The term independent of x in the expansion of (x + 1/x2)9 is
- Tr + 1 = 9Cr x9 − r (x−2)r = 9Cr x9 − 3r.
- Independent of x means the power is 0: 9 − 3r = 0, so r = 3.
- 9C3 = (9 × 8 × 7)/(3 × 2 × 1) = 504/6 = 84.
Answer: 84 (the 4th term)
Binomial theorem and partial fractions
NUST NET (Engineering) · Maths · Sequences, series and the binomial theorem. A short explanation of the idea, the rules to remember, the mistake to avoid, a worked example and practice questions with answers.
The idea
The expansion of (a + b)n for a positive whole number n has n + 1 terms. You almost never need the whole expansion: write the general term, collect the powers of x, and choose r to give the power you want.
When n is negative or a fraction, the binomial series goes on for ever and is only valid for small x. The bracket must start with 1, so take out a factor first if it does not. Its first two terms give quick approximations.
Partial fractions reverse the adding of algebraic fractions. The form you write depends on the factors of the denominator, and the degree of the top must be lower than that of the bottom before you start.
Rules to remember
- General term of (a + b)n: Tr + 1 = nCr an − r br
- Middle term: T(n/2) + 1 if n is even; two middle terms T(n + 1)/2 and T(n + 3)/2 if n is odd
- Sum of the coefficients of an expansion: put x = 1; sum of the binomial coefficients of (1 + x)n is 2n
- Series: (1 + x)n = 1 + nx + n(n − 1)x2/2! + n(n − 1)(n − 2)x3/3! + ..., valid for |x| < 1; for small x, (1 + x)n ≈ 1 + nx
- (a + bx)n = an(1 + bx/a)n, valid for |bx/a| < 1, that is |x| < |a/b|
- Partial fraction forms: (x − a) gives A/(x − a); (x − a)2 gives A/(x − a) + B/(x − a)2; a quadratic that will not factorise gives (Ax + B)/(that quadratic)
- If the degree of the top is not lower than the bottom, divide first; for a non-repeated factor (x − a), find A by covering that factor and putting x = a in the rest
Common mistake
Using r as the term number. nCr belongs to term number r + 1, so r = 3 gives the 4th term. Also keep the minus sign inside b: in (1 − x)n, b = −x.
Worked example
Practice questions
Try each one, then open the answer.
1. If x is so small that x2 and higher powers can be neglected, then (1 + x)4/(1 − x)2 is approximately
- A1 + 6x
- B1 + 2x
- C1 − 6x
- D1 + 8x
Show answer
Answer: A. (1 + x)4 ≈ 1 + 4x and (1 − x)−2 ≈ 1 + 2x, so the product ≈ 1 + 6x. Treating the division as (1 + x)4−2 gives the slip 1 + 2x.
2. The coefficient of x5 in the expansion of (1 − x)8 is
- A56
- B−70
- C−56
- D28
Show answer
Answer: C. The term is 8C5(−x)5 = −56x5, since 8C5 = 56 and an odd power of −x is negative.
3. For |x| < 1/3, the first three terms of the expansion of (1 + 3x)1/3 are
- A1 + x − x2
- B1 + x + x2
- C1 + x − 2x2
- D1 − x + x2
Show answer
Answer: A. (1 + y)1/3 = 1 + y/3 + [(1/3)(−2/3)/2!]y2 = 1 + y/3 − y2/9. With y = 3x this is 1 + x − x2; forgetting the 2! gives −2x2.
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