2 Jan 1999

efinition

Binomial means the sum (or difference) of two terms.

Expansion of (x + y)n, where n is any rational number, is called the binomial expansion.

ositive Integral Index

If n is a positive integer, then

(x + y)n = xn + nxn - 1y + 
n(n - 1)
¾¾¾¾
2!
  xn - 2y2 + 
n(n - 1)(n - 2)
¾¾¾¾¾¾¾
3!
  xn - 3y3 + ¼ + yn.

Notes:

  1. There are n + 1 terms.
  2. The (r + 1)th term is nCr xn - ryr.
  3. The expansion of (x - y)n may be obtain by replacing y by -y.

on-positive Integral Index

If n is not a positive integer, then we have

(1 + x)n = 1 + nx + 
n(n - 1)
¾¾¾¾
2!
  x2 + 
n(n - 1)(n - 2)
¾¾¾¾¾¾¾
3!
  x3 + ¼.

Notes:

  1. If n is NOT a positive integer, the series is infinite.
  2. The above expansion is in ascending powers of x, and is only valid (ie meaningful) when |x| < 1 (small x).
  3. To expand (a + x)n in ascending powers of x, it must first be written as an(1 + x/a)n.

pecial Series

(1 - x)-1 = 1 + x + x2 + ¼ + xr + ¼.

(1 - x)-2 = 1 + 2x + 3x2 + ¼ + (r + 1)xr + ¼.

Replace x by -x to get the corresponding expansions for (1 + x)-1 and (1 + x)-2.

xamples

Example 1: Find the first four terms of (2 - x)-2, stating the set of values of x for which the expansion is valid.

Solution:

(2 - x)-2  =  2-2(1 - x/2)-2
 
 = 
1
  é
-2(-3)
-2(-3)(-4)
 ù
-
  ê 1 + (-2)(- x/2) +  ¾¾¾  (- x/2)2 +  ¾¾¾¾¾  (- x/2)3 + ¼ ú
4
  ë
2!
3!
 û
 
 =  1/4 + (1/4)x + (3/16)x2 + (1/8)x3 + ¼.

The expansion is valid provided |x/2| < 1, ie, |x| < 2.

Example 2:  Give the binomial expansion, for small x, of (1 + x)1/4 up to and including the term in x2, and simplify the coefficients.
By putting x = 1/16 in your expression, show that  4Ö17 » 8317/4096.

Solution:

1
(1/4)(-3/4)
(1 + x)1/4 = 
 1 +  ¾  x +  ¾¾¾¾¾  x2 + ¼
4
2!
 
= 
 1 + x/4 - 3x2/32 + ¼.

Putting x = 1/16,

1/16
3(1/16)2
(1 + 1/16)1/4 = 
 1 +  ¾¾  -  ¾¾¾¾  + ¼
4
32
 
(17/16)1/4 » 
 8317/8192
 
\  4Ö17 » 
 8317/4096.

5
Example 3:  Expand  ¾¾¾¾¾¾¾  as a series
(1 + 3x)(1 - 2x)
of ascending powers of x giving:
(a)    the first four terms,
(b)    the range of values of x for which the expansion is valid.

Solution:

5
Expressing  ¾¾¾¾¾¾¾   as partial fractions gives
(1 + 3x)(1 - 2x)

5
3
2
¾¾¾¾¾¾¾  = 
¾¾¾
  + 
¾¾¾
  
(1 + 3x)(1 - 2x)
1 + 3x
1 - 2x
 
 =  3(1 + 3x)-1 + 2(1 - 2x)-1.

(1 + 3x)-1  =  1 - 3x + (3x)2 - (3x)3 + ¼
 =  1 - 3x + 9x2 - 27x3 + ¼.

(1 - 2x)-1  =  1 + 2x + (2x)2 + (2x)3 + ¼
 =  1 + 2x + 4x2 + 8x3 + ¼.

5
\  ¾¾¾¾¾¾¾  =  3[1 - 3x + 9x2 - 27x3 + ¼]
(1 + 3x)(1 - 2x)
     + 2[1 + 2x + 4x2 + 8x3 + ¼]
 
 =  5 - 5x + 35x2 - 65x3 + ¼.

Expansion is valid when

|3x| < 1   and  |2x| < 1
|x| < 1/3   and  |x| < 1/2
Þ  |x| < 1/3.

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