The Binomial Theorem (VCE SSCE Mathematical Methods): Revision Notes

The Binomial Theorem

Introduction to binomial expansions

A binomial expression is an algebraic expression containing two terms, such as $(x + b)$. When we raise a binomial to a power, we create a binomial expansion. The binomial theorem provides a systematic way to expand expressions of the form $(x + b)^n$ without having to multiply the binomial by itself repeatedly.

Let's first look at the pattern that emerges when we expand binomial powers for small values of $n$:

$(x + b)^0 = 1$

$(x + b)^1 = 1x + 1b$

$(x + b)^2 = 1x^2 + 2xb + 1b^2$

$(x + b)^3 = 1x^3 + 3x^2b + 3xb^2 + 1b^3$

$(x + b)^4 = 1x^4 + 4x^3b + 6x^2b^2 + 4xb^3 + 1b^4$

$(x + b)^5 = 1x^5 + 5x^4b + 10x^3b^2 + 10x^2b^3 + 5xb^4 + 1b^5$

Pascal's triangle

The coefficients in binomial expansions can be arranged in a triangular pattern known as Pascal's triangle:

                1
              1   1
            1   2   1
          1   3   3   1
        1   4   6   4   1
      1   5  10  10   5   1

Each row of Pascal's triangle corresponds to the coefficients in the expansion of $(x + b)^n$, where $n$ is the row number (starting from row 0 at the top). The triangle has a beautiful property: each number is the sum of the two numbers directly above it. For example, in row 4, the number $6$ is obtained by adding $3 + 3$ from row 3.

This pattern continues indefinitely, allowing us to find coefficients for any binomial expansion. However, for large values of $n$, Pascal's triangle becomes impractical. Fortunately, there's a mathematical formula that gives us these coefficients directly.

Combinations and binomial coefficients

Pascal's triangle can be expressed using combination notation. The combination symbol $\binom{n}{r}$ (read as "n choose r") represents the number of ways to choose $r$ objects from $n$ objects, and it's calculated using the formula:

$$\binom{n}{r} = \frac{n!}{r!(n-r)!}$$

We can write Pascal's triangle using this notation:

Row 0: $\binom{0}{0}$

Row 1: $\binom{1}{0}$ $\binom{1}{1}$

Row 2: $\binom{2}{0}$ $\binom{2}{1}$ $\binom{2}{2}$

Row 3: $\binom{3}{0}$ $\binom{3}{1}$ $\binom{3}{2}$ $\binom{3}{3}$

Row 4: $\binom{4}{0}$ $\binom{4}{1}$ $\binom{4}{2}$ $\binom{4}{3}$ $\binom{4}{4}$

Row 5: $\binom{5}{0}$ $\binom{5}{1}$ $\binom{5}{2}$ $\binom{5}{3}$ $\binom{5}{4}$ $\binom{5}{5}$

This notation provides a direct way to calculate any coefficient in a binomial expansion without constructing the entire triangle.

The general binomial theorem

Using the combination notation, we can express the expansion of $(x + b)^6$ as:

$$(x + b)^6 = \binom{6}{0}x^6 + \binom{6}{1}x^5b + \binom{6}{2}x^4b^2 + \binom{6}{3}x^3b^3 + \binom{6}{4}x^2b^4 + \binom{6}{5}xb^5 + \binom{6}{6}b^6$$

This can be written more compactly using summation notation:

$$(x + b)^6 = \sum_{k=0}^{6} \binom{6}{k} x^{6-k} b^k$$

The general formula works for any positive integer $n$:

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

When the first term has a coefficient (not just $x$ but $ax$), the formula becomes:

$$(ax + b)^n = \sum_{k=0}^{n} \binom{n}{k} (ax)^{n-k} b^k$$

These formulas tell us that:

• The first term is $\binom{n}{0}(ax)^n$
• The second term is $\binom{n}{1}(ax)^{n-1}b$
• Each subsequent term follows the pattern where $k$ increases and the powers adjust accordingly

Finding a specific term in an expansion

Often we need to find just one specific term in a binomial expansion rather than expanding the entire expression. The formula for finding a specific term is:

Finding the coefficient of a specific power of $x$

Sometimes we need to find the coefficient of a particular power of $x$ in an expansion. The strategy is to:

1. Write the general term formula
2. Determine which value of $r$ gives the desired power of $x$
3. Calculate the coefficient for that term

Expanding binomial expressions

When expanding a complete binomial expression, we apply the binomial theorem formula and simplify each term. Let's look at a complete expansion.

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