The Binomial Theorem

Learning goals

Expand $(x+y)^n$ by choosing an $x$ or $y$ from each factor
Read $\binom{n}{k}$ as a count of those choices
State the theorem as a sum of $n + 1$ terms
Build Pascal's triangle from the addition rule
Alternate the signs for $(x - y)^n$
Pick one term by solving for $k$ from the exponent

Multiplying out is really choosing

A power like $(x + y)^7$ is not a new kind of object. It is shorthand for a product of seven identical factors:

(x + y)^7 = \underbrace{(x + y)(x + y)\cdots(x + y)}_{7 \text{ factors}}.

So expanding it is just multiplying out a product of sums, and the distributive law has told you how to do that since the beginning. Take one term from each factor, multiply your picks together, and add up the result over every possible way of picking. Nothing else appears in the product, and nothing is left out. You read a product exactly this way when you built Vieta’s formulas, where each factor $(x - r_i)$ offered you an $x$ or a $-r_i$ .

Here every factor is the same, and each one offers exactly two things: an $x$ or a $y$ . How many ways are there to pick? Build the list of picks one factor at a time. Each way of picking from the first $n - 1$ factors turns into exactly two ways once an $n$ -th factor arrives, one taking its $x$ and one taking its $y$ . No other way is possible, so every new factor doubles the list. Beginning from the single empty pick, $n$ doublings give $2^n$ ways in all. The expansion of $(x + y)^n$ is therefore a sum of $2^n$ products before a single like term is collected.

What does one of those products look like? Suppose your picks took the $y$ from exactly $k$ of the factors. Then they took the $x$ from the other $n - k$ , and since multiplication does not care about order, the product is

x^{n-k} y^{k}.

That is the whole shape of the answer. Every one of the $2^n$ products is $x^{n-k}y^k$ for some whole number $k$ between $0$ and $n$ . The exponents always add up to $n$ , and no other kind of term can possibly appear. What is left to find is how many products land on each pile.

Watch it happen for $n = 2$ . The four picks are $xx$ , $xy$ , $yx$ , and $yy$ , so

(x + y)^2 = x^2 + xy + yx + y^2 = x^2 + 2xy + y^2.

That $2$ is not decoration. It counts the two picks that took a single $y$ , one from the first factor and one from the second. Now do $n = 3$ , where there are $2^3 = 8$ picks:

Every product in the expansion of a cube is decided by which of the three factors donate a y. Three of the eight products come out as x squared times y, and that is exactly why its coefficient is 3.

Collecting the eight products by pile gives

(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3,

and the $3$ in front of $x^2y$ is not a mystery number. It is the count of ways to pick which one of the three factors donated its $y$ , and there are three factors to pick from.

That observation deserves a name, because it is the entire lesson.

Definition. For whole numbers $k$ and $n$ with $0 \le k \le n$ , write $\binom{n}{k}$ , read ” $n$ choose $k$ ”, for the number of ways to choose which $k$ of the $n$ factors donate their $y$ .

Notice what this definition is and is not. It is not a formula you have to trust. It is a description of something you can count, and you have already counted four of them by hand:

\binom{3}{0} = 1, \qquad \binom{3}{1} = 3, \qquad \binom{3}{2} = 3, \qquad \binom{3}{3} = 1.

A single pick is completely described by the set of factors that gave a $y$ . So counting the picks with exactly $k$ of the $y$ ‘s and counting the sets of $k$ factors are the same job. That is why one number answers both questions.

The Binomial Theorem

Everything from the last section now assembles into one statement.

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

The expansion of $(x+y)^n$ #

Expanding $(x + y)^n$ by the distributive law produces one product for every way of choosing an $x$ or a $y$ from each of the $n$ factors. The expansion produces each such product exactly once. So the expansion is the sum of those $2^n$ products, and the only work left is to collect the ones that are equal.

Fix a whole number $k$ with $0 \le k \le n$ . If a particular choice takes the $y$ from exactly $k$ of the factors, it must take the $x$ from the remaining $n - k$ , so its product is $x^{n-k}y^k$ . The converse holds too: a choice that takes the $y$ from $j$ factors produces $x^{n-j}y^j$ , and since $x$ and $y$ are independent variables, $x^{n-j}y^j = x^{n-k}y^k$ forces $j = k$ . So the products equal to $x^{n-k}y^k$ are precisely the choices that take a $y$ from exactly $k$ factors, no more and no fewer.

By definition there are $\binom{n}{k}$ such choices, so collecting them contributes exactly $\binom{n}{k}x^{n-k}y^k$ to the expansion.

Finally, every choice takes a $y$ from some whole number of factors between $0$ and $n$ . So letting $k$ run from $0$ to $n$ sweeps up every one of the $2^n$ products once and none of them twice. Adding the collected piles gives the stated identity.

Written out term by term, the theorem says

(x + y)^n = \binom{n}{0}x^n + \binom{n}{1}x^{n-1}y + \binom{n}{2}x^{n-2}y^2 + \cdots + \binom{n}{n}y^n,

which is $n + 1$ terms, one for each value of $k$ from $0$ to $n$ .

Three consequences come free, and each one is worth having.

The coefficients read the same backwards. Choosing which $k$ factors donate a $y$ is the same act as choosing which $n - k$ factors donate an $x$ : fix one selection and the other is decided. Two descriptions of the same act must have the same count, so

\binom{n}{k} = \binom{n}{n-k}.

The coefficients add up to $2^n$ . The theorem is an identity, true for every $x$ and $y$ , so you may substitute whatever you like. Put $x = y = 1$ :

2^n = (1 + 1)^n = \sum_{k=0}^{n} \binom{n}{k}.

That is a second proof of something you already knew, since sorting all $2^n$ picks into piles by their number of $y$ ‘s must account for all $2^n$ of them.

Substitution is a free sanity check. Setting every variable to $1$ turns an expansion into the sum of its coefficients, while the unexpanded side stays easy to evaluate. For $(2x + 3)^4$ that sum has to be $(2 + 3)^4 = 625$ . If your expanded coefficients do not add up to $625$ , you made an arithmetic slip.

Pascal’s triangle, built from addition alone

The theorem is only as useful as your ability to compute $\binom{n}{k}$ , and so far the only method on offer is to list every selection by hand. For $(x + y)^{10}$ that means writing out $2^{10} = 1024$ strings, which is not a method, it is a punishment. One short argument replaces the listing with addition.

Start with the two easy counts. There is exactly one way for no factor to donate a $y$ (every factor donates its $x$ ), and exactly one way for all of them to donate a $y$ . So for every $n$ ,

\binom{n}{0} = 1 \qquad\text{and}\qquad \binom{n}{n} = 1.

Everything in between follows from a single rule.

Pascal's rule: $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$ #

Take any $n \ge 2$ and any interior index $k$ with $1 \le k \le n - 1$ . Number the factors $1$ through $n$ and stare at the last one, factor $n$ . Any selection of $k$ factors to donate a $y$ either uses factor $n$ or it does not, and it cannot do both. Those two cases therefore split every selection into two groups that overlap nowhere and leave nothing out.

Suppose the selection does use factor $n$ . Then factor $n$ supplies one of the $k$ y’s, and the other $k - 1$ have to come from among the first $n - 1$ factors. Each such selection is built from a choice of $k - 1$ factors out of those $n - 1$ , and different choices give different selections. Every choice of $k - 1$ from the first $n - 1$ extends to exactly one selection in this group. So the group has $\binom{n-1}{k-1}$ members.

Suppose instead the selection does not use factor $n$ . Then all $k$ of its y’s come from the first $n - 1$ factors. So this group corresponds exactly to the ways of choosing $k$ factors out of $n - 1$ , and the group has $\binom{n-1}{k}$ members.

The two groups are disjoint and together they are every selection counted by $\binom{n}{k}$ , so the total is the sum of the two counts, which is the rule as stated. Both symbols on the right stay legal: $1 \le k$ makes $k - 1 \ge 0$ , and $k \le n - 1$ keeps $\binom{n-1}{k}$ inside its own range $0 \le k \le n - 1$ .

Pascal’s rule is a recurrence, and you know from the sequences lesson that a recurrence plus its starting values pins down every term. Here the starting values are the $1$ ‘s at both ends of every row. Stack the counts $\binom{n}{0}, \binom{n}{1}, \ldots, \binom{n}{n}$ as row $n$ of a triangle, and the rule says each interior entry is the sum of the two entries directly above it.

Pascal's triangle, rows 0 through 6. The edges are 1 because there is exactly one way to take no y at all and one way to take a y from every factor. Every interior entry is the sum of the two above it, so 4 + 6 = 10.

Row $n$ of the triangle is the complete list of coefficients for $(x + y)^n$ , and row $3$ is $1,\ 3,\ 3,\ 1$ , exactly what the hand count produced. The rows you will reach for most often are

\begin{aligned} n = 4:&\quad 1,\ 4,\ 6,\ 4,\ 1\\ n = 5:&\quad 1,\ 5,\ 10,\ 10,\ 5,\ 1\\ n = 6:&\quad 1,\ 6,\ 15,\ 20,\ 15,\ 6,\ 1\\ n = 7:&\quad 1,\ 7,\ 21,\ 35,\ 35,\ 21,\ 7,\ 1\\ n = 8:&\quad 1,\ 8,\ 28,\ 56,\ 70,\ 56,\ 28,\ 8,\ 1 \end{aligned}

Two habits keep you on the right row. Row $n$ always has $n + 1$ entries, and its second entry is always $n$ , since $\binom{n}{1} = n$ counts the $n$ single factors that could have donated the lone $y$ .

Check your understanding

Use Pascal's triangle. What is the coefficient of $x^2y^3$ in the expansion of $(x + y)^5$ ?

5

10

20

15

Expanding a power

To expand $(A + B)^n$ , write down row $n$ of the triangle and march $k$ from $0$ to $n$ , giving $A$ the exponent $n - k$ and $B$ the exponent $k$ . The one thing to guard is that $A$ and $B$ are whole terms, coefficients included: if $A = 2a$ , then $A^3$ is $(2a)^3 = 8a^3$ , not $2a^3$ .

Worked example 1 Expand $(2a + 3)^4$

Here $A = 2a$ , $B = 3$ , and $n = 4$ , so the coefficients come from row $4$ , which is $1,\ 4,\ 6,\ 4,\ 1$ . Write the five terms with $(2a)$ losing a power and $3$ gaining one:

(2a + 3)^4 = (2a)^4 + 4(2a)^3(3) + 6(2a)^2(3)^2 + 4(2a)(3)^3 + (3)^4.

Now evaluate each piece, keeping the whole of $2a$ inside its power:

(2a)^4 = 16a^4, \qquad 4(8a^3)(3) = 96a^3, \qquad 6(4a^2)(9) = 216a^2,

4(2a)(27) = 216a, \qquad 3^4 = 81.

So the expansion is

(2a + 3)^4 = 16a^4 + 96a^3 + 216a^2 + 216a + 81.

Check it by setting $a = 1$ . The left side is $(2 + 3)^4 = 625$ , and the right side is $16 + 96 + 216 + 216 + 81 = 625$ . The two agree, so no coefficient went missing.

Powers of a difference

Nothing new is needed for $(x - y)^n$ , because a difference is a sum in disguise. Write $x - y$ as $x + (-y)$ and let the theorem do the work:

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

The only thing to simplify is $(-y)^k = (-1)^k y^k$ , which drags one minus sign along for each factor that donated a $y$ . So

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

and since $\binom{n}{k}$ is a count it is never zero or negative. The sign of the coefficient of $x^{n-k}y^k$ is therefore the sign of $(-1)^k$ alone: the coefficient is negative exactly when $k$ is odd, and positive exactly when $k$ is even. The signs alternate $+, -, +, -$ , starting from $+$ on the pure $x^n$ term.

Worked example 2 Expand $(x - 2)^4$

Take $A = x$ and $B = -2$ , so that the theorem applies to the sum $x + (-2)$ . Row $4$ is $1,\ 4,\ 6,\ 4,\ 1$ , and the powers of $-2$ climb from $k = 0$ to $k = 4$ :

(x - 2)^4 = x^4 + 4x^3(-2) + 6x^2(-2)^2 + 4x(-2)^3 + (-2)^4.

Evaluate the powers of $-2$ , which alternate in sign: $(-2)^1 = -2$ , $(-2)^2 = 4$ , $(-2)^3 = -8$ , and $(-2)^4 = 16$ . That gives

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

The signs alternate because the odd powers of $-2$ are negative and the even ones are positive, which is the $(-1)^k$ from the derivation showing up in the arithmetic.

Check with $x = 1$ . The left side is $(1 - 2)^4 = 1$ , and the right side is $1 - 8 + 24 - 32 + 16 = 1$ . They agree, so the signs are where they belong.

Check your understanding

What is the coefficient of $x^4$ in the expansion of $(x - 1)^6$ ?

-20

20

15

-15

A closed form for the coefficients

Pascal’s triangle gives every coefficient you could want, but it makes you build all the rows below the one you need. If a problem asks for a coefficient in $(1 + x)^{20}$ , you do not want to write twenty rows. There is a formula, and you can prove it with the tools already in hand.

First, notation. For a positive whole number $m$ , the factorial $m!$ is the product of every whole number from $m$ down to $1$ :

m! = m \times (m-1) \times \cdots \times 2 \times 1, \qquad\text{so}\qquad 4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24.

The empty case is defined to be $0! = 1$ . That is a convention, chosen because it is the value that makes the formula below come out right at the edges. You will see that convention earn its keep in a moment.

The counting number $\binom{n}{k}$ equals $\dfrac{n!}{k!\,(n-k)!}$ #

Write $f(n, k) = \dfrac{n!}{k!\,(n-k)!}$ for $0 \le k \le n$ . We never assume $f$ counts anything. We show that $f$ obeys the same two rules that build the triangle, and that those rules leave no freedom at all.

The edges match. Using $0! = 1$ ,

f(n, 0) = \frac{n!}{0!\,n!} = 1 \qquad\text{and}\qquad f(n, n) = \frac{n!}{n!\,0!} = 1,

which are exactly the values of $\binom{n}{0}$ and $\binom{n}{n}$ .

Pascal’s rule holds for $f$ . Take an interior index, $1 \le k \le n-1$ , so that every factorial below is defined. Add the two entries from the row above and put them over the common denominator $k!\,(n-k)!$ , multiplying the first fraction above and below by $k$ and the second by $n - k$ :

\begin{aligned} f(n-1, k-1) + f(n-1, k) &= \frac{(n-1)!}{(k-1)!\,(n-k)!} + \frac{(n-1)!}{k!\,(n-1-k)!} \\ &= \frac{k\,(n-1)!}{k!\,(n-k)!} + \frac{(n-k)\,(n-1)!}{k!\,(n-k)!}. \end{aligned}

The two numerators now combine, and $k + (n - k) = n$ collapses them:

\frac{(n-1)!\,\big[k + (n-k)\big]}{k!\,(n-k)!} = \frac{n \cdot (n-1)!}{k!\,(n-k)!} = \frac{n!}{k!\,(n-k)!} = f(n, k).

The two rules determine everything. Row $0$ has the single entry $\binom{0}{0} = 1$ , and $f(0, 0) = 1$ as well, so the two agree there. Now suppose they agree on row $n - 1$ . In row $n$ the two end entries are $1$ for both, and every interior entry of both is the sum of the same two entries from row $n - 1$ . So the two agree on row $n$ as well. Working down one row at a time, they agree on every row, and therefore

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

for all $0 \le k \le n$ . The counting definition and the formula are the same number.

In practice you never expand the factorials in full. Cancel $(n-k)!$ against the tail of $n!$ first, which leaves $k$ factors on top and $k$ factors on the bottom.

Worked example 3 Find the coefficient of $x^3$ in $(1 + x)^{20}$

Take $A = 1$ and $B = x$ . Since $1^{20-k} = 1$ for every $k$ , the term holding $x^3$ is simply $\binom{20}{3}x^3$ , and the coefficient is $\binom{20}{3}$ .

Twenty rows of Pascal’s triangle would settle it, and so would the formula, in one line. Write out the factorials and cancel $17!$ against the tail of $20!$ :

\binom{20}{3} = \frac{20!}{3!\,17!} = \frac{20 \times 19 \times 18 \times 17!}{(3 \times 2 \times 1) \times 17!} = \frac{20 \times 19 \times 18}{6}.

Cancel before multiplying: $18 \div 6 = 3$ , so the product is $20 \times 19 \times 3$ .

\binom{20}{3} = 1140.

The coefficient of $x^3$ is $1140$ . Notice the shape of the shortcut: three factors on top counting down from $20$ , and $3!$ underneath. That pattern is what the cancellation always produces.

The same formula also explains the symmetry you spotted earlier, since swapping $k$ and $n - k$ swaps the two factorials in the denominator and leaves the value alone. Later, in the counting chapter, you will meet $\binom{n}{k}$ again under the name combinations, where this formula is proved a second time straight from counting principles. Nothing there is needed here: everything above came out of the multiplication itself.

Picking out a single term

Expanding a whole power is rarely the point. The question that actually gets asked is “what is the coefficient of $x^{12}$ ”, and expanding twenty terms to read off one of them is wasted work. The theorem lets you go straight to it, because it hands you the general term:

T_k = \binom{n}{k}\, A^{\,n-k} B^{\,k}, \qquad k = 0, 1, 2, \ldots, n.

The method is always the same three steps. Write the general term for your $A$ and $B$ . Collect all the powers of the variable into a single exponent. Set that exponent equal to the one you want, solve for $k$ , and only then evaluate.

Be careful with the counting. The term $T_k$ is the $(k+1)$ st term of the expansion, not the $k$ th, because the list starts at $k = 0$ . The fifth term of $(x + y)^{10}$ has $k = 4$ .

Worked example 4 Find the coefficient of $a^3b^5$ in $(2a - b)^8$

Here $A = 2a$ and $B = -b$ , with $n = 8$ . The general term is

T_k = \binom{8}{k} (2a)^{8-k} (-b)^{k} = \binom{8}{k}\, 2^{8-k} (-1)^k\, a^{8-k} b^{k}.

You want $b^5$ , so $k = 5$ . That also fixes the power of $a$ at $8 - 5 = 3$ , which matches the $a^3$ asked for, so such a term does exist. Now evaluate the three pieces. Row $8$ gives $\binom{8}{5} = 56$ , the power of $2$ is $2^{8-5} = 2^3 = 8$ , and $(-1)^5 = -1$ because $5$ is odd:

T_5 = 56 \cdot 8 \cdot (-1)\, a^3 b^5 = -448\,a^3b^5.

The coefficient is $-448$ . Two traps sat in that problem: the $2$ inside $2a$ had to be cubed along with the $a$ , and the odd $k$ made the sign negative.

Worked example 5 Find the constant term of $\left(x^2 + \dfrac{3}{x}\right)^{6}$

A constant term is the term whose power of $x$ is $x^0$ . You do not know in advance which $k$ produces it, so let the exponent tell you. With $A = x^2$ and $B = \dfrac{3}{x} = 3x^{-1}$ and $n = 6$ , the general term is

T_k = \binom{6}{k} (x^2)^{6-k} \left(3x^{-1}\right)^{k} = \binom{6}{k}\, 3^{k}\, x^{12 - 2k}\, x^{-k}.

Combine the two powers of $x$ by adding exponents, which is the whole reason for writing $\dfrac{3}{x}$ as $3x^{-1}$ :

T_k = \binom{6}{k}\, 3^{k}\, x^{12 - 3k}.

The term is constant when its exponent is zero:

12 - 3k = 0 \quad\Longrightarrow\quad k = 4.

Only $k = 4$ works, so there is exactly one constant term. Evaluate it with $\binom{6}{4} = 15$ from row $6$ and $3^4 = 81$ :

T_4 = 15 \cdot 81 = 1215.

The constant term is $1215$ . Nothing was expanded: one equation in $k$ replaced all seven terms.

Check your understanding

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

10

20

40

80

Common mistakes

Raising the variable but not its coefficient. In $(2a + 3)^4$ the term with $k = 1$ is $4(2a)^3(3) = 96a^3$ , not $4a^3(3)$ . Whatever sits inside the parentheses is the whole of $A$ , and the exponent lands on all of it.
Losing the alternating sign. In $(x - y)^n$ the coefficient of $x^{n-k}y^k$ carries a factor of $(-1)^k$ , so it is negative exactly when $k$ is odd. Rewrite the difference as $x + (-y)$ and let the minus sign travel inside the power, rather than trying to remember which terms it lands on.
Confusing the term number with $k$ . The list of terms starts at $k = 0$ , so the term $T_k$ is the $(k+1)$ st term. The fourth term of $(x + y)^7$ has $k = 3$ , giving $\binom{7}{3}x^4y^3$ .
Reading the wrong row. The row for $(x + y)^n$ has $n + 1$ entries and its second entry is $n$ . The coefficients of $(x + y)^6$ are $1,\ 6,\ 15,\ 20,\ 15,\ 6,\ 1$ , not the shorter row above it.
Inventing terms whose exponents do not add to $n$ . Every term of $(x + y)^7$ has its two exponents summing to $7$ , because each of the seven factors donates exactly one letter. There is no $x^4y^2$ term to find, so a search for one has gone wrong before it started.

Key takeaways

$(x + y)^n$ is a product of $n$ identical factors, and expanding it means taking an $x$ or a $y$ from each factor. So every term looks like $x^{n-k}y^k$ with the exponents summing to $n$ .
The coefficient $\binom{n}{k}$ counts the ways of choosing which $k$ of the $n$ factors donate their $y$ . That count is the definition, and everything else follows from it.
The Binomial Theorem: $(x + y)^n = \displaystyle\sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k$ , an expansion with $n + 1$ terms.
Pascal’s rule: $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$ for an interior index $1 \le k \le n-1$ , with $\binom{n}{0} = \binom{n}{n} = 1$ holding the two ends. It builds the whole triangle from addition, and row $n$ lists the coefficients of $(x + y)^n$ .
For a difference: $(x - y)^n = \sum_{k=0}^{n} (-1)^k \binom{n}{k} x^{n-k} y^k$ , so the signs alternate and the coefficient is negative exactly when $k$ is odd.
To pick out one term: write the general term $\binom{n}{k}A^{n-k}B^k$ , collect the powers of the variable into one exponent, set it equal to the target exponent, and solve for $k$ .
Closed form: $\binom{n}{k} = \dfrac{n!}{k!\,(n-k)!}$ with $0! = 1$ . Also $\binom{n}{k} = \binom{n}{n-k}$ , and the coefficients of row $n$ sum to $2^n$ .

The Binomial Theorem

Learning goals

Multiplying out is really choosing

The Binomial Theorem

Pascal’s triangle, built from addition alone

Expanding a power

Powers of a difference

A closed form for the coefficients

Picking out a single term

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)