Binomial Expansion for the TMUA

The binomial theorem is mathematically rich, and while there are many ways to approach it, for examinations like the TMUA, it is best to understand the underlying patterns rather than just memorising formulae. You are expected to be able to calculate the value of $\binom{n}{r}$ and expand expressions of the form $(1 + x)^n$ or $(a + f(x))^n$ for positive integers $n$.

Using the Binomial Expansion

For simple cases where $n$ is small, you may have used Pascal's triangle. However, for the TMUA, this is often too slow. If you need the first five terms of $(2 + 3x)^{17}$, calculating the 17th row of the triangle is inefficient. Instead, we use the general formula:

$(a + f(x))^n = \sum_{k=0}^{n} \binom{n}{k} a^k [f(x)]^{n-k}$

Note that this can also be written as $(a + f(x))^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} [f(x)]^k$. Both versions are identical because of the symmetry of the binomial coefficients. The pattern to remember is that the powers of the two components in any term must always sum to the total power $n$.

Practical Application: Finding a Specific Term

To find a specific term quickly, you should build the expression in three distinct stages.

Example 1: Find the term involving $x^7$ in the expansion of $(3 + 2x)^8$.

1. Identify the power of the $x$ component. Since we need $x^7$, we must use $(2x)^7$. It is vital to wrap this in brackets, as both the 2 and the $x$ are raised to the power of 7.
2. Identify the power of the constant. Because the total power is 8, and we have used a power of 7 for the $x$ term, the constant 3 must be raised to the power of $8 - 7 = 1$. Thus, we have $3^1(2x)^7$.
3. Apply the combination coefficient. The top number $n$ is the total power of the bracket, which is 8. The bottom number $r$ can be the power of either component (either 1 or 7), because $\binom{8}{1} = \binom{8}{7}$.

The final term is $\binom{8}{7} 3^1 (2x)^7$.

Example 2: Find the coefficient of $x^5$ in $(2 - 3x)^7$.

Following the same rules, we identify that we need $(-3x)^5$. This leads to the term $\binom{7}{5} 2^2 (-3x)^5$. A common error here is forgetting the minus sign or failing to raise $-3$ to the power of 5. The coefficient is the numerical part of this term, so the final answer would not include $x^5$.

How the Expansion Works: Combinatorics

The symbol $\binom{n}{r}$ represents 'n choose r', which is the number of ways to select a collection of $r$ objects from $n$ distinct objects without regard to order.

Consider five letters: A, B, C, D, E. If we want to choose 3 letters, we have 5 choices for the first, 4 for the second, and 3 for the third, giving $5 \times 4 \times 3 = 60$ permutations. However, since order does not matter, a set like {A, B, C} is the same as {B, A, C}. There are $3 \times 2 \times 1 = 6$ ways to order any 3 letters, so we divide our 60 choices by 6 to get 10 unique combinations. This is written as:

$\binom{5}{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = \frac{5!}{3!(5-3)!} = 10$

In general, the number of ways to fill $r$ boxes from a set of $n$ is shown conceptually here:

The formula for this is $\binom{n}{r} = \frac{n!}{r!(n-r)!}$.

Symmetry and Brackets

One useful property is symmetry: $\binom{n}{r} = \binom{n}{n-r}$. Choosing $r$ objects to keep is identical to choosing $n-r$ objects to throw away.

When we multiply out brackets like $(2 + 3x)^5$, we are effectively choosing either a 2 or a $3x$ from each of the five brackets. To find the term for $x^2$, we must choose $3x$ from exactly two brackets and 2 from the remaining three. The number of ways to choose which two brackets provide the $3x$ is $\binom{5}{2}$, which explains why the coefficient appears in the expansion.