Find the coefficient of x2 in the expansion of (3 − 2x)5.
- The x2 term is 5C2 × 33 × (−2x)2.
- 5C2 = 10, 33 = 27 and (−2x)2 = 4x2.
- 10 × 27 × 4 = 1080.
Answer: The coefficient is 1080
Series
AS Level Mathematics · Pure Mathematics 1. A short explanation of the idea, the rules to remember, the mistake to avoid, a worked example and practice questions with answers.
The idea
This topic has two parts. The first is the binomial expansion of (a + b)n for a positive whole number n. You rarely need every term. Pick out the one term that gives the power of x you want.
The second part is progressions. In an arithmetic progression (AP) you add the same number d each time. In a geometric progression (GP) you multiply by the same number r each time. Each has a formula for the nth term and for the sum of the first n terms.
If the common ratio of a GP lies between −1 and 1, the terms shrink towards zero and the sum settles at a fixed value, called the sum to infinity.
Rules to remember
- In (a + b)n the term containing br is nCr × an − r × br, where nCr = n!/(r!(n − r)!)
- AP nth term = a + (n − 1)d
- AP sum: Sn = (n/2)(2a + (n − 1)d) = (n/2)(a + l), where l is the last term
- GP nth term = arn − 1
- GP sum: Sn = a(1 − rn)/(1 − r)
- Sum to infinity: S∞ = a/(1 − r), only when −1 < r < 1
- For any series, nth term = Sn − Sn − 1
Common mistake
In a binomial term, students raise only the x to the power and forget the number and sign with it. In (3 − 2x)5 the x2 term uses (−2x)2 = 4x2, not −2x2. Keep the whole term in brackets.
Worked example
Practice questions
Try each one, then open the answer.
1. An arithmetic progression has first term 3 and common difference 5. Find the sum of the first 20 terms.
- A1060
- B1010
- C980
- D505
Show answer
Answer: B. Sn = (n/2)(2a + (n − 1)d) = 10 × (6 + 19 × 5) = 10 × 101 = 1010.
2. Find the 20th term of the arithmetic progression 5, 9, 13, ...
- A85
- B77
- C81
- D80
Show answer
Answer: C. a = 5, d = 4, so u20 = a + 19d = 5 + 76 = 81. Using 20d instead of 19d gives 85.
3. Find the term independent of x in the expansion of (x + 2/x2)6.
- A15
- B240
- C20
- D60
Show answer
Answer: D. General term C(6, r) x6−r (2/x2)r = C(6, r) 2r x6−3r. The power is zero when r = 2, giving C(6, 2) × 22 = 15 × 4 = 60.
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