Calculating Coefficientswithout Pascal’s Triangle
Calculating Coefficientswithout Pascal’s Triangle                      n!         n             Ck                   k!n...
Calculating Coefficients without Pascal’s Triangle                       n!          n              Ck                   ...
Calculating Coefficients  without Pascal’s Triangle                                            n!                         ...
Calculating Coefficients  without Pascal’s Triangle                                            n!                         ...
Calculating Coefficients  without Pascal’s Triangle                                            n!                         ...
Calculating Coefficients  without Pascal’s Triangle                                            n!                         ...
Calculating Coefficients  without Pascal’s Triangle                                            n!                         ...
Step 2 : Assume the result is true for k  r where r is an integer  0
Step 2 : Assume the result is true for k  r where r is an integer  0                                     n!             ...
Step 2 : Assume the result is true for k  r where r is an integer  0                                     n!             ...
Step 2 : Assume the result is true for k  r where r is an integer  0                                     n!             ...
Step 2 : Assume the result is true for k  r where r is an integer  0                                     n!             ...
Step 2 : Assume the result is true for k  r where r is an integer  0                                     n!             ...
Step 2 : Assume the result is true for k  r where r is an integer  0                                     n!             ...
Step 2 : Assume the result is true for k  r where r is an integer  0                                     n!             ...
Step 2 : Assume the result is true for k  r where r is an integer  0                                     n!             ...
Step 2 : Assume the result is true for k  r where r is an integer  0                                     n!             ...
n!n  1  r  1    r  1!n  r !
n!n  1  r  1  r  1!n  r !     n!n  r   r  1!n  r !
n!n  1  r  1  r  1!n  r !     n!n  r   r  1!n  r !             n!    r  1!n  r  1!
n!n  1  r  1  r  1!n  r !     n!n  r   r  1!n  r !             n!    r  1!n  r  1!Hence the ...
n!n  1  r  1         r  1!n  r !         n!n  r           r  1!n  r !                 n!           ...
The Binomial Theorem
The Binomial Theorem  1  x n  nC0  nC1 x  nC2 x 2   nCk x k   nCn x n
The Binomial Theorem  1  x n  nC0  nC1 x  nC2 x 2   nCk x k   nCn x n             n            nCk x k      ...
The Binomial Theorem  1  x n  nC0  nC1 x  nC2 x 2   nCk x k   nCn x n              n            nCk x k     ...
The Binomial Theorem    1  x n  nC0  nC1 x  nC2 x 2   nCk x k   nCn x n                n              nCk x ...
The Binomial Theorem          1  x n  nC0  nC1 x  nC2 x 2   nCk x k   nCn x n                        n        ...
The Binomial Theorem           1  x n  nC0  nC1 x  nC2 x 2   nCk x k   nCn x n                        n       ...
The Binomial Theorem          1  x n  nC0  nC1 x  nC2 x 2   nCk x k   nCn x n                        n        ...
The Binomial Theorem          1  x n  nC0  nC1 x  nC2 x 2   nCk x k   nCn x n                        n        ...
The Binomial Theorem          1  x n  nC0  nC1 x  nC2 x 2   nCk x k   nCn x n                        n        ...
The Binomial Theorem          1  x n  nC0  nC1 x  nC2 x 2   nCk x k   nCn x n                        n        ...
The Binomial Theorem          1  x n  nC0  nC1 x  nC2 x 2   nCk x k   nCn x n                        n        ...
(ii) Find the value of n so that;
(ii) Find the value of n so that; a) nC5  nC8
(ii) Find the value of n so that; a) nC5  nC8     n         Ck  nCnk
(ii) Find the value of n so that; a) nC5  nC8     n         Ck  nCnk     8  n  5      n  13
(ii) Find the value of n so that; a) nC5  nC8                       b) nC7  nC8  20C8     n         Ck  nCnk     8 ...
(ii) Find the value of n so that; a) nC5  nC8                       b) nC7  nC8  20C8                                  ...
(ii) Find the value of n so that; a) nC5  nC8                       b) nC7  nC8  20C8                                  ...
(ii) Find the value of n so that; a) nC5  nC8                            b) nC7  nC8  20C8                             ...
(ii) Find the value of n so that; a) nC5  nC8                            b) nC7  nC8  20C8                             ...
(ii) Find the value of n so that; a) nC5  nC8                            b) nC7  nC8  20C8                             ...
(ii) Find the value of n so that; a) nC5  nC8                            b) nC7  nC8  20C8                             ...
(ii) Find the value of n so that; a) nC5  nC8                            b) nC7  nC8  20C8                             ...
(ii) Find the value of n so that; a) nC5  nC8                            b) nC7  nC8  20C8                             ...
(ii) Find the value of n so that; a) nC5  nC8                            b) nC7  nC8  20C8                             ...
9                                           x2  1 iv  Obtain the term independent of x in  3                       ...
9iv  Obtain the term independent of x in  3 x2  1                                                                  ...
9iv  Obtain the term independent of x in  3 x2  1                                                                  ...
9iv  Obtain the term independent of x in  3 x2  1                                                                  ...
9iv  Obtain the term independent of x in  3 x2  1                                                                  ...
9iv  Obtain the term independent of x in  3 x2  1                                                                  ...
9iv  Obtain the term independent of x in  3 x2  1                                                                  ...
9iv  Obtain the term independent of x in  3 x2  1                                                                  ...
Upcoming SlideShare
Loading in …5
×

12 x1 t08 03 binomial theorem (12)

1,186 views

Published on

Published in: Education, Technology
0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total views
1,186
On SlideShare
0
From Embeds
0
Number of Embeds
537
Actions
Shares
0
Downloads
24
Comments
0
Likes
0
Embeds 0
No embeds

No notes for slide

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

×