Mastering Binomial Expansion: Your A-Level Maths Guide

What the examiner is testing

The examiner assesses your ability to accurately expand expressions of the form \( (a + bx)^n \) or \( (1 + x)^n \), including cases where \( n \) is negative or fractional. Marks are awarded for correct application of the binomial series formula, accurate calculation of coefficients, and determining the valid range of \( x \) for non-integer \( n \).

The method

1. Identify the form: Determine if the expression is \( (1 + x)^n \) or \( (a + bx)^n \). If it's the latter, factor out \( a \) to get \( a^n (1 + \frac{b}{a}x)^n \). This is crucial for using the standard binomial series.
2. Identify \( n \) and \( x \) (or the 'chunk'): From your \( (1 + \text{chunk})^n \) form, clearly identify the value of \( n \) and the entire 'chunk' that replaces \( x \) in the formula \( 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots \).
3. Substitute into the formula: Carefully substitute \( n \) and the 'chunk' into the binomial series expansion. Write out the terms one by one, paying close attention to signs and powers.
4. Simplify each term: Calculate the numerical coefficients and powers of \( x \) for each term. If you factored out \( a^n \) in step 1, multiply each term by this factor.
5. Determine the range of validity (if \( n \) is not a positive integer): For non-positive integer values of \( n \), the expansion is only valid for \( |\text{chunk}| < 1 \). Solve this inequality to find the range of \( x \) for which the expansion holds.

Worked example

Expand \( (2 + 3x)^{-2} \) up to and including the term in \( x^3 \), and state the range of values of \( x \) for which the expansion is valid.

$$ \begin{aligned} (2 + 3x)^{-2} &= \left( 2 \left( 1 + \frac{3}{2}x \right) \right)^{-2} \\ &= 2^{-2} \left( 1 + \frac{3}{2}x \right)^{-2} \\ &= \frac{1}{4} \left( 1 + (-2)\left(\frac{3}{2}x\right) + \frac{(-2)(-2-1)}{2!}\left(\frac{3}{2}x\right)^2 + \frac{(-2)(-2-1)(-2-2)}{3!}\left(\frac{3}{2}x\right)^3 + \dots \right) \\ &= \frac{1}{4} \left( 1 - 3x + \frac{(-2)(-3)}{2}\left(\frac{9}{4}x^2\right) + \frac{(-2)(-3)(-4)}{6}\left(\frac{27}{8}x^3\right) + \dots \right) \\ &= \frac{1}{4} \left( 1 - 3x + 3\left(\frac{9}{4}x^2\right) - 4\left(\frac{27}{8}x^3\right) + \dots \right) \\ &= \frac{1}{4} \left( 1 - 3x + \frac{27}{4}x^2 - \frac{27}{2}x^3 + \dots \right) \\ &= \frac{1}{4} - \frac{3}{4}x + \frac{27}{16}x^2 - \frac{27}{8}x^3 + \dots \end{aligned} $$
For the expansion to be valid, we need \( \left|\frac{3}{2}x\right| < 1 \).
$$ \begin{aligned} \left|\frac{3}{2}x\right| &< 1 \\ |x| &< \frac{2}{3} \\ -\frac{2}{3} &< x < \frac{2}{3} \end{aligned} $$
Sanity check: For small \( x \), \( (2+3x)^{-2} \approx (2)^{-2} = \frac{1}{4} \). The first term matches. The second term is negative, which makes sense as \( (2+3x)^{-2} \) is a decreasing function for small positive \( x \).

Worked example: a harder one

Given that \( f(x) = \frac{1+x}{\sqrt{4-x}} \), find the series expansion of \( f(x) \) in ascending powers of \( x \) up to and including the term in \( x^2 \). State the set of values of \( x \) for which the expansion is valid.

First attempt: Try to expand \( (1+x)(4-x)^{-1/2} \) directly without factoring. This often leads to errors because the standard formula requires the form \( (1 + \text{chunk})^n \). It's easy to forget to factor out the 4.

Correct approach:
$$ \begin{aligned} f(x) &= (1+x)(4-x)^{-1/2} \\ &= (1+x) \left( 4 \left( 1 - \frac{x}{4} \right) \right)^{-1/2} \\ &= (1+x) 4^{-1/2} \left( 1 - \frac{x}{4} \right)^{-1/2} \\ &= (1+x) \frac{1}{2} \left( 1 - \frac{x}{4} \right)^{-1/2} \end{aligned} $$
Now, expand \( \left( 1 - \frac{x}{4} \right)^{-1/2} \) up to \( x^2 \):
Let \( n = -\frac{1}{2} \) and 'chunk' \( = -\frac{x}{4} \).
$$ \begin{aligned} \left( 1 - \frac{x}{4} \right)^{-1/2} &= 1 + \left(-\frac{1}{2}\right)\left(-\frac{x}{4}\right) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)}{2!}\left(-\frac{x}{4}\right)^2 + \dots \\ &= 1 + \frac{x}{8} + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2}\left(\frac{x^2}{16}\right) + \dots \\ &= 1 + \frac{x}{8} + \frac{3}{8}\left(\frac{x^2}{16}\right) + \dots \\ &= 1 + \frac{x}{8} + \frac{3}{128}x^2 + \dots \end{aligned} $$
Now substitute this back into the expression for \( f(x) \):
$$ \begin{aligned} f(x) &= \frac{1}{2}(1+x) \left( 1 + \frac{x}{8} + \frac{3}{128}x^2 + \dots \right) \\ &= \frac{1}{2} \left( 1\left(1 + \frac{x}{8} + \frac{3}{128}x^2\right) + x\left(1 + \frac{x}{8} + \dots\right) \right) \\ &= \frac{1}{2} \left( 1 + \frac{x}{8} + \frac{3}{128}x^2 + x + \frac{x^2}{8} + \dots \right) \\ &= \frac{1}{2} \left( 1 + \left(\frac{1}{8}+1\right)x + \left(\frac{3}{128}+\frac{1}{8}\right)x^2 + \dots \right) \\ &= \frac{1}{2} \left( 1 + \frac{9}{8}x + \left(\frac{3}{128}+\frac{16}{128}\right)x^2 + \dots \right) \\ &= \frac{1}{2} \left( 1 + \frac{9}{8}x + \frac{19}{128}x^2 + \dots \right) \\ &= \frac{1}{2} + \frac{9}{16}x + \frac{19}{256}x^2 + \dots \end{aligned} $$
The expansion is valid for \( \left|-\frac{x}{4}\right| < 1 \), which means \( |x| < 4 \), so \( -4 < x < 4 \).

Practice

1. Expand \( (1 - 2x)^5 \) in ascending powers of \( x \).
2. Find the first three terms, in ascending powers of \( x \), of the binomial expansion of \( (4 + x)^{-1/2} \). State the range of values of \( x \) for which the expansion is valid.
3. Find the coefficient of \( x^3 \) in the expansion of \( (2 - 3x)^7 \).
4. Given that \( (1 + ax)(1 - 2x)^5 = 1 - 9x + bx^2 + \dots \), find the values of the constants \( a \) and \( b \).

Answers

1. \( 1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5 \)
2. \( \frac{1}{2} - \frac{1}{16}x + \frac{3}{256}x^2 \), valid for \( -4 < x < 4 \)
3. \( -15120 \)
4. First, expand \( (1 - 2x)^5 \):
   $$ \begin{aligned} (1 - 2x)^5 &= 1^5 + 5(1)^4(-2x) + \frac{5 \times 4}{2!}(1)^3(-2x)^2 + \frac{5 \times 4 \times 3}{3!}(1)^2(-2x)^3 + \dots \\ &= 1 - 10x + 10(4x^2) + 10(-8x^3) + \dots \\ &= 1 - 10x + 40x^2 - 80x^3 + \dots \end{aligned} $$
   Now multiply by \( (1 + ax) \):
   $$ \begin{aligned} (1 + ax)(1 - 10x + 40x^2 + \dots) &= 1(1 - 10x + 40x^2) + ax(1 - 10x) + \dots \\ &= 1 - 10x + 40x^2 + ax - 10ax^2 + \dots \\ &= 1 + (-10 + a)x + (40 - 10a)x^2 + \dots \end{aligned} $$
   Comparing coefficients with \( 1 - 9x + bx^2 \):
   Coefficient of \( x \): \( -10 + a = -9 \implies a = 1 \)
   Coefficient of \( x^2 \): \( b = 40 - 10a \)
   Substitute \( a=1 \): \( b = 40 - 10(1) = 30 \)
   So, \( a = 1 \) and \( b = 30 \).

The three mistakes that lose marks

1. Forgetting to factor out \( a \): When expanding \( (a + bx)^n \) for non-integer \( n \), failing to write it as \( a^n (1 + \frac{b}{a}x)^n \) before applying the formula.
   • Wrong answer produced: Using \( n \) as the power and \( bx \) as the 'chunk' directly in \( (1 + \text{chunk})^n \) formula, leading to incorrect coefficients. For example, expanding \( (2+x)^{-1} \) as \( 1 + (-1)x + \dots \) instead of \( \frac{1}{2}(1 + \frac{x}{2})^{-1} \).
2. Incorrectly calculating factorials or signs: Especially with negative or fractional \( n \), miscalculating \( \frac{n(n-1)}{2!} \) or \( \frac{n(n-1)(n-2)}{3!} \) or dropping a negative sign.
   • Wrong answer produced: \( (1-x)^{-1/2} = 1 - \frac{1}{2}x + \frac{(-\frac{1}{2})(-\frac{3}{2})}{2}x^2 = 1 - \frac{1}{2}x - \frac{3}{8}x^2 \) (should be \( + \frac{3}{8}x^2 \)).
3. Stating the wrong range of validity: Forgetting that the condition \( |x| < 1 \) applies to the entire chunk within the \( (1 + \text{chunk})^n \) form, not just \( x \).
   • Wrong answer produced: For \( (1 + 2x)^{-1} \), stating the range of validity is \( |x| < 1 \) instead of \( |2x| < 1 \implies |x| < \frac{1}{2} \).

30-second recap

Binomial expansion extends \( (1+x)^n \) to non-integer \( n \). Always convert \( (a+bx)^n \) to \( a^n(1 + \frac{b}{a}x)^n \). The expansion is \( 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots \). For non-positive integer \( n \), the expansion is valid when \( |\text{chunk}| < 1 \).