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O-Level Additional Mathematics · Secondary 3 and 4

Binomial Theorem

The binomial theorem gives a quick way to expand (a + b)^n for a positive integer n without multiplying the bracket out n times. In A-Math you expand such expressions, find a particular term such as the term in x⁵ or the term independent of x, and use expansions to approximate values. The coefficients come from Pascal's triangle or the nCr function on your calculator. The theorem returns in H2 Mathematics, where it is extended to other powers, and nCr is also used to count selections in probability.

What the syllabus expects

  • Use the notation nCr and evaluate it, including with a calculator
  • Expand (a + b)^n for a positive integer n using the binomial theorem
  • Use the general term T(r+1) = nCr a^(n-r) b^r to find a specified term or coefficient
  • Find the term independent of x in expansions such as (x² + 2/x)^6
  • Find a coefficient in the product of a binomial expansion with another expression, such as (1 + 2x)(1 - x)^6
  • Use an expansion to estimate a value such as (0.98)^5 by choosing a suitable value of x

The binomial theorem and nCr

For a positive integer n, (a + b)^n = a^n + nC1 a^(n-1) b + nC2 a^(n-2) b² + ... + b^n. There are n + 1 terms. In each term the powers of a and b add up to n, with the power of a falling by one each time and the power of b rising by one. The coefficient nCr is the number of ways to choose r items from n, and it equals n(n - 1)(n - 2)...(n - r + 1) divided by r(r - 1)...(2)(1). For example, 6C2 = (6 × 5)/(2 × 1) = 15 and 7C3 = (7 × 6 × 5)/(3 × 2 × 1) = 35. Useful facts are nC0 = nCn = 1, nC1 = n and nCr = nC(n - r). The coefficients also form the rows of Pascal's triangle, such as 1, 4, 6, 4, 1 for n = 4. So (1 + x)⁴ = 1 + 4x + 6x² + 4x³ + x⁴.

The general term

The term containing b^r is T(r+1) = nCr a^(n-r) b^r. It is the (r + 1)th term, because the first term has r = 0. This lets you find one term without writing the whole expansion. Put the whole of each part of the bracket, including its sign and coefficient, in place of a and b. For (2 - 3x)^5, a = 2 and b = -3x, so the term in x² is 5C2 (2)³(-3x)² = 10 × 8 × 9x² = 720x². Notice that (-3x)² = 9x², which is positive, while (-3x)³ = -27x³ is negative. The coefficient is the number in front of the power of x, so the coefficient of x² here is 720.

Terms independent of x

When the bracket contains both x and 1/x, the powers of x combine. In (x² + 2/x)^6, the general term is 6Cr (x²)^(6-r) (2/x)^r = 6Cr 2^r x^(12-2r) x^(-r) = 6Cr 2^r x^(12-3r). For the term independent of x, the power of x must be zero, so 12 - 3r = 0 and r = 4. The term is 6C4 × 2⁴ = 15 × 16 = 240. The same method finds any power. For the term in x³, set 12 - 3r = 3, which gives r = 3 and the term 6C3 × 2³ x³ = 20 × 8x³ = 160x³. If solving for r gives a value that is not a whole number from 0 to n, there is no such term in the expansion.

Products and approximations

To find a coefficient in a product such as (1 + 2x)(1 - x)^6, expand the binomial only as far as you need, then pick out the pairs of terms that give the required power. Here (1 - x)^6 = 1 - 6x + 15x² - 20x³ + ..., and the x² term of the product comes from 1 × 15x² and 2x × (-6x), giving 15x² - 12x² = 3x². Expansions also give approximations. To estimate (0.98)^5, write 0.98 = 1 - 0.02 and use (1 - x)^5 = 1 - 5x + 10x² - ... with x = 0.02. This gives (0.98)^5 ≈ 1 - 0.1 + 0.004 = 0.904. The true value is 0.90392 to 5 decimal places, so three terms are accurate enough here because x is small.

Worked example

(a) Write down the first three terms in the expansion of (1 - 2x)^7 in ascending powers of x. (b) Hence find the coefficient of x² in the expansion of (3 + x)(1 - 2x)^7.

  1. Use the binomial theorem with a = 1, b = -2x and n = 7.
  2. First term: 7C0 (1)^7 = 1.
  3. Second term: 7C1 (1)^6 (-2x) = 7 × (-2x) = -14x.
  4. Third term: 7C2 (1)^5 (-2x)² = 21 × 4x² = 84x², so (1 - 2x)^7 = 1 - 14x + 84x² + ...
  5. For (3 + x)(1 - 14x + 84x² + ...), the x² terms come from 3 × 84x² and x × (-14x).
  6. Add them: 252x² - 14x² = 238x².

Answer: (a) 1 - 14x + 84x². (b) The coefficient of x² is 238.

Common mistakes

  • Leaving out the sign or the coefficient inside the bracket, for example writing the x² term of (1 - 2x)^7 as 21x² instead of 21(-2x)² = 84x². Put brackets around the whole of b.
  • Using r = 3 for the third term. In the general term T(r+1), the third term has r = 2.
  • Writing (-3x)³ as 27x³. An odd power of a negative quantity is negative, so (-3x)³ = -27x³.
  • Giving the whole term when the question asks for the coefficient, or the other way round. The coefficient of x² in 720x² is 720.
  • Forgetting that (k/x)^r = k^r x^(-r) when finding the term independent of x. Combine all the powers of x into one index before setting it equal to zero.
  • Missing one of the pairs of terms that produce the required power in a product such as (3 + x)(1 - 2x)^7. List every pair whose powers add up to the target before multiplying.

Exam tips

  • Write the general term with the powers of x simplified into a single index, such as x^(12-3r), before you look for a particular power. This one line does most of the work.
  • Check nCr values with the calculator, but show the general term in your working so the method is clear even if a number slips.
  • In 'hence' questions, expand only as far as the power you need. Extra terms take time and add chances to make a sign error.

Practice questions

1. Find the coefficient of x³ in the expansion of (2 + x)^6.

Show answer

160. The term in x³ is 6C3 (2)³ x³ = 20 × 8x³ = 160x³.

2. Find the term independent of x in the expansion of (x - 3/x)^6.

Show answer

-540. The general term is 6Cr x^(6-r) (-3/x)^r = 6Cr (-3)^r x^(6-2r). Setting 6 - 2r = 0 gives r = 3, so the term is 6C3 (-3)³ = 20 × (-27) = -540.

3. The first three terms in the expansion of (1 + kx)^n in ascending powers of x are 1 + 20x + 180x². Find the values of n and k.

Show answer

n = 10 and k = 2. From the second term, nk = 20. From the third term, nC2 k² = n(n - 1)k²/2 = 180, so n(n - 1)k² = 360. Dividing by (nk)² = 400 gives (n - 1)/n = 0.9, so n = 10 and k = 2. Check: 10C2 × 2² = 45 × 4 = 180.

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