Binomial Theorem
Advanced Level Pure Mathematics
Advanced Level Pure Mathematics
5
Algebra
Chapter 5
Appendix 1
Binomial Theorem
Summation Sign and Product Sign
2
5.1
Introduction
6
5.3
Binomial Theorem for Positive Integral Index
10
5.4
The Greatest Coefficient and
The Term with The Greatest Absolute Value
14
5.5
More Properties of Binomial Coefficients
16
5.6*
Binomial Series
23
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1
Binomial Theorem
Advanced Level Pure Mathematics
Appendix 1 Summation Sign and Product Sign
n
1.
 ar = a
The sigma notation :
1
+ a2 +  + an 1 + an
r 1
5
9
 ar = a
Example
1
+ a2 + a3 + a4 + a5
r3
;
r 1
2.
3.
n
n
n
k 0
l 0
i 0
=
r6
 a k   a l   ai
n
n
n
k 1
k 1
k 1
 (a k  bk )   a k   bk
n
4.
If c is a constant , then
 (ca k )
=
k 1
n
5.
If c is a constant , then
c
=
k 1
n
6.
 (a k  a k 1 )
=
k 1
7.
n
n 1
k 1
k 0
 a k   a k 1 =
2
8.
n
n
 n 
  ai     ai a j
 i 1 
i 1 j 1
n
9.
The product notation :
ak

k 1
= a1  a2    an 1  an
5
Example
3k

k 1
n
2
=
3

k 1
;
=
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Binomial Theorem
Advanced Level Pure Mathematics

n
10.

n
n

 (a k bk )    a k    bk 
k 1
k 1
k 1
n
11.
(ca k )

k 1
n
12.
a
a k
k 1
r 
r 1
n
14*.
=
k 1
n
13*.
=
r2
n(n  1)
2

r 1
n(n  1)(2n  1)
6
n 2 (n  1) 2  n 
15*.  r 
   r
 r 1 
4
r 1
n
n
16.
a
k 0
17.
n
k
= a0 +
a
k 1
n
n
k 0
l 0
k
= an +
 kr k   lr l
n 1
n
18.
2
3
 ar
k 1
k
  ar
k 0
k 1
n 1
  ar k 1
k 2
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Binomial Theorem
Advanced Level Pure Mathematics
19.
n 1
?
k 3
?
 kp k 1   ? p k
n3
20.
C
k 1
n 1
k 1
?
x   Ck? x ?
k
?
n
Example 1
Show that
1
n 1
1
 k 1   k .
r 1
k 2
n
Hence express
1 n 1

in term of n .

k 1 k
k 1 k  1
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Binomial Theorem
Advanced Level Pure Mathematics
Example 2
(a) Find the constants A and B in the following identity
x3
A
B


2
x  x x 1 x
n
(b) Hence evaluate lim
n 
k3 
.
2
 k
 1
  k  1  k
k 2
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Binomial Theorem
Advanced Level Pure Mathematics
5.1
Introduction
1.
Binomial Theorem : (1 + x)n = C0  C1 x  C2 x ...Cn x .
2.
The term Cr xr is called the general term.
3.
Crn is called the binomial coefficients.
4.
(1  x)n = C0  C1 x  C2 x ...( 1) Cn x
5.
Factorial Notation :
n
n
n
2
n
n
n
n
Example
N.B.
n
3! = 321
n
;
0! = 1
(2)
n! = n(n 1)!
(3)
n! =
Definition
n
n
n
9! = 987654321
(n  1)!
n 1
C r , Crn , n Cr
Let n be a positive integer and r non-negative integer with r  n.
The symbol C
Example
n
n! = n(n 1)(n 2)321
(1)
Combinations :
2
C39 
n
r
Prn
n!
n(n  1) (n  r  1)


is defined as C 
r ! r !(n  r )!
r!
n
r
,
C0n  Cnn 
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Binomial Theorem
Advanced Level Pure Mathematics
Example 3
Let C
n
r
denote the binomial coefficient, show that
C1n 2C2n 3C3n
nCnn n(n  1)
.
 n  n  n 
2
C0n
C1
C2
Cn 1
Pascal Triangle :
n=0
1
n=1
1
n=2
1
n=3
1
n=4
1
C0n
n=n
Example
1
2
3
4
1
3
6
1
4
 
C1n
C14  C34
and
1
C nn1
Cnn
C34  C33  C23
Theorem 5.1 Let n be a positive integers , then
(a)
Crn  Cnn r
n 1
(b) Cr  Cr
n
r = 0 , 1 , 2 , ... , n
 Crn11
r = 0 , 1 , 2 , ... , n1.
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Binomial Theorem
Advanced Level Pure Mathematics
Properties :
(1)
Cok = C0k 1 =
(2)
Ckk = Ckk11 =
(3)
Crn11 =
n 1 n
Cr
r 1
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Binomial Theorem
Advanced Level Pure Mathematics
Example 4
Prove that
(C0n  C1n )(C1n  C2n ) (Cnn1  Cnn ) 
1
(n  1) n C1n C2n  Cnn .
n!
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Binomial Theorem
Advanced Level Pure Mathematics
5.3
Binomial Theorem for Positive Integral Index
n
n
Theorem 5.2 For any positive integer n , (1 + x)
=
 Crn x r .
r 0
i.e.
(1 + x)n = C0  C1 x  C2 x ...Cn x
n
n
n
2
n
n
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Binomial Theorem
Advanced Level Pure Mathematics
Definition
i.e.
(a + b)n =
n
n
r 0
r 0
 Crn a r b nr   Crn a nr b r
(a + b)n = C0 a b  C1 a
Example 5
n
n
0
n
n 1 1
b Cnn1a 1b n 1  Cnn a 0 b n
If the coefficients of three consecutive terms in the expansion of (1 + x)n are 120 , 210
and 252 respectively, find the value of n .
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Binomial Theorem
Advanced Level Pure Mathematics
n 1
Example 6
Prove that
 Crn1 x r
r 0

1
[(1 + x)n  1 ]
x
n 1
Example 7
Prove that
 rCrn x r = nx [(1 + x)
n 1
 x n 1 ].
r 0
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Binomial Theorem
Advanced Level Pure Mathematics
Example 8
(a) Prove that for any positive integer n ,
(1+
1 n
) =1+
n
k 
 1 r 1
  r !  (1  n )  .
k 0
r 1 

n
(b) Hence , or otherwise , show that for n  2 ,
(i)
(1+
1 n
1 n +1
) < (1+
) ,
n
n1
(ii)
(1+
1 n
) <3.
n
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Binomial Theorem
Advanced Level Pure Mathematics
5.4
The Greatest Coefficient and The Term with The Greatest
Absolute Value
n
For (1 + x)n , when n is even , the greatest coefficient is C n , when n is odd ,
2
n
n
2
2
the greatest coefficients are C n 1 and C n 1 .
Example 9
Find the greatest coefficient in the expansion of (3 + 5x)18 and the greatest term when x =
Example 10 Let the kth in the binomial expansion of (1  x) 2 n in ascending powers of x be
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2
.
3
Binomial Theorem
Advanced Level Pure Mathematics
denoted by Tk , i.e. Tk  C
2n
k 1
x
k 1
.
1
(a) If x  , find the range of values of k such that Tk 1  Tk .
3
(b) Find the greatest term in the expansion if x 
1
and n  15 .
3
[HKAL2001]
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Binomial Theorem
Advanced Level Pure Mathematics
5.5
More Properties of Binomial Coefficients
Consider (1  x ) n  C 0n  C1n x  C n2 x 2    C nr x r    C nn x n
(a) Change x to
 (*)
1
, we have
x
2
r
1
1
1
1
1
(1  ) n  C 0n  C1n    C n2      C nr      C nn  
x
x
x
x
x
n
1
1
But (1  ) n  n (1  x ) n
x
x

1
(1  x ) n  x n (1  ) n
x
(1  x )
n
=
2
r
n
 n

n 1 
n 1 
n 1 
n 1  

x C 0  C1    C 2      C r      C n  

x
x
x
 x  

=
C 0n x n  C1n x n 1  C n2 x n  2    C nr x n  r    C nn
=
C nn  C nn 1 x  C nn  2 x 2    C nn r x r    C 0n x n
n
Compared coefficient with (*)
C 0n  C nn
We have
C1n  C nn 1
C n2  C nn 2  C nn  C 0n
(b) Put x  1 into (*)
2 n  C 0n  C1n  C n2    C nr    C nn
(c) Put x   x
(1  x ) n  C 0n  C1n x  C n2 x 2    (1) r C nr x r    (1) n C nn x n
If x  1 , (1  1) n  0  C 0n  C1n  C n2    (1) r C nr    (1) n C nn
We have C 0n  C n2  C n4    C1n  C 3n  C 5n  

Sum of even coefficient is equal to the sum of odd coefficient
By (b),
C 0n  C n2  C n4    C1n  C 3n  C 5n   
2n
2
(d) Differentiate (*) w.r.t.x
n(1  x) n 1  C1n  2C n2 x    rC nr x r 1  nC nn x n 1
Put x  1 ,
n (2) n 1  C1n  2C n2    rC nr  nC nn
(e) Integrate both sides w.r.t.x from 0 to 1
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Binomial Theorem
Advanced Level Pure Mathematics
1
 (1  x)
1
n
dx   (C 0n  C1n x  C n2 x 2    C nr x r    C nn x n )dx
0
0
(1  x ) n 1
n 1
1
C1n x 2
C x 
0
2
n
0
0
1
1
0
1
C n2 x 3

3
0
1
C nr x r 1
C nn x n 1


r 1 0
n 1
1
0
Cn
Cn
Cn
2 n 1  1
 C 0n  1    r    n
n 1
2
r 1
n 1
(f)
Consider (1  x ) m n  (1  x ) m (1  x ) n , where m, n are positive integers.
C 0m  n  C1m  n x  C m2  n x 2    C mr  n x r    C mm  nn x m  n
=
(C 0m  C1m x  C m2 x 2    C mr x r    C mm x m )(C 0n  C1n x  C n2 x 2    C nr x r    C nn x n )
Compared coefficient of x r
C mr  n
(Equating coefficient of x r )
=
Example 11 Consider the binomial expansion (1 + x)2n . Prove that
(a)
C02n  C12n  C22n C22nn = 22n
(b) C0  C1  C2 C2n = 0
2n
(c)
2n
2n
2n
C02n  C22n  C42n C22nn  C12n  C32n  C52n C22nn1 = 2n 1
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Binomial Theorem
Advanced Level Pure Mathematics
Example 12 Prove that
(a)
C1n  2C2n  3C3n  4C4n nCnn = n2n 1
(b) C1  2C2  3C3  4C4 ( 1)
n
(c)
C0n 
n
n
n
n 1
nCnn = 0
1 n 1 n 1 n
1
2 n 1  1
C1  C 2  C3 
Cnn 
2
3
4
n 1
n 1
Example 13 Prove that
(a) 1  2  C2  2  3  C3  3  4  C4 (n  1)nCn = n(n  1)2n 2
n
n
n
n
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Binomial Theorem
Advanced Level Pure Mathematics
(b) C  2C  3C  4C (n  1)C
n
0
Example 14
n
1
n
2
n
3
n
n
n 1
= (n + 2)2
Prove that if n is even ,
1
1
1
1
22n  1




(1)!( 2n)! ( 2)!( 2n  1)! (3)!( 2n  2)!
(n)!(n  1)! ( 2n  1)!
Example 15 Prove that Cn  (C0 )  (C1 )  (C2 ) (Cn ) .
2n
n
2
n
2
n
2
n
2
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Binomial Theorem
Advanced Level Pure Mathematics
Example 16
Let f (r )  C 0n C nr  C1n C nr 1    C nn  r C nn
(a) Show that C nr  C nn r for r  0,1,2,, n .
(b) By considering the expansion of (1  x) 2 n , prove that
f (r ) 
(2n )!
(n  r )!(n  r )!
(c) Show that C 0n f (0)  C1n f (1)    C nn f (n ) 
(3n )!
.
n!(2n )!
Example 17 (a) Show that
{1 + (n + 1) x}(1 + x)n 1 = C0  2C1 x  3C2 x ( n  1)Cn x
n
n
n
2
n
n
.
(b) Hence show that
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Binomial Theorem
Advanced Level Pure Mathematics
(C0n ) 2  2(C1n ) 2  3(C2n ) 2 (n  1)(Cnn ) 2 
Example 18
(n  2)( 2n  1)!
.
n !(n  1)!
Let m and n be two positive integers with m  n ．
n 1
(a) Show that
C
r 1
2 n 1
nr
 2 2n ．
n
(b) By considering the coefficient of x
k
 1
in 1  x  1   , show that
x

m
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Binomial Theorem
Advanced Level Pure Mathematics
mk
C
r 0
m
k r
C rn  C nmkn ﹐ m  n  k  m ．
[HKAL86]
Example 19 Let n be an integer and n > 1 . By considering the binomial expansion of (1 + x)n , or
otherwise ,
(a) show that C1  2C2  3C3  nCn = 2n 1 n;
n
n
n
n
1
2
3
( 1) n 1 n



(b) evaluate
.
(n  1)! 2 !(n  2)! 3!(n  3)!
n!
[HKAL95]
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Binomial Theorem
Advanced Level Pure Mathematics
Example 20 Let k and n be non-negative integers.
Prove that
(a)
Ckn 
k  1 n 1
Ck 1 , where 0  k  n ;
n1
n 1
(b)
 ( 1) k Ckn 1 = 0 ;
k 0
(1) k n
1
Ck 
(c) 
.
n 1
k 0 k  1
n
5.6*
[HKAL90]
Binomial Series
Recall : For r < 1 ,
1
= 1 + r + r2 + r3 + r4 + ... and
1 r
1
= 1  r + r2  r3 + r4  ...
1 r
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Binomial Theorem
Advanced Level Pure Mathematics
For a real number x with | x | < 1 and a rational number n ,
 n
r 0  r 

(1+x)n =    x , where
r
 n n( n  1)( n  2) ( n  r  1)
 
 r
r!
i.e. (1  x ) n  1  nx  n (n  1) x 2    n (n  1)( n  2)  (n  r  1) x r  
2!
r!
I
Some examples of binomial series :
(1  x ) 1 
1

1 x
(b) (1  x ) 1 
1

1 x
(a)
(c)
(1  x ) 2 
(d) (1  x ) 2 
N.B..
 n
 r
Sometimes we use the symbol   for the binomial coefficient Cr .
n
Example 21 Given that C0 , C1 , C2 , ... , Cn denote the coefficients in the expansion of (1 + x)n .
(a) By considering the coefficient of xn in the expansion of (1 + x)2n ,
show that C02 + C12 + C22 + ... + Cn 2 =
(2n)!
(n !) 2
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Binomial Theorem
Advanced Level Pure Mathematics
(b) By considering the coefficient of xn in the expansion of (1  x2)n , show that
C0  C1 + C2  ... + (1) Cn
2
Example 22
2
2
n
2
if n is odd
0 n

n!
= 
( 1) 2 n n
if n is even .

( 2 )!( 2 )!
Given two positive integers n and r ﹐let Px x r  x  1    x  n 
r
r
r
(a) When Px is written in the form Px    at x t ﹐show that

r t
a r  n  1 , at  C 1
r
t
2
r t
 n
r t

t 0
for t  0 , 1 , 2 ,  , r  1 ．
Prepared by K. F. Ngai
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Binomial Theorem
Advanced Level Pure Mathematics
n
(b) Let S 0 , n  n  1 and S t , n    m t ﹐where t  1 , 2 ,  ,
m 1
r 1
n  1r   Ctr S t , n  ．
t 0
Use (b) to find S 1 , n ﹐ S 2 , n and S 3 , n.
Show that
[HKAL 1987]
Example 23 (a) Let m and n be positive integers. Using the identity
n
n +1
(1 + x) + (1 + x)
n +m
+ ... + (1 + x)
(1  x ) n  m 1  (1  x ) n
=
,
x
Prepared by K. F. Ngai
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Binomial Theorem
Advanced Level Pure Mathematics
where x  0 , show that C  C
n
n
n 1
n
C
nm
n
C
n  m 1
.
n 1
(b) Using (a) , or otherwise , show that
m 4
 r (r  1)(r  2)(r  3)  24C5m5  1
.
r 5
k
Hence evaluate
 r (r  1)(r  2)(r  3)
for k  4 .
r 0
[HKAL94]
Prepared by K. F. Ngai
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