1. Calculating Coefficients
without Pascal’s Triangle

2. Calculating Coefficients
without Pascal’s Triangle
n!
n
Ck 
k!n  k !

3. Calculating Coefficients
without Pascal’s Triangle
n!
n
Ck 
k!n  k !
Proof

4. Calculating Coefficients
without Pascal’s Triangle
n!
n
Ck 
k!n  k !
Proof
Step 1: Prove true for k = 0

5. Calculating Coefficients
without Pascal’s Triangle
n!
n
Ck 
k!n  k !
Proof
Step 1: Prove true for k = 0
LHS  nC0
1

6. Calculating Coefficients
without Pascal’s Triangle
n!
n
Ck 
k!n  k !
Proof
Step 1: Prove true for k = 0
n!
LHS  nC0 RHS 
0!n!
1
n!

n!
1

7. Calculating Coefficients
without Pascal’s Triangle
n!
n
Ck 
k!n  k !
Proof
Step 1: Prove true for k = 0
n!
LHS  nC0 RHS 
0!n!
1
n!

n!
 LHS  RHS 1

8. Calculating Coefficients
without Pascal’s Triangle
n!
n
Ck 
k!n  k !
Proof
Step 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. Step 2 : Assume the result is true for k  r where r is an integer  0

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. 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. 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. 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. 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. 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. 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. 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. 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. n!n  1  r  1

r  1!n  r !

20. n!n  1  r  1

r  1!n  r !
n!n  r 

r  1!n  r !

21. n!n  1  r  1

r  1!n  r !
n!n  r 

r  1!n  r !
n!

r  1!n  r  1!

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. 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
Step 4: Hence the result is true for all positive integral values of n
by induction

24. The Binomial Theorem

25. The Binomial Theorem
1  x n  nC0  nC1 x  nC2 x 2   nCk x k   nCn x n

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. 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. 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. 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
This extends to; n
a  b    n C k a n  k b k
n
k 0

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) terms
This extends to; n
a  b    n C k a n  k b k
n
k 0
e.g . Evaluate 11C4

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) terms
This extends to; n
a  b    n C k a n  k b k
n
k 0
11!
C4 
11 11
e.g . Evaluate C4
4!7!

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) terms
This extends to; n
a  b    n C k a n  k b k
n
k 0
11!
C4 
11 11
e.g . Evaluate C4
4!7!
1110  9  8

4  3  2 1

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) terms
This extends to; n
a  b    n C k a n  k b k
n
k 0
11!
C4 
11 11
e.g . Evaluate C4
4!7!
1110  9  8

4  3  2 1

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) terms
This extends to; n
a  b    n C k a n  k b k
n
k 0
11!
C4 
11 11
e.g . Evaluate C4
4!7! 3
1110  9  8

4  3  2 1

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) terms
This extends to; n
a  b    n C k a n  k b k
n
k 0
11!
C4 
11 11
e.g . Evaluate C4
4!7! 3
1110  9  8

4  3  2 1
 330

36. (ii) Find the value of n so that;

37. (ii) Find the value of n so that;
a) nC5  nC8

38. (ii) Find the value of n so that;
a) nC5  nC8
n
Ck  nCnk

39. (ii) Find the value of n so that;
a) nC5  nC8
n
Ck  nCnk
8  n  5
n  13

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. (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. (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. (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. (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. (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. (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. (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. (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. (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. 9
 x2  1 
iv  Obtain the term independent of x in  3 
 2x 

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. 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. 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. 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. 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 6
T7  C6 3 x
9

2 3  1 
 
 2x 

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 6
T7  C6 3 x
9

2 3  1 
 
 2x 
9
C6 33
 6
2

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 6
T7  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