For positive integer values of $\ n$, the binomial expansion is given by: $(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$

Where:

• $\ \binom{n}{k}$ is the binomial coefficient, calculated as $\ \binom{n}{k} = \frac{n!}{k!(n-k)!} ,$
• $\ a^{n-k} \ and \ b^k$ are the terms of the expansion.

When $\ n$ is not a positive integer, the binomial expansion still applies but with an infinite series: $(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots$

This can be written more generally as:

$\ (1 + x)^n = \sum_{k=0}^{\infty} \binom{n}{k} x^k$

Where the generalised binomial coefficient $\ \binom{n}{k}$ is defined as:

$\binom{n}{k} = \frac{n(n-1)(n-2)\dots(n-k+1)}{k!}$

Example:

Let's expand $\ (1 + x)^{\frac{1}{2}}$using the general binomial expansion:

$(1 + x)^{\frac{1}{2}} = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \cdots$

Here:

• The first term is $\ 1$,
• The second term is $\ \frac{1}{2}x ,$
• The third term is $\ \frac{1}{2} \times \frac{-1}{2} \times \frac{1}{2!} x^2 = -\frac{1}{8}x^2 ,$
• And so on.

Example with Multiple Terms:

For an expression such as $\ (1 + x + y)^n$, the expansion is given by:

$(1 + x + y)^n = \sum_{k=0}^{n} \sum_{l=0}^{k} \binom{n}{k}\binom{k}{l} x^l y^{k-l}$

Here, $\ \binom{n}{k}$ is the binomial coefficient and$\ \binom{k}{l}$ is another binomial coefficient.

Example Exam Question:

Question: Expand $\ (2 - 3x)^{\frac{1}{2}}$ in ascending powers of $\ x$ up to and including the term in$\ x^2 .$

[4 marks]

Solution:

Step 1: Express in the form suitable for binomial expansion:

$(2 - 3x)^{\frac{1}{2}} = 2^{\frac{1}{2}} \left( 1 - \frac{3x}{2} \right)^{\frac{1}{2}}$

Step 2: Expand using the general binomial series for $\ (1 + u)^n :$

$\left( 1 - \frac{3x}{2} \right)^{\frac{1}{2}} = 1 + \frac{\frac{1}{2} \left(-\frac{3x}{2}\right)}{1!} + \frac{\frac{1}{2} \left(\frac{1}{2} - 1\right)\left(-\frac{3x}{2}\right)^2}{2!} + \cdots$

Simplifying:

$= 1 - \frac{3x}{4} + \frac{9x^2}{32} + \cdots$

Step 3: Multiply the expansion by $\ 2^{\frac{1}{2}} = \sqrt{2} :$

$(2 - 3x)^{\frac{1}{2}} = \sqrt{2} \left(1 - \frac{3x}{4} + \frac{9x^2}{32} \right)$

Expanding:

$= \sqrt{2} - \frac{3\sqrt{2}}{4}x + \frac{9\sqrt{2}}{32}x^2$

Final Answer:

$(2 - 3x)^{\frac{1}{2}} = \sqrt{2} - \frac{3\sqrt{2}}{4}x + \frac{9\sqrt{2}}{32}x^2$