THE BINOMIAL THEOREM
Robert Yen
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
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
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
Aim to understand the theory TODAY, not later

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
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INTRODUCTION

Binomial theorem = expanding (a + x)n

Difficult topic at end of Ext 1 course: high-level algebra

Targeted at better Extension 1 students aiming for Band E4

Master this topic to get ahead in the exam

No shortcuts for this topic: must learn the theory
BINOMIAL EXPANSIONS AND PASCAL’S TRIANGLE
Coefficients
(a + x)1 = a + x
11
(a + x)2 = a2 + 2ax + x2
(a + x)3 = a3 + 3a2x + 3ax2 + x3
(a + x)4 = a4 + 4a3x + 6a2x2 + 4ax3 + x4
121
1331
14641
(a + x)5 = a5 + 5a4x + 10a3x2 + 10a2x3 + 5ax4 + x5 1 5 10 10 5 1
 For (a + x)n, the powers of a are ↓ and the powers of x are ↑
 The sum of the powers in each term is always n
nC ,
k
A FORMULA FOR PASCAL’S TRIANGLE
1
11
121
1331
14641
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
nC
k
0C
0
1C 1C
0
1
2C 2C 2C
0
1
2
3C 3C 3C 3C
0
1
2
3
4C 4C 4C 4C 4C
0
1
2
3
4
5C 5C 5C 5C 5C 5C
0
1
2
3
4
5
6C 6C 6C 6C 6C 6 C 6C
0
1
2
3
4
5
6
7 C 7C 7C 7C 7C 7C 7C 7C
0
1
2
3
4
5
6
7
8C 8C 8C 8C 8C 8C 8C 8C 8C
0
1
2
3
4
5
6
7
8
gives the value of row n, term k,
if we start numbering the rows and terms from 0
nC ,
k
A FORMULA FOR PASCAL’S TRIANGLE
n
n


Ck 
 
k 
C stands for coefficient as
well as combination
CALCULATING
Mentally
5
C3
5  4  3 20
C3 

 10
3  2 1 2
5
CALCULATING
5
C3
Mentally
5  4  3 20
C3 

 10
3  2 1 2
5
Formula
5!
120
C3 

 10
3!2! 6  2
5
n
n! 
 Ck 

k!n  k ! 

CALCULATING
5
C3
Mentally
5  4  3 20
C3 

 10
3  2 1 2
5
Formula
5!
120
C3 

 10
3!2! 6  2
5
n
n! 
 Ck 

k!n  k !

because ...
5  4  3 5  4  3 2 1
5!
C3 



3  2  1 3  2  1 2  1 3!  2!
5
THE BINOMIAL THEOREM
(a + x)n = nC0 an + nC1 an-1 x + nC2 an-2 x2 + nC3 an-3 x3
+ nC4 an-4 x4 + ... + nCn xn
n
=
 Ck a
n
k 0
The sum of terms
from k = 0 to n
nk k
x
Don’t worry too much about
writing in  notation: just have a
good idea of the general term
PROPERTIES OF nCk
1. nC0 = nCn = 1
1st and last
2. nC1 = nCn-1 = n
2nd and 2nd-last
1
11
121
1331
14641
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
0C
0
1C 1C
0
1
2C 2C 2C
0
1
2
3C 3C 3C 3C
0
1
2
3
4C 4C 4C 4C 4C
0
1
2
3
4
5C 5C 5C 5C 5C 5C
0
1
2
3
4
5
6C 6C 6C 6C 6C 6 C 6C
0
1
2
3
4
5
6
7 C 7C 7C 7C 7C 7C 7C 7C
0
1
2
3
4
5
6
7
8C 8C 8C 8C 8C 8C 8C 8C 8C
0
1
2
3
4
5
6
7
8
PROPERTIES OF nCk
3. nCk = nCn-k
Symmetry
4. n+1Ck = nCk-1 + nCk
Pascal’s triangle result: each
coefficient is the sum of the two
coefficients in the row above it
1
11
121
1331
14641
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
0C
0
1C 1C
0
1
2C 2C 2C
0
1
2
3C 3C 3C 3C
0
1
2
3
4C 4C 4C 4C 4C
0
1
2
3
4
5C 5C 5C 5C 5C 5C
0
1
2
3
4
5
6C 6C 6C 6C 6C 6 C 6C
0
1
2
3
4
5
6
7 C 7C 7C 7C 7C 7C 7C 7C
0
1
2
3
4
5
6
7
8C 8C 8C 8C 8C 8C 8C 8C 8C
0
1
2
3
4
5
6
7
8
n+1C
k
= nCk-1 + nCk
1
11
121
1331
14641
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
Pascal’s triangle result
0C
0
1C 1C
0
1
2C 2C 2C
0
1
2
3C 3C 3C 3C
0
1
2
3
4C 4C 4C 4C 4C
0
1
2
3
4
5C 5C 5C 5C 5C 5C
0
1
2
3
4
5
6C 6C 6C 6C 6C 6 C 6C
0
1
2
3
4
5
6
7 C 7C 7C 7C 7C 7C 7C 7C
0
1
2
3
4
5
6
7
8C 8C 8C 8C 8C 8C 8C 8C 8C
0
1
2
3
4
5
6
7
8
15 = 10 + 5
6C = 5C + 5C
4
3
4
Example 1
(a) (a + 3)5 =
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 =
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
Example 2 (2008 HSC, Question 1(d), 2 marks)
Find an expression for the coefficient of x8y4
in the expansion of (2x + 3y)12.
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
For coefficient of x8y4, substitute k = ?
 = 1 mark
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 + ...
 = 1 mark
General term Tk = 12Ck (2x)12-k(3y)k
For coefficient of x8y4, substitute k = 4:
T4 = 12C4 (2x)8(3y)4
= 12C4 28 34 x8 y4
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’.

Find an expression for the
coefficient of x8y4
in the expansion of (2x + 3y)12.
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 calling it ‘the kth term’: too confusing

HSC questions ask for ‘the term that contains x8’ rather than
‘the 9th term’
Example 3 (2011 HSC, Question 2(c), 2 marks)
Find an expression for the coefficient of x2
in the expansion of
8
4

 3x   .
x

Example 3 (2011 HSC, Question 2(c), 2 marks)
8
1
2
4
4 8
4
8
7
6

8
8
 3 x    C0  3x   C1  3x      C2  3x      ...
x

 x
 x
General term Tk = 8Ck (3x)8-k   4 
 x
k
= 8Ck 38-k x8-k (-4)k x-k
= 8Ck 38-k x8-2k (-4)k 
For the coefficient of x2:
8 – 2k = 2
-2k = -6
k=3
T3 = 8C3 38-3 x8-2×3 (-4)3
= 8C3 35 (-4)3 x2
= -870 192 x2
Coefficient is -870 192 or 8C3 35 (-4)3 
FINDING THE GREATEST COEFFICIENT
(1 + 2x)8 = 1 + 16x + 112x2 + 448x3 + 1120x4
+ 1792x5 + 1792x6 + 1024x7 + 256x8
(1 + 2x)8 = 1 + 16x + 112x2 + 448x3 + 1120x4
+ 1792x5 + 1792x6 + 1024x7 + 256x8
Example 4
8
Suppose (1 +
2x)8
k
t
x
=  k .
k 0
(a) Find an expression for tk, the coefficient of xk.
tk 1 2 8  k 
(b) Show that

.
tk
k 1
(c) Show that the greatest coefficient is 1792.
Example 4
(a) (1 + 2x)8 = 8C0 18 + 8C1 17 (2x)1 + 8C2 16 (2x)2 + 8C3 15 (2x)3 +...
General term Tk = 8Ck 18-k (2x)k
= 8Ck 1 (2k) xk
= 8Ck 2k xk
 tk = 8Ck 2k
Leave out xk as we are only
interested in the coefficient
Example 4
(a) (1 + 2x)8 = 8C0 18 + 8C1 17 (2x)1 + 8C2 16 (2x)2 + 8C3 15 (2x)3 +...
General term Tk = 8Ck 18-k (2x)k
= 8Ck 1 (2k) xk
= 8Ck 2k xk
 tk = 8Ck 2k
tk 1 2 8  k 
(b) Show that

.
tk
k 1
Leave out xk as we are only
interested in the coefficient
Example 4
(a) (1 + 2x)8 = 8C0 18 + 8C1 17 (2x)1 + 8C2 16 (2x)2 + 8C3 15 (2x)3 +...
General term Tk = 8Ck 18-k (2x)k
= 8Ck 1 (2k) xk
= 8Ck 2k xk
 tk = 8Ck 2k
tk 1 2 8  k 
(b) Show that

.
tk
k 1
tk 1 8 Ck 1 2k 1
 8
tk
Ck 2 k
Leave out xk as we are only
interested in the coefficient
Example 4
tk 1 8 Ck 1 2k 1
(b)
 8
tk
Ck 2 k
tk 1 2 8  k 

.
(b) Show that
tk
k 1
8!
8!


 21
 k  1!8   k  1! k !8  k !
8! k !  8  k  !
 .
.
.2
8!  k  1 !  7  k !
1 8k

.
.2
k 1 1
2 8  k 

k 1
Ratio of consecutive factorials
n  1! 
n!
Example 4
8
k 1
tk 1 2 8  k 
t
C
2
k

1
k

1

.
(b)
(b) Show that
 8
tk
k 1
tk
Ck 2k
8!
8!


 21
 k  1!8   k  1! k !8  k !
8! k !  8  k  !
 .
.
.2
8!  k  1 !  7  k !
1 8k

.
.2
k 1 1
2 8  k 

k 1
Ratio of consecutive factorials
n  1!  n  1
n!
8! 8  7  6  5 ... 1
eg

8
7!
7  6  5 ... 1
Example 4
(c) Show that the greatest coefficient is 1792.
For the greatest coefficient tk+1, we want:
tk+1 > tk
for the largest
tk 1
1
tk
possible integer
value of k
28  k 
1
k 1
16 – 2k > k + 1
-3k > -15
k<5
k=4
k must be
a whole number
Example 4
(c) Show that the greatest coefficient is 1792.
Greatest coefficient tk+1 = t5
= 8C5 25
= 56  32
= 1792
tk = 8Ck 2k
THE BINOMIAL THEOREM FOR (1 + x)n
(1 + x)n = nC0 + nC1 x + nC2 x2 + nC3 x3 + nC4 x4 +
... + nCn xn
n
=
 Ck x
k 0
n
k
Example 5 (similar to 2010 HSC, Question 7(b)(i), 1 mark)
Expand (1 + x)n and substitute
an appropriate value of x to prove that
n

k 0
n
Ck  2 n
Example 5
(1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn
Sub x = ?
n
[Aiming to prove:

k 0
n
Ck  2 n ]
Example 5
(1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn
This will make the x’s disappear
and make the LHS become 2n
Sub x = 1:
(1 + 1)n = nC0 + nC1 (1) + nC2 (12) + ... + nCn (1n)
2n = nC0 + nC1 + nC2 + ... + nCn
n
2   n Ck
n
k 0
Example 6
By considering that (1 + x)2n = (1 + x)n(1 + x)n
and examining the coefficient of xn on each side, prove that

n
k 0
n
Ck   2 nCn .
2
Example 6
(1 + x)2n = (1 + x)n(1 + x)n
(1 + x)2n = 2nC0 + 2nC1 x + 2nC2 x2 + ... + 2nC2n x2n
Term with xn = 2nCn xn
Coefficient of xn = 2nCn
 C 
n
[Aiming to prove
2
n
k 0
k
 2 nCn ]
Example 6
(1 + x)2n = (1 + x)n(1 + x)n
(1 + x)2n = 2nC0 + 2nC1 x + 2nC2 x2 + ... + 2nC2n x2n
Term with xn = 2nCn xn
Coefficient of xn = 2nCn
(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
 C 
n
[Aiming to prove
2
n
k 0
k
 2 nCn ]
Example 6
(1 + x)2n = (1 + x)n(1 + x)n
(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-1 xn-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
nC
 C 
n
[Aiming to prove
2
n
k 0
k
 2 nCn ]
k=
nC
n-k
(1 + x)2n = (1 + x)n(1 + x)n
Example 6
 By equating coefficients of xn on both sides of
(1 + x)2n = (1 + x)n.(1 + x)n
2nC
n
= (nC0)2 + (nC1)2 + (nC2)2 + ... + (nCn)2
n
2n
 
Cn   Ck
n
k 0
2
Example 7 (2006 HSC, Question 2(b), 2 marks)
(i) By applying the binomial theorem to (1 + x)n and
differentiating, show that
n 1  x 
n 1
 n  n
 n  r 1
 n  n1
    2   x  ...  r   x  ...  n   x .
 1  2
r
 n
(ii) Hence deduce that
n 1
n3
 n
 n  r 1
 n  n1
    ...  r   2  ...  n   2 .
1
r
 n
Example 7 (2006 HSC, Question 2(b), 2 marks)
[Need to prove n 1  x n1   n   2  n  x  ...  r  n  xr 1  ...  n  n  xn1 ]
1
 2
r
(i) (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn
Differentiating both sides:
 n
Example 7 (2006 HSC, Question 2(b), 2 marks)
[Need to prove n 1  x n1   n   2  n  x  ...  r  n  xr 1  ...  n  n  xn1 ]
1
 2
r
 n
(i) (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn
The general term
Differentiating both sides:
n(1 + x)n-1 = 0 + nC1 + 2 nC2 x + ... + r nCr xr-1 + ... + n nCn xn-1
= nC1 + 2 nC2 x + ... + r nCr xr-1 + ... + n nCn xn-1 
Example 7 (2006 HSC, Question 2(b), 2 marks)
[Need to prove n 1  x n1   n   2  n  x  ...  r  n  xr 1  ...  n  n  xn1 ]
1
 2
r
 n
(i) (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn
Differentiating both sides:
n(1 + x)n-1 = 0 + nC1 + 2 nC2 x + ... + r nCr xr-1 + ... + n nCn xn-1
= nC1 + 2 nC2 x + ... + r nCr xr-1 + ... + n nCn xn-1 
n 1
(ii) Substitute x = ? to prove n3
 n
 n  r 1
 n  n1
    ...  r   2  ...  n   2 .
1
r
 n
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 + ... + r nCr xr-1 + ... + n nCn xn-1
= nC1 + 2 nC2 x + ... + r nCr xr-1 + ... + n nCn xn-1 
 n
 n
 n
 
 
 
(ii) Substitute x = 2 to prove n3n1     ...  r   2r 1  ...  n   2n1.
1
r
n
n(1 + 2)n-1 = nC1 + 2 nC2 2 + ... + r nCr 2r-1 + ... + n nCn 2n-1
n 3n-1 = nC1 + 4 nC2 + ... + r nCr 2r-1 + ... + n nCn 2n-1 
Example 8 (2008 HSC, Question 6(c), 5 marks)
Let p and q be positive integers with p ≤ q.
(i) Use the binomial theorem to expand (1 + x)p+q , and hence
write down the term of
1 x
(ii) Given that  q
x
pq
1  x 
xq
 1  x 
pq
which is independent of x.
q
p
 1  apply the binomial
1   ,
 x
theorem and the result of part (i) to find a simpler expression
 p  q   p  q 
 p  q 
for 1          ...     .
 1  1   2  2 
 p  p 
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
1  x p  q
x
q
independent of x will be
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
1  x p  q
xq
pq
Cq x q
x
q
independent of x will be
, that is,
pq
Cq

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
1  x p  q
xq
pq
Cq x q
x
1 x

(ii) Given that
xq
pq
q
independent of x will be
, that is,
pq
Cq

q
 1
 1  x  1   , apply the binomial
 x
p
theorem and the result of part (i) to find a simpler expression
for
 p  q   p  q 
 p  q 
1          ...     .
 1  1   2  2 
 p  p 
Example 8 (2008 HSC, Question 6(c), 5 marks)
(ii) (1 + x)p = pC0 + pC1 x + pC2 x2 + ... + pCp xp
q
2
q
 1
1
1
1
1    qC0  qC1    qC2    ...  qCq  
x

x
x
x
 1 
 1 
q
q 1
q
q
 C0  C1    C2  2   ...  Cq  q  
x
x 
x 
q
1
p
 1  x  1   
x

C
p
0
 pC1 x  pC2 x 2  ...  pC p x p

2
q
q
1
1
1






  C0  qC1    qC2    ...  qCq   

x
x
 x  
1 x
 p q  p q
 p q 

Find 1            ...      using
q
1
1
2
2
p
p
x
     
  
pq
 1  x 
p
 1
1  
 x
q
1 q  1 
 1 
q
 C0  C1    C2  2   ...  Cq  q 
Example 8 (2008 HSC,
 x Question

 x  6(c), 5 marks)
x 
q
q
q
1
p
(ii)  1  x  1   
 x
C
p
p
p
2
p
p

C
x

C
x

...

C
x
0
1
2
p


1
 1 
 1 
  qC0  qC1    qC2  2   ...  qCq  q 
 x
x 
 x 

If we expanded the RHS, there would be many terms
Terms independent of x
1 x
 p q  p q
 p q 

Find 1            ...      using
q
1
1
2
2
p
p
x
     
  
pq
 1  x 
p
 1
1  
 x
q
1 q  1 
 1 
q
 C0  C1    C2  2   ...  Cq  q 
Example 8 (2008 HSC,
 x Question

 x  6(c), 5 marks)
x 
q
q
p
(ii)  1  x  1     pC0  pC1 x  pC2 x 2  ...  pC p x p 


1
x
q

1
 1 
 1 
  qC0  qC1    qC2  2   ...  qCq  q 
 x
x 
 x 

If we expanded the RHS, there would be many terms
Terms independent of x
=
=
 1 
 + ... + pCp xp qCp  1 
p

x 
pC qC + pC x qC  1  + pC x2 qC 
0
0
1
1
2
2 x2

x
 
p
q
p
q
p
q
1 + C1 C1 + C2 C2 + ... + Cp Cp 
p≤q
1 x
 p q  p q
 p q 

Find 1            ...      using
q
1
1
2
2
p
p
x
     
  
pq
 1  x 
p
 1
1  
 x
q
Example 8 (2008 HSC, Question 6(c), 5 marks)
By equating the terms independent of x on both sides of
From
(i)
1  x p  q  1  x p 1  1 q
q
x

x
p+qC = 1 + pC qC + pC qC + ... + pC qC
q
1
1
2
2
p
p
1 + pC1 qC1 + pC2 qC2 + ... + pCp qCp = p+qCq 
1 x
 p q  p q
 p q 

Find 1            ...      using
q
1
1
2
2
p
p
x
     
  
pq
 1  x 
p
 1
1  
 x
q
And now ...
A very hard identity to prove:
Example 9 from 2002 HSC
Question 7(b), 6 marks
Example 9 (2002 HSC, Question 7(b), 6 marks)
The coefficient of xk in (1 + x)n, where n is a positive integer, is
denoted by ck (so nck = ck).
To save time, this question
asks us to abbreviate nCk to ck
(i) Show that c0 + 2c1 + 3c2 + ... + (n + 1)cn = (n + 2) 2n-1.
c0
c1
c2
(ii) Find the sum



1.2 2.3 3.4
  1
n
cn
 n  1 n  2 
.
Write your answer as a simple expression in terms of n.
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]
To save time, this
question asks us to
abbreviate nCk to ck
Identity to be proved involves (n + 2) 2n-1 so try ...
[Aiming to prove: c0 + 2c1 + 3c2 + ... + (n + 1)cn = (n + 2) 2n-1 ]
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 ]
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]
To prove result, we must add c0 + c1 + c2 + ... + cn
To get this, sub x = ? into [1] above:
[Aiming to prove: c0 + 2c1 + 3c2 + ... + (n + 1)cn = (n + 2) 2n-1 ]
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]
Now add [2] and [3]
to prove result
To get this, sub x = 1 into [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 ]
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 = c0 + c1 + c1 + 2c2 + c2 + ... + ncn + cn
2n-1 (n + 2) = c0 + 2c1 + 3c2 + ... + (n + 1)cn 
Result proved: each
coefficient of ck increased
by 1 as required
[Aiming to prove: c0 + 2c1 + 3c2 + ... + (n + 1)cn = (n + 2) 2n-1 ]
Example 9 (2002 HSC, Question 7(b), 6 marks)
c0
cn
c1
c2
n





1
.
 
(ii) Find the sum 1.2 2.3 3.4
 n  1 n  2 
Write your answer as a simple expression in terms of n.
Answer involves dividing by (k + 1)(k + 2) and alternating
–/+ pattern so try integrating (1 + x)n from (i) twice and
substituting x = -1.
(1 + x)n = c0 + c1 x + c2 x2 + ... + cn xn [1]
Example 9 (2002 HSC, Question 7(b), 6 marks)
(ii) (1 + x)n = c0 + c1 x + c2 x2 + ... + cn xn [1]
Integrate both sides:
1
c1 2 c2 3
cn n 1
n 1
1  x   c0 x  x  x  
x k
n 1
2
3
n 1
To find k, sub x = 0:
Don’t forget the
1 n 1
1  0  0  0   0  k
constant of integration
n 1
1
k 
n 1
1
c1 2 c2 3
cn n 1
1
n 1
1  x   c0 x  x  x  

x 
n 1
2
3
n 1
n 1 
c0
cn
c1
c2
n


    1
[Aiming to find
]
n  1n  2
1.2 2.3 3.4
1
 k  9 (2002 HSC, Question 7(b), 6 marks)
Example
n 1
1
c1 2 c2 3
cn n 1
1
n 1

(ii) 
1  x   c0 x  x  x  
x 
n 1
2
3
n 1
n 1
Integrate again to get 1.2, 2.3, 3.4 denominators:
1
c0 x 2 c1 x3 c2 x 4
cn x n 1
x
n2
1  x  


 

d
n  1n  2
n  1n  2 n  1
2
2.3
3.4
To find d, sub x = 0:
1
1n  2  0  0  0    0  0  d
n  1n  2
1
d 
n  1n  2
2
3
4
n2
1
c
x
c
x
c
x
c
x
n
c0 x n c21  c02  1  n 2 c


1
n 
[Aiming
n  1ton find
2  1.2  2.3  31..42  2.3 1 3.n4 1n  2n]  1n  2 
x
1
1
 d  9 (2002 HSC, Question 7(b), 6 marks)
Example
n  1n  2
(ii) 
cn x n  2
c0 x 2 c1 x3 c2 x 4
1
n2
1  x  
 


n  1n  2
n  1n  2
3. 4
2.3
1.2
1
x



n  1 n  1n  2 
Sub x = ? for –/+ pattern:
c0
cn
c1
c2
n








1
[Aiming to find
]
n  1n  2
1.2 2.3 3.4
1
 d  9 (2002 HSC, Question 7(b), 6 marks)
Example
n  1n  2
cn x n  2
c0 x 2 c1 x3 c2 x 4
1
n2
1  x  
 


(ii) 
n  1n  2
n  1n  2
3. 4
2.3
1.2
1
x


n  1 n  1n  2 
Sub x = -1 for –/+ pattern:
2
3
4
n2
1
c
(

1
)
c
(

1
)
c
(

1
)
c
(

1
)
0n  2  0
 1
 2
  n
n  1n  2
n  1n  2
1.2
2.3
3.4
1
1


n  1 n  1n  2 
c0  c1 c2
cn (1) n  1
1
1
0


 


n  1n  2 n  1 n  1n  2
1.2 2.3 3.4
c0
cn 1
cc1 (1c)2n
n
c0[Aiming
c1 tocfind
1
 n
   1 
]


 2 




1
.
2
2
.
3
3
.
4
n

1
n

2
n  1n  2 n  1 n  1n  2
1.2 2.3 3.4
2
Example 9 (2002 HSC, Question 7(b), 6 marks)
c0  c1 c2
cn (1) n  12
1
1


 


(ii) 0 
n  1n  2 n  1 n  1n  2
1.2 2.3 3.4
c0
c1
c2
cn
1
1
n



  (1)


n  1n  2 n  1 n  1n  2
1.2 2.3 3.4
n  2  1

n  1n  2
n 1

n  1n  2
1

n2 
c0
cn
c1
c2
n


    1
[Aiming to find
]
n  1n  2
1.2 2.3 3.4
Example 10 (2007 HSC, Question 4(a), 4 marks)
In a large city, 10% of the population has green eyes.
(i) What is the probability that two randomly chosen people have
green eyes?
(ii) What is the probability that exactly two of a group of 20
randomly chosen people have green eyes? Give your answer
to three decimal places.
(iii) What is the probability that more than two of a group of 20
randomly chosen people have green eyes? Give your answer
to two decimal places.
Example 10 (2007 HSC, Question 4(a), 4 marks)
(i) P(both green) = 0.1  0.1
= 0.01 
Example 10 (2007 HSC, Question 4(a), 4 marks)
(i) P(both green) = 0.1  0.1
= 0.01 
(ii) What is the probability that exactly two of a group of 20
randomly chosen people have green eyes? Give your answer
to three decimal places.
Example 10 (2007 HSC, Question 4(a), 4 marks)
(i) P(both green) = 0.1  0.1
= 0.01 
(ii) p = 0.1, q = 1 – 0.1 = 0.9, n = 20
P(X = 2) =
Example 10 (2007 HSC, Question 4(a), 4 marks)
(i) P(both green) = 0.1  0.1
= 0.01 
(ii) p = 0.1, q = 1 – 0.1 = 0.9, n = 20
P(X = 2) = 20C2 0.12 0.918
= 0.28517 ...
 0.285 
Example 10 (2007 HSC, Question 4(a), 4 marks)
(i) P(both green) = 0.1  0.1
= 0.01 
(ii) p = 0.1, q = 1 – 0.1 = 0.9, n = 20
P(X = 2) = 20C2 0.12 0.918
= 0.28517 ...
 0.285 
(iii) What is the probability that more than two of a group of 20
randomly chosen people have green eyes? Give your answer
to two decimal places.
Example 10 (2007 HSC, Question 4(a), 4 marks)
(i) P(both green) = 0.1  0.1
= 0.01 
(ii) p = 0.1, q = 1 – 0.1 = 0.9, n = 20
P(X = 2) = 20C2 0.12 0.918
= 0.28517 ...
 0.285 
(iii) P(X > 2) = 1 – P(X = 0) – P(X = 1) – P(X = 2)
= 1 – 20C0 0.10 0.920 – 20C1 0.110.919 – 0.285 from (ii) 
= 1 – 0.920 – 20(0.1)0.919 – 0.285
= 0.32325 ...
 0.32 
HOW TO STUDY FOR MATHS (P-R-A-C)
1. Practise your maths
2. Rewrite your maths
3. Attack your maths
4. Check your maths
WORK HARD AND BEST OF LUCK
FOR YOUR HSC EXAMS!
robert.yen@cengage.com