7.3 Binomial theorem

Definition and purpose

The binomial theorem gives you a formula for expanding expressions like $(x + y)^n$ without multiplying the binomial by itself over and over. For small exponents you can expand by hand, but once $n$ gets large, the theorem saves enormous effort and reveals patterns that connect algebra to combinatorics.

Historical context

Early forms of the binomial theorem appeared in ancient India, with the mathematician Pingala (c. 200 BCE) exploring combinatorial ideas related to it. Blaise Pascal formalized the connection to his famous triangle in the 17th century. Isaac Newton later generalized the theorem to handle negative and fractional exponents, opening the door to infinite series and calculus applications.

Applications in mathematics

• Algebra: Expanding and simplifying polynomial expressions efficiently
• Probability: Calculating the likelihood of outcomes in binomial experiments (coin flips, yes/no trials)
• Combinatorics: Counting selections and arrangements, since the coefficients are literally "n choose k" values
• Calculus: Building Taylor series expansions through Newton's generalized version

Binomial expansion formula

The formula expresses $(x + y)^n$ as a sum of $n + 1$ terms. Each term pairs a binomial coefficient with specific powers of $x$ and $y$, and the exponents in every term always add up to $n$.

General form

$(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k$

The binomial coefficient $\binom{n}{k}$ is calculated as:

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

For example, expanding $(x + y)^3$:

$\binom{3}{0}x^3 + \binom{3}{1}x^2y + \binom{3}{2}xy^2 + \binom{3}{3}y^3 = x^3 + 3x^2y + 3xy^2 + y^3$

Notice how the powers of $x$ decrease from 3 to 0 while the powers of $y$ increase from 0 to 3, and every term's exponents sum to 3.

Pascal's triangle connection

Pascal's triangle provides a quick way to read off binomial coefficients without computing factorials. Row $n$ of the triangle (starting from row 0) gives the coefficients for $(x + y)^n$:

• 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

Each entry is the sum of the two entries directly above it. This reflects Pascal's rule: $\binom{n+1}{k} = \binom{n}{k-1} + \binom{n}{k}$. The symmetry you see in each row (reading the same forwards and backwards) corresponds to the property $\binom{n}{k} = \binom{n}{n-k}$.

Expanding binomial expressions

Positive integer exponents

Here's a step-by-step process for expanding $(x + y)^n$:

1. Write out $n + 1$ terms (for $(x+y)^4$, you'll have 5 terms)
2. For each term $k = 0, 1, 2, \ldots, n$, compute the binomial coefficient $\binom{n}{k}$
3. Attach $x^{n-k}$ and $y^k$ to each coefficient
4. Simplify and combine

Worked example: Expand $(2a + b)^4$

$\binom{4}{0}(2a)^4 + \binom{4}{1}(2a)^3 b + \binom{4}{2}(2a)^2 b^2 + \binom{4}{3}(2a) b^3 + \binom{4}{4} b^4$

$= 16a^4 + 32a^3b + 24a^2b^2 + 8ab^3 + b^4$

Negative and fractional exponents

When $n$ is negative or fractional, Newton's generalized binomial theorem applies. The expansion becomes an infinite series rather than a finite sum, and convergence matters: $(1 + x)^r$ converges when $|x| < 1$ for non-integer $r$. This version is used heavily in calculus for approximations and Taylor series.

Coefficients in binomial expansions

Combinatorial interpretation

The coefficient $\binom{n}{k}$ counts the number of ways to choose $k$ items from a set of $n$ items. This isn't a coincidence. When you expand $(x + y)^n$, you're choosing $x$ or $y$ from each of $n$ factors. The coefficient of $x^{n-k}y^k$ counts exactly how many ways you can pick $y$ from $k$ of those factors (and $x$ from the rest).

For instance, $\binom{5}{2} = 10$ tells you there are 10 ways to choose 2 items from 5, and it's also the coefficient of $x^3y^2$ in $(x+y)^5$.

Properties of coefficients

These properties are worth memorizing because they show up repeatedly in proofs and problem-solving:

• Symmetry: $\binom{n}{k} = \binom{n}{n-k}$ (choosing what to include is the same as choosing what to leave out)
• Sum of coefficients: $\sum_{k=0}^{n} \binom{n}{k} = 2^n$ (set $x = y = 1$ in the theorem)
• Alternating sum: $\sum_{k=0}^{n} (-1)^k \binom{n}{k} = 0$ (set $x = 1, y = -1$)
• Pascal's rule: $\binom{n+1}{k} = \binom{n}{k-1} + \binom{n}{k}$ (the recursive relationship behind Pascal's triangle)

Binomial theorem vs polynomial expansion

Efficiency comparison

Expanding $(x + y)^{10}$ by repeated multiplication requires multiplying polynomials together 9 times, with each step producing more terms to track. The binomial theorem jumps straight to the answer: 11 terms, each computed independently using $\binom{10}{k}$. This reduces the work from roughly $O(n^2)$ operations to $O(n)$ for a binomial expansion.

Limitations of binomial theorem

The theorem only applies to two-term expressions $(x + y)^n$. If you need to expand something like $(x + y + z)^n$, you'll need the multinomial theorem instead. You can sometimes work around this by grouping terms, for example treating $(x + (y + z))^n$ as a binomial and then expanding the inner binomial separately.

Applications in probability

Bernoulli trials

A Bernoulli trial is a single experiment with exactly two outcomes: success (probability $p$) or failure (probability $1 - p$). When you repeat this experiment $n$ independent times, the probability of getting exactly $k$ successes is:

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$

This formula comes directly from the binomial theorem. For example, the probability of getting exactly 3 heads in 5 fair coin flips is:

$P(X = 3) = \binom{5}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^2 = 10 \cdot \frac{1}{32} = \frac{10}{32} = 0.3125$

Binomial distribution

The full set of probabilities for $k = 0, 1, 2, \ldots, n$ successes forms the binomial distribution. Two key summary statistics:

• Mean: $\mu = np$
• Variance: $\sigma^2 = np(1-p)$

This distribution appears across many fields. In quality control, it models the number of defective items in a batch. In medicine, it models the number of patients responding to a treatment.

Generalizations and extensions

Multinomial theorem

The multinomial theorem generalizes the binomial theorem to sums of $m$ terms:

$(x_1 + x_2 + \cdots + x_m)^n = \sum_{k_1 + k_2 + \cdots + k_m = n} \frac{n!}{k_1! k_2! \cdots k_m!} x_1^{k_1} x_2^{k_2} \cdots x_m^{k_m}$

The sum runs over all ways to partition $n$ into $m$ non-negative integers. The coefficients $\frac{n!}{k_1! k_2! \cdots k_m!}$ are called multinomial coefficients and count the number of ways to arrange objects with repeated types.

Newton's generalized binomial theorem

For any real number $r$ (not just positive integers):

$(1 + x)^r = 1 + rx + \frac{r(r-1)}{2!}x^2 + \frac{r(r-1)(r-2)}{3!}x^3 + \cdots$

When $r$ is a positive integer, this series terminates after $r + 1$ terms and matches the standard theorem. For all other values of $r$, the series is infinite and converges when $|x| < 1$. This is the foundation for many approximation techniques in calculus.

Computational techniques

Manual calculation methods

1. Use Pascal's triangle for small $n$: build the triangle to row $n$ and read off coefficients
2. Write a template: list placeholders with decreasing powers of $x$ and increasing powers of $y$
3. Fill in coefficients using $\binom{n}{k}$ or Pascal's triangle
4. Simplify each term (especially when $x$ or $y$ are expressions like $2a$ or $-3b$)

For specific shortcuts: squaring a binomial gives $(x+y)^2 = x^2 + 2xy + y^2$, and cubing gives $(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$. These are worth memorizing since they come up constantly.

Software tools for expansion

Computer algebra systems like Mathematica, Maple, and SymPy can handle binomial expansions instantly, which is useful for checking your work or dealing with very large exponents. Python's math.comb(n, k) function computes binomial coefficients directly. These tools are helpful for verification, but you should be comfortable doing expansions by hand for exams.

Proofs and derivations

Algebraic proof (by induction)

The standard proof uses mathematical induction:

1. Base case: Verify $(x + y)^1 = \binom{1}{0}x + \binom{1}{1}y = x + y$. This holds.
2. Inductive hypothesis: Assume $(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k$ is true for some $n$.
3. Inductive step: Multiply both sides by $(x + y)$ and use Pascal's rule $\binom{n}{k-1} + \binom{n}{k} = \binom{n+1}{k}$ to show the formula holds for $n + 1$.

Combinatorial proof

This proof interprets $(x + y)^n$ as a product of $n$ identical factors $(x + y)$. To expand this product, you choose either $x$ or $y$ from each factor. The term $x^{n-k}y^k$ arises whenever you choose $y$ from exactly $k$ of the $n$ factors. The number of ways to make that selection is $\binom{n}{k}$, so the coefficient of $x^{n-k}y^k$ must be $\binom{n}{k}$. This argument directly establishes the theorem without induction.

Common mistakes and misconceptions

Exponent confusion

The most common error is assuming $(x + y)^n = x^n + y^n$. This is only true when $n = 1$. A quick check: $(1 + 1)^2 = 4$, but $1^2 + 1^2 = 2$. Always use the full expansion. For $n = 2$ alone, the correct result is $x^2 + 2xy + y^2$, and that middle term $2xy$ is exactly what people forget.

Term order and sign errors

When expanding $(x - y)^n$, the signs alternate because $(-y)^k$ is negative for odd $k$ and positive for even $k$:

$(x - y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} (-y)^k = \sum_{k=0}^{n} (-1)^k \binom{n}{k} x^{n-k} y^k$

For example, $(x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3$. Watch the signs carefully. Also make sure your terms are written in descending powers of $x$ (or whichever variable comes first), as scrambled ordering makes it easy to miss or duplicate terms.