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 n1Cr 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 n1Cr 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 n1Cr 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 n1Cr 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 !
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
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!
1110 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!
1110 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
1110 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
1110 9 8
4 3 2 1
330
38. (ii) Find the value of n so that;
a) nC5 nC8
n
Ck nCnk
39. (ii) Find the value of n so that;
a) nC5 nC8
n
Ck nCnk
8 n 5
n 13
40. (ii) Find the value of n so that;
a) nC5 nC8 b) nC7 nC8 20C8
n
Ck nCnk
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 Cnk
n
Ck n1Ck 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 Cnk
n
Ck n1Ck 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 Cnk
n
Ck n1Ck 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 Cnk
n
Ck n1Ck 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 Cnk
n
Ck n1Ck 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 Cnk
n
Ck n1Ck 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 Cnk
n
Ck n1Ck 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 Cnk
n
Ck n1Ck 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 Cnk
n
Ck n1Ck 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
x182 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
x182 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
x182 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
x182 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