Math · College algebra · Concept

The binomial theorem and Pascal’s triangle

The binomial theorem expands ${(a + b)}^{n}$ into $n + 1$ terms: the term with $b^{k}$ is $C (n, k) a^{n - k} b^{k}$ . The coefficients $C (n, k)$ are row $n$ of Pascal’s triangle, and they count the ways to choose $b$ from $k$ of the $n$ factors.

A concrete case

Multiply it out, then count

${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ . Multiply once more by $(a + b)$ and the terms collect into $a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$ . The coefficients $1$ , $2$ , $1$ have become $1$ , $3$ , $3$ , $1$ .

Where does the $3$ in $3 a^{2} b$ come from? ${(a + b)}^{3}$ is three factors, $(a + b) (a + b) (a + b)$ . Each term of the product takes $a$ or $b$ from every factor. A term with exactly one $b$ takes it from the first, the second or the third factor. There are three ways to do that, and each gives $a^{2} b$ , so together they make $3 a^{2} b$ .

{(a + b)}^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}

The three ways to take one $b$ from $(a + b) (a + b) (a + b)$
First factor	Second factor	Third factor	Term
$b$	$a$	$a$	$a^{2} b$
$a$	$b$	$a$	$a^{2} b$
$a$	$a$	$b$	$a^{2} b$

The rule

The binomial theorem

For any numbers $a$ and $b$ and any whole number $n$ , ${(a + b)}^{n}$ has $n + 1$ terms. The powers of a fall from $n$ to $0$ while the powers of $b$ rise from $0$ to $n$ , so the exponents in every term add to $n$ . The coefficient of $a^{n - k} b^{k}$ is the binomial coefficient $C (n, k)$ , read “ $n$ choose $k$ ” and also written with $n$ above $k$ in parentheses.

{(a + b)}^{n} = \sum_{k = 0}^{n} (\begin{matrix} n \\ k \end{matrix}) a^{n - k} b^{k}

Binomial coefficients and Pascal’s triangle

$C (n, k)$ $= n! / (k! (n - k)!)$ , where $n! = n \cdot (n - 1) \dots 2 \cdot 1$ and $0! = 1$ . For example $C (4, 2) = 24 / (2 \cdot 2) = 6$ .

The same numbers fill Pascal’s triangle. Row $n$ lists $C (n, 0)$ through $C (n, n)$ , and the top row is row 0. Each row begins and ends with $1$ , and each inner entry is the sum of the two entries above it.

(\begin{matrix} n \\ k \end{matrix}) = \frac{n!}{k! (n - k)!}

The first rows of Pascal’s triangle
$n$	Row $n$
$0$	$1$
$1$	$1$ $1$
$2$	$1$ $2$ $1$
$3$	$1$ $3$ $3$ $1$
$4$	$1$ $4$ $6$ $4$ $1$
$5$	$1$ $5$ $10$ $10$ $5$ $1$

Why it works

Why the coefficients count choices

Expand ${(a + b)}^{n}$ without collecting anything. Every term is a product of $n$ letters, one from each factor. A term equal to $a^{n - k} b^{k}$ takes $b$ from exactly $k$ of the $n$ factors and $a$ from the rest. The number of such terms is the number of ways to choose which $k$ factors give $b$ , and that count is $C (n, k)$ .

Pascal’s rule follows from the same count. To choose $k$ of $n$ factors, either the last factor is chosen or it is not. If it is, the other $k - 1$ come from the first $n - 1$ factors, in $C (n - 1, k - 1)$ ways. If it is not, all $k$ come from the first $n - 1$ , in $C (n - 1, k)$ ways. Together they give $C (n, k)$ , which is why each entry is the sum of the two above it.

(\begin{matrix} n \\ k \end{matrix}) = (\begin{matrix} n - 1 \\ k - 1 \end{matrix}) + (\begin{matrix} n - 1 \\ k \end{matrix})

The method

Expanding a binomial

Name $a$ , $b$ and $n$ first, with every sign and coefficient inside $a$ and $b$ . The table works through ${(x - 2)}^{3}$ , the first worked example below.

The steps on ${(x - 2)}^{3}$
Step	What to do	For ${(x - 2)}^{3}$
1	Name $a$ , $b$ and $n$ , keeping signs	$a = x$ , $b = - 2$ , $n = 3$
2	Read the coefficients from row $n$	$1$ , $3$ , $3$ , $1$
3	Write each term with $a$ and $b$ in parentheses	$x^{3} + 3 x^{2} (- 2) + 3 x {(- 2)}^{2} + {(- 2)}^{3}$
4	Simplify each term	$x^{3} - 6 x^{2} + 12 x - 8$
5	Check with a number	At $x = 3$ both sides give $1$

Finding one term

The term with $b^{k}$ is $C (n, k) a^{n - k} b^{k}$ . Find $k$ from the power you want, then compute only that term. In ${(2 x + 1)}^{5}$ the power of $x$ in a term is $5 - k$ , so $x^{3}$ needs $k = 2$ , and the term is $C (5, 2) {(2 x)}^{3} {(1)}^{2}$ $= 10 \cdot 8 x^{3}$ $= 80 x^{3}$ . Pascal’s triangle is quicker for a whole expansion, and the formula is quicker for one term.

(\begin{matrix} 5 \\ 2 \end{matrix}) {(2 x)}^{3} {(1)}^{2} = 80 x^{3}

Where it breaks

${(a + b)}^{n}$ is not $a^{n} + b^{n}$ once $n$ is $2$ or more. ${(1 + 1)}^{3} = 8$ , while $1^{3} + 1^{3} = 2$ . The missing $6$ is the middle terms, $3 + 3$ .

A coefficient inside a term is raised with it. In ${(2 x + 1)}^{5}$ , $a = 2 x$ , so $a^{3} = 8 x^{3}$ , not $2 x^{3}$ . A minus sign belongs to $b$ . In ${(x - 2)}^{3}$ , $b = - 2$ , and its odd powers are negative, so the signs alternate.

The theorem as written needs a whole-number power. ${(1 + x)}^{1 / 2}$ has no last term. For small $x$ it is approximated by an endless sum, $1 + x / 2 - x^{2} / 8 + \dots$ , which a later course builds.

$C (n, k)$ needs $0 \leq k \leq n$ . Row 5 has six entries, $C (5, 0)$ to $C (5, 5)$ , so there is no seventh term to ask for.

Check your answer

An expansion holds for every $x$ , so the easy values of $x$ test it.

Add the coefficients. Setting every variable to $1$ turns ${(x + 1)}^{n}$ into $2^{n}$ , so the entries of row $n$ add to $2^{n}$ . Row 4 gives $1 + 4 + 6 + 4 + 1$ $= 16$ $= 2^{4}$ .

Put in a value that makes the binomial easy. $x = 3$ makes ${(x - 2)}^{3} = 1$ , so the expansion must give $27 - 54 + 36 - 8 = 1$ too.

Count the terms and the exponents. There are $n + 1$ terms, and the exponents in each add to $n$ .

To have Chalk check the work, write the binomial power on the first row of a Math box, the terms with $a$ and $b$ substituted on the next, and the simplified expansion below. The Checker compares each row with the one above. A lost sign reads ✗ with the values that disagree, such as “Not equivalent. At x = 2, the lines give 0 and 64.” For a single term, start the box at the term itself, such as $C (5, 2) {(2 x)}^{3} {(1)}^{2}$ , and simplify it row by row.

Counting, chance and other powers

The same coefficients count choices anywhere. The number of 5-card hands from a 52-card deck is $C (52, 5) = 2, 598, 960$ . The chance of exactly $k$ heads in $n$ fair coin tosses is $C (n, k) / 2^{n}$ , the binomial distribution. Exactly $2$ heads in $4$ tosses has chance $6 / 16 = 0.375$ .

Newton found that the expansion of ${(1 + x)}^{r}$ works for any real power $r$ when $x$ is between $- 1$ and $1$ . Its coefficients are $r (r - 1) \dots (r - k + 1) / k!$ . For a whole number $r$ they become zero once $k$ passes $r$ , so the sum stops, and the result is the theorem on this page.

Common mistakes

Dropping the middle terms. ${(a + b)}^{2}$ is $a^{2} + 2 a b + b^{2}$ , not $a^{2} + b^{2}$ . At $a = b = 1$ the first gives $4$ and the second $2$ , so the $2 a b$ must be there.
Losing the sign of $b$ . In ${(x - 2)}^{3}$ , $b = - 2$ , so the terms alternate, $x^{3} - 6 x^{2} + 12 x - 8$ . Writing every term positive gives $x^{3} + 6 x^{2} + 12 x + 8$ , which is ${(x + 2)}^{3}$ .
Forgetting to raise a coefficient. ${(2 x)}^{3}$ is $8 x^{3}$ , not $2 x^{3}$ , because the cube applies to the $2$ as well as the $x$ .
Using the wrong row. The power $3$ uses row 3, which is $1$ , $3$ , $3$ , $1$ . Row 3 is the fourth line of the triangle, because the top line is row 0.
Matching $k$ to the wrong power. In ${(2 x + 1)}^{5}$ the $x^{3}$ term has $k = 2$ , since the power of $x$ is $5 - k$ . Choosing $k = 3$ gives the $x^{2}$ term, $40 x^{2}$ , instead.
Reporting the term when the coefficient is asked for. The $x^{3}$ term of ${(2 x + 1)}^{5}$ is $80 x^{3}$ , and its coefficient is $80$ . Read which one the question wants.

Key terms

Binomial theorem: For a whole number $n$ , ${(a + b)}^{n}$ $= Σ C (n, k) a^{n - k} b^{k}$ for $k = 0$ to $n$ : each term chooses $b$ from $k$ of the $n$ factors, in $C (n, k)$ ways.
Pascal’s triangle: A triangular array with 1s down both edges in which every other entry is the sum of the two above it. Counting the top row as row 0, row $n$ lists the binomial coefficients $C (n, 0)$ to $C (n, n)$ .
Combination count: The number of ways to choose $k$ objects from $n$ when order doesn’t matter: $C (n, k)$ $= n! / (k! (n - k)!)$ . For example, $2$ people can be chosen from $4$ in $6$ ways.
Factorial: For a positive integer $n$ , $n!$ is the product of the integers from $1$ through $n$ , and $0!$ is defined as $1$ . Factorials count arrangements of distinct objects.
Exponent: The raised number in a power, telling how many times to multiply the base: $2^{3} = 2 \cdot 2 \cdot 2$ . A negative exponent means a reciprocal $(2^{- 3} = 1 / 8)$ , and a fractional one means a root.

Work through an example

Expand ${(x - 2)}^{3}$ .

Expand

{(x - 2)}^{3}

with the binomial theorem →

Find one term of a binomial expansion →

Sources and scope

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After reading

Study it, then try it

Practice the key terms as flashcards, then open the example in its tool, change a value and keep a useful result on your board.

Study the 5 key terms Explore Pascal’s Triangle Check the expansion in Math See row 3 as paths in Binomial & the Galton Board Open worked example on a board The binomial theorem in Math Reference

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