Binomial Expansion for Advanced Mathematics

The binomial theorem is a powerful mathematical tool. For the ESAT, you must be able to calculate values of $\binom{n}{r}$ and determine any specific term in expressions of the form $(1 + x)^n$ or $(a + f(x))^n$. For example, you might be asked to find the constant term in the expansion of $(\frac{1}{x^2} - 3x^3)^{10}$. While Pascal's triangle is useful for small values of $n$, it becomes inefficient for larger powers such as $n=17$. In such cases, using the binomial expansion formula directly is far more effective.

The Binomial Expansion Formula

The expansion can be written in full as:

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

Note that an alternative form is $(a + f(x))^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} [f(x)]^k$. Both versions are identical in result because of the symmetry of the coefficients. When using this formula, focus on the following patterns:

1. The powers of the two terms in any given part of the expansion always add up to the total power $n$.
2. The value $n$ at the top of the $\binom{n}{r}$ notation is always the power to which the bracket is raised.
3. The value $r$ at the bottom can be either of the powers appearing in that specific term.

Example 1: Finding a Specific Term

Find the term with $x^7$ in the expansion of $(3 + 2x)^8$.

To build this term, we follow three steps:

First, since we need $x^7$, we identify the part of the expression involving $x$. This is $(2x)$, which must be raised to the power of 7. It is vital to remember the brackets: $(2x)^7$. This ensures both the coefficient 2 and the variable $x$ are raised to the power.

Second, we determine the power for the other term, 3. Since the sum of the powers must be 8, and we already have a power of 7, the power of 3 must be $8 - 7 = 1$. Thus, we have $3^1(2x)^7$.

Third, we apply the combination coefficient $\binom{n}{r}$. Here $n=8$. For $r$, we can use either the power of the first term (1) or the power of the second term (7). Both $\binom{8}{1}$ and $\binom{8}{7}$ give the same value. The final expression for the term is:

$\binom{8}{1} 3^1(2x)^7$

Example 2: Handling Negative Terms

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

A common mistake is to ignore the negative sign. We must treat the second term as $(-3x)$. Following our rules, for $x^5$, we need $(-3x)^5$. The power for the term 2 must be $7 - 5 = 2$. The coefficient is $\binom{7}{5}$.

The correct term is $\binom{7}{5} 2^2(-3x)^5$. Expanding this gives $\binom{7}{5} 2^2 (-3)^5 x^5$. The coefficient is the numerical part of this result.

How the Binomial Expansion Works

To understand why these coefficients appear, consider the logic of combinations. If you have five letters A, B, C, D, and E, and you want to choose a collection of 3 without regard for order, you are calculating '5 choose 3'.

Initially, you have 5 choices for the first box, 4 for the second, and 3 for the third, giving $5 \times 4 \times 3 = 60$ ways. However, this counts different orders of the same letters (like ABC and CBA) as distinct. Since there are $3! = 3 \times 2 \times 1 = 6$ ways to order any 3 letters, we divide 60 by 6 to get 10 unique collections. This is $\frac{5 \times 4 \times 3}{3 \times 2 \times 1} = \frac{5!}{3!2!} = \binom{5}{3}$.

In the general case, to choose $r$ objects from $n$, we fill $r$ boxes and leave $n-r$ empty.

The number of ways to arrange the objects is $\frac{n!}{(n-r)!r!}$. This explains the symmetry $\binom{n}{r} = \binom{n}{n-r}$: choosing $r$ objects to keep is the same as choosing $n-r$ objects to discard.

When expanding $(2 + 3x)^5$, we are effectively multiplying five brackets: $(2 + 3x)(2 + 3x)(2 + 3x)(2 + 3x)(2 + 3x)$. To get the $x^2$ term, we must choose the $(3x)$ term from 2 brackets and the 2 from the remaining 3 brackets. The number of ways to choose which 2 brackets provide the $(3x)$ is $\binom{5}{2}$, leading to the term $\binom{5}{2} 2^3 (3x)^2$.