Binomial Expansion (Integer Powers)

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1. Pascal's Triangle

For small powers like $(a+b)^3$ or $(a+b)^4$, the coefficients follow a simple pattern.

Interactive Challenge: Complete the 4th row of Pascal's Triangle.

1
1    1
1    2    1
1    3    3    1
1 1

Correct! The row is 1, 4, 6, 4, 1. These are the coefficients for $(a+b)^4$.

2. The $\binom{n}{r}$ Formula

For high powers (e.g., power 12), drawing the triangle is too slow. We use the formula:

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

Where $n$ is the power, and $r$ is the term number (starting from 0).

Practice: Calculate $\binom{5}{2}$ (The coefficient of the $x^3y^2$ term).

Formula: $\frac{5!}{2!3!} = \frac{120}{2 \times 6}$

Answer:

Exactly. $\binom{5}{2} = 10$. Your calculator button usually says nCr.

3. Expanding $(a+b)^n$

Let's expand: $$ (x + 2)^3 $$

Remember: Powers of $x$ go down (3, 2, 1, 0). Powers of 2 go up (0, 1, 2, 3).

$$ 1(x)^3 + \mathbf{A}(x)^2(2) + \mathbf{B}(x)(2)^2 + 1(2)^3 $$

Step 1: What are coefficients $\mathbf{A}$ and $\mathbf{B}$ from Pascal's triangle (row 3)?

$A =$ $B =$

Step 2: Simplify the numbers.

Term 2: $3 \times x^2 \times 2 = $ $x^2$

Term 3: $3 \times x \times 2^2 = 3 \times 4x = $ $x$

🎉 Final Answer

$$ x^3 + 6x^2 + 12x + 8 $$

4. Exam Style: Specific Terms

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

General Term: $\binom{4}{r} (2x)^{4-r} (-3)^r$

Step 1: Find $r$ We need $x^2$. The $x$ part is $(2x)^{4-r}$. So we need power $4-r = 2$. Therefore, $\mathbf{r = 2}$.
Step 2: Setup the calculation $$ \binom{4}{2} \times (2)^{2} \times (-3)^2 $$ Note: We separated the coefficients from the $x$.
Step 3: Calculate parts
$\binom{4}{2}$
$2^2$
$(-3)^2$
Step 4: Multiply them all $$ 6 \times 4 \times 9 = 216 $$ The coefficient is 216.