# Binomial Theorem

> IB Mathematics Analysis and Approaches HL · Number & Algebra
> Source: https://www.owlsprep.com/study/ib-math-aa-hl-u1-binomial-theorem/

This module covers the binomial theorem for positive integer exponents, including calculating binomial coefficients, finding specific terms in expansions, and common applications to approximation and combinatorial problems for IB AA HL.

**Prerequisites:** [Combinations and permutations](https://www.owlsprep.com/study/ib-math-aa-hl-u1-combinations-permutations/)

## Learning objectives

- Expand binomial expressions of the form $(a + b)^n$ for positive integer $n$
- Calculate binomial coefficients using combinations and Pascal's triangle
- Find specific terms and coefficients in a binomial expansion
- Apply the binomial theorem to approximation problems

## Binomial Theorem Basics and Binomial Coefficients

**Binomial Theorem** — For any positive integer $n$, the expansion of $(a + b)^n$ is the sum of terms of the form $\binom{n}{k} a^{n-k} b^k$ for $k$ from $0$ to $n$.

*Notation:* (a + b)^n, n \in \mathbb{Z}^+

*Example:* For $n=2$, $(a + b)^2 = a^2 + 2ab + b^2$

> **Pascal's Triangle Rule**
>
> Row $n$ (starting from 0) of Pascal's Triangle gives all binomial coefficients for $(a + b)^n$. First and last entry of every row is 1, each entry = sum of the two entries above it.

**Worked example:** Calculate $\binom{5}{2}$ using both the combination formula and Pascal's triangle.

1. Recall the combination formula for binomial coefficients:
2. $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$
3. Substitute $n=5, k=2$:
4. $$\binom{5}{2} = \frac{5!}{2! 3!} = \frac{120}{2 \times 6} = 10$$
5. For Pascal's triangle, row 5 (starting at $n=0$) is $[1, 5, 10, 10, 5, 1]$. The entry at index 2 is 10, matching the formula result.

> **Exam tip:** Always check that the sum of the exponents of $a$ and $b$ in any term equals $n$ for a quick error check.

## Expanding Full Binomial Expressions

For small $n$, Pascal's triangle gives quick coefficients, while for larger $n$, the combination formula is more reliable. Always include the sign of any negative term when expanding.

**Worked example:** Expand $(2x - 3y)^4$ fully.

1. Identify $a = 2x$, $b = -3y$, $n=4$. Binomial coefficients for $n=4$ are $1, 4, 6, 4, 1$.
2. Calculate each term individually:
3. $$k=0: 1(2x)^4(-3y)^0 = 16x^4$$
4. $$k=1: 4(2x)^3(-3y)^1 = -96x^3y$$
5. $$k=2: 6(2x)^2(-3y)^2 = 216x^2y^2$$
6. $$k=3: 4(2x)^1(-3y)^3 = -216xy^3$$
7. $$k=4: 1(2x)^0(-3y)^4 = 81y^4$$
8. Combine all terms to get the full expansion:
9. $$(2x - 3y)^4 = 16x^4 - 96x^3y + 216x^2y^2 - 216xy^3 + 81y^4$$

**Check your understanding**

What is the coefficient of $x^3$ in $(x + 2)^5$?

1. What is the coefficient of $x^3$ in $(x + 2)^5$?

   - 10
   - 40
   - 80
   - 20

   *Why:* The term is $\binom{5}{3}x^3(2)^2 = 10 \times 4 x^3 = 40x^3$, so the coefficient is 40.

## Finding a Specific Term or Coefficient

Most IB exam questions do not require a full expansion. Instead, they ask for the coefficient of a specific power of $x$, or the constant term, which we can find using the general term formula.

**General Term** — The $(k+1)$th term in the expansion of $(a + b)^n$ is $T_{k+1} = \binom{n}{k} a^{n-k} b^k$, where $k$ ranges from $0$ to $n$.

**Worked example:** Find the constant term in the expansion of $\left(x^2 + \frac{2}{x}\right)^6$.

1. Write the general term for this expansion:
2. $$T_{k+1} = \binom{6}{k} (x^2)^{6 - k} \left(\frac{2}{x}\right)^k$$
3. Simplify the exponent of $x$:
4. $$\text{exponent} = 2(6 - k) - k = 12 - 3k$$
5. We want the constant term, so set the exponent of $x$ equal to 0:
6. $$12 - 3k = 0 \implies k = 4$$
7. Substitute $k=4$ back to calculate the constant term:
8. $$T_5 = \binom{6}{4} (x^2)^2 \left(\frac{2}{x}\right)^4 = 15 \times x^4 \times \frac{16}{x^4} = 240$$

> **Exam tip:** Term number is $k+1$, not $k$. If asked for the 5th term, use $k=4$, not $k=5$.

## Approximation Applications

The binomial theorem is commonly used to approximate values of the form $(1 + x)^n$ when $x$ is very small, so higher power terms become negligible.

**Worked example:** Approximate $(1.02)^8$ correct to 3 decimal places using the binomial theorem.

1. Rewrite $1.02$ as $1 + 0.02$, so we expand $(1 + 0.02)^8$:
2. $$k=0: 1(1)^8(0.02)^0 = 1$$
3. $$k=1: 8(1)^7(0.02)^1 = 0.16$$
4. $$k=2: 28(1)^6(0.02)^2 = 0.0112$$
5. $$k=3: 56(1)^5(0.02)^3 = 0.000448$$
6. Terms for $k \geq 4$ are smaller than 0.000001, negligible for 3 decimal place accuracy. Sum the terms:
7. $$1 + 0.16 + 0.0112 + 0.000448 = 1.171648 \approx 1.172$$

## Common pitfalls

- **Wrong:** Forgetting to include the negative sign when expanding $(a - b)^n$, treating $b$ as positive.
  - Why it fails: This leads to incorrect signs on alternating terms in the expansion.
  - Correct: Always write $b$ as $-|b|$, so $(-|b|)^k$ gives the correct sign automatically.
- **Wrong:** Taking $k=5$ for the 5th term, instead of $k=4$.
  - Why it fails: The general term starts at $k=0$ for the first term, not $k=1$.
  - Correct: Remember $T_{k+1}$ is the $(k+1)$th term: subtract 1 from the term number to get $k$.
- **Wrong:** Incorrect exponent simplification, e.g. $(x^2)^{n-k} = x^{2 + n -k}$ instead of $x^{2(n-k)}$.
  - Why it fails: Misapplication of the exponent power rule leads to wrong $k$ and wrong coefficient.
  - Correct: Always write out the exponent simplification step by step, use $(x^m)^n = x^{mn}$.
- **Wrong:** Using the finite binomial formula for non-positive integer $n$.
  - Why it fails: Confusing the positive integer binomial theorem with the infinite general binomial series.
  - Correct: Only use the sum from $k=0$ to $n$ when $n$ is a positive integer, use the general series for other exponents.

## Cheatsheet

| Concept | Formula / Rule |
| --- | --- |
| Binomial Theorem | $(a+b)^n = \sum_{k=0}^n \binom{n}{k}a^{n-k}b^k, n \in \mathbb{Z}^+$ |
| Binomial Coefficient | $\binom{n}{k} = \frac{n!}{k!(n-k)!}, \binom{n}{0} = 1$ |
| General Term | $T_{k+1} = \binom{n}{k}a^{n-k}b^k$, $k=0$ → first term |
| Constant Term | Set exponent of $x$ = 0, solve for $k$ |

## What's next

Mastering the binomial theorem for positive integer exponents is foundational for combinatorics, probability, and the general binomial series for non-integer exponents, which is used for approximation problems in calculus and series. This topic is frequently combined with combinations to solve counting problems in IB exams. Building on this knowledge, you can explore further topics in algebra and series that build on core binomial concepts.

- [Proof techniques (HL only)](https://www.owlsprep.com/study/ib-math-aa-hl-u1-proof-techniques/)
- [Complex numbers fundamentals](https://www.owlsprep.com/study/ib-math-aa-hl-u1-complex-numbers-fundamentals/)
- [Polar and exponential form of complex numbers](https://www.owlsprep.com/study/ib-math-aa-hl-u1-polar-and-exponential-form-of/)

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