Understanding the Binomial Theorem

The Binomial Theorem provides a quick and efficient way to expand a binomial expression raised to any power, such as $(a + b)^n$. Instead of manually multiplying the terms over and over, you can use a formula based on combinations.

The Binomial Formula

The general formula for expanding $(a + b)^n$ is:

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

Here's what the pieces mean:

• $n$ is the power to which the binomial is raised.
• $k$ is the index of the term, starting from $0$ and going up to $n$.
• $\binom{n}{k}$ (read as "n choose k") is the binomial coefficient, calculated as $\frac{n!}{k!(n-k)!}$.

Pascal's Triangle Connection

You don't always have to calculate $\binom{n}{k}$ manually. The coefficients for any power $n$ directly correspond to the $n$-th row of Pascal's Triangle (starting with row $0$ at the top).

Example 1: Expanding a Binomial

Problem: Expand $(2x - 3)^4$.

Solution: Here, $a = 2x$, $b = -3$, and $n = 4$. The coefficients for $n=4$ (from Pascal's Triangle or $\binom{4}{k}$) are $1, 4, 6, 4, 1$.

Let's build each term:

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

Combine them to get the final expansion: $16x^4 - 96x^3 + 216x^2 - 216x + 81$

Example 2: Finding a Specific Coefficient

Sometimes you only need a single term from the expansion, not the whole thing.

Problem: Find the coefficient of $x^3$ in $(x + 2)^7$.

Solution: The general term in the expansion is given by $\binom{n}{k} a^{n-k} b^k$. For $(x + 2)^7$, $a = x$, $b = 2$, and $n = 7$. The general term is: $\binom{7}{k} (x)^{7-k} (2)^k$

We want the term where the power of $x$ is $3$. Set the exponent equal to $3$: $7 - k = 3 \implies k = 4$

Now, plug $k = 4$ into the general term: $\binom{7}{4} (x)^3 (2)^4$

Calculate the parts:

• $\binom{7}{4} = \frac{7!}{4!3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$
• $2^4 = 16$

Multiply them together: $35 \cdot x^3 \cdot 16 = 560x^3$

The coefficient of $x^3$ is $560$.