Binomial Theorem

Q: How is the Binomial Theorem related to Pascal's Triangle?

The coefficients in the expansion of $(a+b)^n$ are exactly the entries in the $n$-th row of Pascal's Triangle. For example, row 4 is 1, 4, 6, 4, 1, which matches the coefficients of $(a+b)^4$. You can read the coefficients directly from the triangle instead of computing factorials.

Q: Does the Binomial Theorem work with subtraction, like $(a - b)^n$?

Yes. Treat $(a - b)^n$ as $(a + (-b))^n$. The theorem applies exactly the same way, but you substitute $-b$ for $b$. This means the terms alternate in sign: positive when $k$ is even, negative when $k$ is odd.

Q: How many terms are in a binomial expansion of $(a+b)^n$?

There are always $n + 1$ terms. For instance, $(a+b)^4$ has 5 terms, and $(a+b)^{10}$ has 11 terms. Each term corresponds to a value of $k$ from 0 through $n$.

Binomial Theorem

A method for distributing powers of binomials as shown below.

Formula:

Example:

(a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³+ b⁴

The coefficients are from the fourth row of Pascal's Triangle as shown below.

Key Formula

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

Where:

$a$ = The first term of the binomial
$b$ = The second term of the binomial
$n$ = The non-negative integer exponent (the power you are raising the binomial to)
$k$ = The index of summation, running from 0 to n
$\binom{n}{k}$ = The binomial coefficient, equal to n! / (k!(n−k)!), which gives the coefficient of each term

Worked Example

Problem: Expand

(x + 2)^4

using the Binomial Theorem.

Step 1: Identify the parts: here

a = x

b = 2

, and

n = 4

. Write the general expansion.

(x + 2)^4 = \sum_{k=0}^{4} \binom{4}{k} x^{4-k} \cdot 2^{k}

Step 2: Find the binomial coefficients from the 4th row of Pascal's Triangle (or compute them). They are 1, 4, 6, 4, 1.

\binom{4}{0}=1,\; \binom{4}{1}=4,\; \binom{4}{2}=6,\; \binom{4}{3}=4,\; \binom{4}{4}=1

Step 3: Write out each term by combining the coefficient, the power of

x

, and the power of 2.

1 \cdot x^4 \cdot 2^0 + 4 \cdot x^3 \cdot 2^1 + 6 \cdot x^2 \cdot 2^2 + 4 \cdot x^1 \cdot 2^3 + 1 \cdot x^0 \cdot 2^4

Step 4: Simplify each term by evaluating the powers of 2 and multiplying.

x^4 + 4 \cdot 2x^3 + 6 \cdot 4x^2 + 4 \cdot 8x + 1 \cdot 16

Step 5: Write the final expanded form.

x^4 + 8x^3 + 24x^2 + 32x + 16

Answer:

(x + 2)^4 = x^4 + 8x^3 + 24x^2 + 32x + 16

Another Example

This example shows how to find a single specific term rather than expanding the entire binomial, and it involves a negative second term, which requires careful handling of signs.

Problem: Find the 4th term in the expansion of

(3x - 1)^5

Step 1: Identify the parts:

a = 3x

b = -1

, and

n = 5

. The

k

-th term (starting from

k = 0

) is

\binom{n}{k} a^{n-k} b^{k}

. The 4th term corresponds to

k = 3

T_4 = \binom{5}{3}(3x)^{5-3}(-1)^{3}

Step 2: Calculate the binomial coefficient.

\binom{5}{3} = \frac{5!}{3! \cdot 2!} = 10

Step 3: Evaluate the powers:

(3x)^2 = 9x^2

and

(-1)^3 = -1

T_4 = 10 \cdot 9x^2 \cdot (-1) = -90x^2

Answer: The 4th term is

-90x^2

Frequently Asked Questions

How is the Binomial Theorem related to Pascal's Triangle?

The coefficients in the expansion of

(a+b)^n

are exactly the entries in the

n

-th row of Pascal's Triangle. For example, row 4 is 1, 4, 6, 4, 1, which matches the coefficients of

(a+b)^4

. You can read the coefficients directly from the triangle instead of computing factorials.

Does the Binomial Theorem work with subtraction, like

(a - b)^n

Yes. Treat

(a - b)^n

(a + (-b))^n

. The theorem applies exactly the same way, but you substitute

-b

for

b

. This means the terms alternate in sign: positive when

k

is even, negative when

k

is odd.

How many terms are in a binomial expansion of

(a+b)^n

There are always

n + 1

terms. For instance,

(a+b)^4

has 5 terms, and

(a+b)^{10}

has 11 terms. Each term corresponds to a value of

k

from 0 through

n

Binomial Theorem vs. Multinomial Theorem

	Binomial Theorem	Multinomial Theorem
What it expands	A two-term expression raised to a power: $(a+b)^n$	An expression with any number of terms raised to a power: $(a_1 + a_2 + \cdots + a_m)^n$
Coefficients	Binomial coefficients $\binom{n}{k}$	Multinomial coefficients $\frac{n!}{k_1! k_2! \cdots k_m!}$
When to use	When the base expression has exactly two terms	When the base expression has three or more terms

Why It Matters

The Binomial Theorem appears throughout algebra, precalculus, and standardized tests like the SAT and ACT whenever you need to expand or work with powers of binomials. It is also fundamental in probability and statistics, where binomial coefficients determine the number of ways to choose outcomes. In calculus and beyond, a generalized version of the theorem extends to non-integer exponents, providing the basis for important series approximations.

Common Mistakes

Mistake: Forgetting to apply the exponent to the entire second term, including its coefficient. For example, in

(x + 3)^5

, writing

3

instead of

3^k

for each term.

Correction: Always raise the entire second term to the power

k

. In

(x+3)^5

, the

k

-th term includes

3^k

, so you must compute powers like

3^2 = 9

3^3 = 27

, etc., and multiply them into the coefficient.

Mistake: Losing track of negative signs when expanding

(a - b)^n

. Students often make all terms positive.

Correction: Rewrite as

(a + (-b))^n

. Then

(-b)^k

is negative when

k

is odd and positive when

k

is even. Check the sign of every term individually.

Related Terms

Binomial — The two-term expression the theorem expands
Binomial Coefficients — The numerical coefficients in each term
Pascal's Triangle — Triangular array that lists binomial coefficients
Binomial Coefficients in Pascal's Triangle — Shows how coefficients map to triangle rows
Sigma Notation — Compact summation notation used in the formula
Power — The exponent applied to the binomial
Coefficient — The numerical factor in front of each term
Distribute — The multiplication process the theorem generalizes