IB AA: Binomial Theorem

The Binomial Theorem provides a powerful, systematic way to expand expressions raised to a power, moving far beyond tedious repeated multiplication. It connects core areas of algebra, combinatorics, and probability, forming an essential tool for solving complex problems in both pure and applied mathematics, from calculating specific terms in an expansion to modeling real-world probabilistic scenarios.

From Patterns to a General Formula

Expanding a simple binomial like $(x+y{)}^2$ is straightforward: $x^2+2xy+y^2$. For $(x+y{)}^3$, we get $x^3+3x^{2}y+3x{y}^2+y^3$. A pattern in the coefficients emerges: 1, 2, 1 and then 1, 3, 3, 1. These are precisely the rows of Pascal's Triangle, a triangular array where each number is the sum of the two numbers directly above it. The top row (row 0) is simply 1.

$$\begin{array}{ccccccccc}& & & 1& & & & & (a+b{)}^0\\ & & 1& & 1& & & & (a+b{)}^1\\ & 1& & 2& & 1& & & (a+b{)}^2\\ 1& & 3& & 3& & 1& & (a+b{)}^3\end{array}$$

This triangle gives us the coefficients for any expansion of $(x+y{)}^n$ where $n$ is a non-negative integer. For example, row 3 gives the coefficients for $(x+y{)}^3$, as we saw. However, relying on the triangle becomes impractical for large $n$, such as finding $(x+y{)}^{10}$. This limitation leads us to a more powerful algebraic representation.

Binomial Coefficients and Combinatorics

The coefficients in the expansion are called binomial coefficients. Instead of constructing Pascal's Triangle, we can calculate any coefficient directly using combinatorial notation. The binomial coefficient, denoted $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$ or ${}^n{C}_r$, represents "the number of ways to choose $r$ objects from $n$ distinct objects" and is calculated as:

$$\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\frac{n!}{r!(n-r)!}$$

Here, $n!$ (n factorial) is the product $n\times (n-1)\times ...\times 2\times 1$. This formula is a direct computation of the entries in Pascal's Triangle. Crucially, $\left(\genfrac{}{}{0ex}{}{n}{0}\right)=\left(\genfrac{}{}{0ex}{}{n}{n}\right)=1$.

This combinatorial interpretation is key. In the expansion of $(x+y{)}^n$, the term containing $x^r{y}^{n-r}$ arises from choosing the '$x$' from $r$ of the $n$ binomial factors and the '$y$' from the remaining $(n-r)$ factors. The number of ways to make that choice is exactly $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$, which becomes the term's coefficient.

The Binomial Theorem Statement

For any positive integer $n$, the Binomial Theorem states:

$$(x+y{)}^n=(\genfrac{}{}{0ex}{}{n}{0})x^n{y}^0+(\genfrac{}{}{0ex}{}{n}{1})x^{n-1}{y}^1+(\genfrac{}{}{0ex}{}{n}{2})x^{n-2}{y}^2+...+(\genfrac{}{}{0ex}{}{n}{n})x^0{y}^n$$

Using summation notation, this is written concisely as:

$$(x+y{)}^n=\sum _{r=0}^n(\genfrac{}{}{0ex}{}{n}{r})x^{n-r}{y}^r$$

Let's apply this. To expand $(2x-3{)}^4$, we identify $a=2x$, $b=-3$, and $n=4$.

$$\begin{array}{rl}(2x-3{)}^4& =\sum _{r=0}^4\left(\genfrac{}{}{0ex}{}{4}{r}\right)(2x{)}^{4-r}(-3{)}^r\\ & =\left(\genfrac{}{}{0ex}{}{4}{0}\right)(2x{)}^4(-3{)}^0+\left(\genfrac{}{}{0ex}{}{4}{1}\right)(2x{)}^3(-3{)}^1+\left(\genfrac{}{}{0ex}{}{4}{2}\right)(2x{)}^2(-3{)}^2+\left(\genfrac{}{}{0ex}{}{4}{3}\right)(2x{)}^1(-3{)}^3+\left(\genfrac{}{}{0ex}{}{4}{4}\right)(2x{)}^0(-3{)}^4\\ & =(1)(16x^4)(1)+(4)(8x^3)(-3)+(6)(4x^2)(9)+(4)(2x)(-27)+(1)(1)(81)\\ & =16x^4-96x^3+216x^2-216x+81\end{array}$$

A common task is finding a specific term without writing the full expansion. The general term $T_{r+1}$ in the expansion of $(x+y{)}^n$ is given by:

$$T_{r+1}=\left(\genfrac{}{}{0ex}{}{n}{r}\right)x^{n-r}{y}^r$$

For example, to find the term in $x^3$ in the expansion of $(x+5{)}^7$, we need the exponent of $x$ to be 3, so $n-r=3$. With $n=7$, this gives $r=4$. Therefore, the term is: $T_5=\left(\genfrac{}{}{0ex}{}{7}{4}\right)x^3(5{)}^4=35\times x^3\times 625=21875x^3$.

Applications to Probability

The Binomial Theorem has a profound connection to probability, specifically in binomial distributions. Consider a simple binomial probability scenario: a coin is flipped $n$ times. The probability of getting exactly $r$ heads, if $p$ is the probability of heads on a single flip, is given by $\left(\genfrac{}{}{0ex}{}{n}{r}\right)p^r(1-p{)}^{n-r}$.

Notice the structure: this is exactly a single term from the binomial expansion of $(p+(1-p){)}^n$. In fact, the sum of all such probabilities from $r=0$ to $r=n$ is $(p+(1-p){)}^n=1^n=1$, confirming the total probability is 1. This direct application allows you to model any process with two independent outcomes (success/failure) over a fixed number of trials.

Extension to Fractional and Negative Exponents (HL)

At Higher Level, the Binomial Theorem is extended to cases where the exponent $n$ is not a positive integer—it can be a negative integer or a rational number (fraction). This allows for expansions like $(1+x{)}^{-2}$ or $\sqrt{1+x}=(1+x{)}^{1/2}$.

The general form for $|x|<1$ (a critical condition for convergence) becomes the infinite series:

$$(1+x{)}^n=1+nx+\frac{n(n-1)}{2!}{x}^2+\frac{n(n-1)(n-2)}{3!}{x}^3+...$$

where $n$ is any real number. The combinatorial coefficient $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$ is generalized using the formula:

$$\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\frac{n(n-1)(n-2)...(n-r+1)}{r!}\text{for }r\ge 1,\text{ and }\left(\genfrac{}{}{0ex}{}{n}{0}\right)=1.$$

For example, to find the first four terms of $(1+x{)}^{-3}$:

$$\begin{array}{rl}(1+x{)}^{-3}& =1+(-3)x+\frac{(-3)(-4)}{2!}{x}^2+\frac{(-3)(-4)(-5)}{3!}{x}^3+...\\ & =1-3x+6x^2-10x^3+...\end{array}$$

This extension is invaluable for approximations and calculus.

Common Pitfalls

1. Incorrect Application of the General Term Formula: A frequent error is misidentifying $r$. Remember, for the term containing $x^{n-r}{y}^r$, the binomial coefficient is $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$, and this is the $(r+1)$th term $T_{r+1}$. If you are asked for the 5th term, then $r+1=5$, so $r=4$.
2. Forgetting the Condition for Infinite Expansions (HL): When expanding $(1+x{)}^n$ for fractional or negative $n$, the expansion is only valid as an infinite series when $|x|<1$. Using it for $|x|>1$ leads to an incorrect, divergent series.
3. Sign Errors in Expansions: When the binomial is of the form $(a-b{)}^n$, you must treat it as $(a+(-b){)}^n$. The sign of each term will depend on $(-b{)}^r$. If $r$ is odd, the term is negative; if $r$ is even, the term is positive. Carefully include parentheses: $(2x-3{)}^4=(2x+(-3){)}^4$.
4. Confusing Combinatorial Notation: Remember that $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$ is not $\frac{n!}{r!}$, but $\frac{n!}{r!(n-r)!}$. A quick check: $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$ should always equal $\left(\genfrac{}{}{0ex}{}{n}{n-r}\right)$.

Summary

• The Binomial Theorem provides a formula for expanding $(x+y{)}^n$: $(x+y{)}^n=\sum _{r=0}^n(\genfrac{}{}{0ex}{}{n}{r})x^{n-r}{y}^r$.
• The binomial coefficients $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$ can be found via Pascal's Triangle or calculated directly using the combinatorial formula $\frac{n!}{r!(n-r)!}$.
• Finding a specific term is efficient using the general term: $T_{r+1}=\left(\genfrac{}{}{0ex}{}{n}{r}\right)x^{n-r}{y}^r$.
• The theorem is directly applied in probability to model binomial distributions, where the probability of $r$ successes in $n$ trials is a single term from a binomial expansion.
• At Higher Level, the theorem extends to fractional and negative exponents via an infinite series, valid for $|x|<1$, enabling expansions and approximations of roots and reciprocals.