Find Binomial Expansion Of Rational Functions :
Here we are going to see some practice questions on finding binomial expansion of rational functions.
(1 + x)n
(1 - x)n
(1 + x)-n
(1 - x)-n
Note :
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 = 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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