13.6 Binomial Theorem

The Binomial Theorem gives you a formula for expanding expressions like $(a + b)^n$ without multiplying the binomial by itself over and over. This matters most when the exponent is large, where manual expansion would take forever and invite errors.

Binomial Theorem and Its Applications

The Binomial Theorem Formula

The theorem expands any binomial $(a + b)^n$, where $n$ is a non-negative integer, into a sum of terms:

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

Each term in the expansion has three pieces:

• A binomial coefficient $\binom{n}{k}$, which tells you the numerical multiplier
• A power of $a$ that decreases from $n$ down to $0$
• A power of $b$ that increases from $0$ up to $n$

The binomial coefficient is calculated with factorials:

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

Recall that $n!$ (read "n factorial") means the product of all positive integers from $1$ to $n$. For example, $4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24$. By definition, $0! = 1$.

Expanding $(2x - 3)^4$ step by step:

1. Identify $a = 2x$, $b = -3$, and $n = 4$. Notice that the subtraction is handled by making $b$ negative.

2. Write the expansion using the formula: $(2x - 3)^4 = \sum_{k=0}^{4} \binom{4}{k} (2x)^{4-k} (-3)^k$

3. Compute each term by plugging in $k = 0, 1, 2, 3, 4$:

   • $k=0$: $\binom{4}{0}(2x)^4(-3)^0 = 1 \cdot 16x^4 \cdot 1 = 16x^4$
   • $k=1$: $\binom{4}{1}(2x)^3(-3)^1 = 4 \cdot 8x^3 \cdot (-3) = -96x^3$
   • $k=2$: $\binom{4}{2}(2x)^2(-3)^2 = 6 \cdot 4x^2 \cdot 9 = 216x^2$
   • $k=3$: $\binom{4}{3}(2x)^1(-3)^3 = 4 \cdot 2x \cdot (-27) = -216x$
   • $k=4$: $\binom{4}{4}(2x)^0(-3)^4 = 1 \cdot 1 \cdot 81 = 81$

4. Combine: $(2x-3)^4 = 16x^4 - 96x^3 + 216x^2 - 216x + 81$

Notice how the signs alternate. That always happens when $b$ is negative, because odd powers of a negative number stay negative while even powers become positive.

Finding a Specific Term

You don't always need the full expansion. To find just one term, use the fact that the $(k+1)$th term is:

$\binom{n}{k} a^{n-k} b^{k}$

The trick is figuring out which value of $k$ gives you the term you want. Since $a$ has exponent $n - k$, you solve for $k$ based on the power you're looking for.

Example: Find the coefficient of $x^3$ in $(2x - 3)^4$.

1. You need the power of $x$ to be $3$. Since $a = 2x$ and its exponent is $4 - k$, set $4 - k = 3$, giving $k = 1$.

2. Calculate the binomial coefficient: $\binom{4}{1} = \frac{4!}{1! \cdot 3!} = \frac{24}{1 \cdot 6} = 4$

3. Build the full term: $4 \cdot (2x)^3 \cdot (-3)^1 = 4 \cdot 8x^3 \cdot (-3) = -96x^3$

So the coefficient of $x^3$ is $-96$.

Pascal's Triangle for Binomial Expansion

Pascal's Triangle is a shortcut for finding binomial coefficients without computing factorials. It's a triangular array where each number is the sum of the two numbers directly above it:

</>Code
Row 0:         1
Row 1:        1  1
Row 2:       1  2  1
Row 3:      1  3  3  1
Row 4:     1  4  6  4  1

The entries in row $n$ give you the binomial coefficients for $(a + b)^n$. For instance, row 3 contains 1, 3, 3, 1, so:

$(a + b)^3 = 1a^3 + 3a^2b + 3ab^2 + 1b^3$

To use Pascal's Triangle for expansion:

1. Find the row that matches your exponent $n$
2. Write each coefficient from that row
3. Attach decreasing powers of $a$ (from $n$ down to $0$) and increasing powers of $b$ (from $0$ up to $n$)

Pascal's Triangle works well for small exponents. For larger values of $n$, the factorial formula is more practical since building the triangle row by row gets tedious.

Related Concepts in Algebraic Expansion

• Factorial ($n!$): The product of all positive integers up to $n$. Factorials are the building blocks of binomial coefficients.
• Permutation: An arrangement of objects in a specific order. Permutations use the same factorial calculations that appear in binomial coefficients.
• Polynomial: A more general expression with multiple terms (binomials are polynomials with exactly two terms). The Binomial Theorem produces a polynomial as its output.
• Algebraic expansion: The process of multiplying out grouped expressions to remove parentheses. The Binomial Theorem is a shortcut that handles this for any $(a + b)^n$.