  9 1   2x  term independent of x means term with x 0
53. 53. 9iv  Obtain the term independent of x in  3 x2  1   k  2x  Tk 1  Ck 3 x     2 9 k  9 1   2x  term independent of x means term with x 0 x  x  2 9 k 1 k  x0
54. 54. 9iv  Obtain the term independent of x in  3 x2  1   k  2x  Tk 1  Ck 3 x     2 9 k  9 1   2x  term independent of x means term with x 0 x  x  2 9 k 1 k  x0 x182 k  x k  x 0 18  3k  0 k 6
55. 55. 9iv  Obtain the term independent of x in  3 x2  1   k  2x  Tk 1  Ck 3 x     2 9 k  9 1   2x  term independent of x means term with x 0 x  x  2 9 k 1 k  x0 x182 k  x k  x 0 18  3k  0 6 k 6T7  C6 3 x 9  2 3  1     2x 
56. 56. 9iv  Obtain the term independent of x in  3 x2  1   k  2x  Tk 1  Ck 3 x     2 9 k  9 1   2x  term independent of x means term with x 0 x  x  2 9 k 1 k  x0 x182 k  x k  x 0 18  3k  0 6 k 6T7  C6 3 x 9  2 3  1     2x  9 C6 33  6 2
57. 57. 9iv  Obtain the term independent of x in  3 x2  1   k  2x  Tk 1  Ck 3 x     2 9 k  9 1   2x  term independent of x means term with x 0 x  x  2 9 k 1 k  x0 x182 k  x k  x 0 18  3k  0 6 k 6T7  C6 3 x 9  2 3  1     2x  9 C6 33  6 2 Exercise 5D; 2, 3, 5, 7, 9, 10ac, 12ac, 13, 14, 15ac, 19, 25