# FIND BINOMIAL EXPANSION OF RATIONAL FUNCTIONS

Find Binomial Expansion Of Rational Functions :

Here we are going to see some practice questions on finding binomial expansion of rational functions.

## Formula for expansion of rational functions

(1 + x)n

(1 - x)n

(1 + x)-n

(1 - x)-n

Note :

• When we have negative signs for either power or in the middle, we have negative signs for alternative terms.
• If we have negative for power, then the formula will change from (n - 1) to (n + 1) and (n - 2) to (n + 2).
• If we have negative signs for both middle term and power, we will have a positive sign for every term.

Observation of the coefficient in each of such expansions will be very helpful in solving problems. Let us list some of them:

1. (1 + x)−1 = 1− x + x2 − x3 + · · ·

2. (1 − x)−1 = 1+x + x2 + x3 + · · ·

3. (1 − x)−2 = 1 + 2x + 3x2 + 4x3 + 5x4 + 6x5 + · · ·

4. (1 + x)−2 = 1− 2x + 3x2 − 4x3 + 5x4 − 6x5 + · · ·

All the above expansions are valid only when |x| < 1.

Question 1 :

Expand the following in ascending powers of x and find the condition on x for which the binomial expansion is valid.

(i)  1/(5 + x)

Solution :

1/(5 + x)  =  (1/5)(1 + (x/5))-1

x  =  x/5     |x| < 5

(1 + x)−1 = 1− x + x2 − x3 + · · ·

=  (1/5) [1 - (x/5) + (x/5)2 - (x/5)3 + .............]

=  (1/5) [1 - (x/5) + (x2/25) - (x3/125) + .............]

(ii)  2/(3 + 4x)2

Solution :

2/(3 + 4x) =  2 (3 + 4x)-2

=  2 (3 + 4x)-2

(1 + x)−2 = 1− 2x + 3x2 − 4x3 + 5x4 − 6x5 + · · ·

=  (2/3)[ (1− 2(4/3) + 3(4x/3)2 − 4(4x/3)3  + ............]

(iii)  (5 + x2)2/3

Solution :

=  52/3 (1 + x2/5)2/3

1st term  =   1

2nd term  = nx  =  (2/3)(x2/5)  =  2x2/15

3rd term  = (n(n-1)/2!)x2  =  [(2/3)(-1/3)/2!](x2/5)2

=  -x4/225

4th term  = (n(n-1)(n-2)/3!)x3

=  [(2/3)(-1/3)(-4/3)/3!](x2/5)3

=  4x6/(81 ⋅ 125)

=  1 + (2x2/15) - (x4/225) + 4x6/(81 ⋅ 125) + ................

Required condition is |x2| < 5

(iv)  (x + 2)-2/3

Solution :

=  (2 + x)-2/3  =  2-2/3 (1 + x/2)-2/3

1st term  =   1

2nd term  = nx  =  (-2/3)(x/2)  =  -x/3

3rd term  = (n(n-1)/2!)x2  =  [(-2/3)(-5/3)/2!](x/2)2

=  5x2/36

4th term  = (n(n-1)(n-2)/3!)x3

=  [(-2/3)(-5/3)(-8/3)/3!](x/2)3

=  -5x3/81

=  1 - (x/3) + (5x2/36) - (5x3/81) + ................

Required condition is |x| < 2

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