Algebra 2 · A.APR.C.5

The binomial theorem

Multiplying out (a + b)^{10} by hand takes nine long multiplications. Why are the numbers in front of its terms exactly the combination counts \binom{10}{r}, and how does that let you write down any one term at once?

You need

Pascal's triangleAlgebra 2 Multiplying two binomialsAlgebra 1

Unlocks

Combinatorial proof: counting one set two waysDiscrete Mathematics and Proof The binomial distributionGrades 11 to 12

Where you are 10 in Counting

Probability4

Counting10

▴ 5 above The fundamental counting principle Pascal's triangle Permutations Counting with restrictions

The binomial theorem

Why this exists

Multiplying two binomials gives (a + b)^2 = a^2 + 2ab + b^2. Multiplying by a + b once more gives

(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3

and each further power costs another, longer multiplication. Reaching (a + b)^{10} this way means nine rounds of multiplying and collecting like terms, with plenty of room for a slip, and a question such as “what is the coefficient of x^6 in (x - 2)^{10}?” would need the whole expansion to answer one number.

The coefficients 1, 2, 1 and 1, 3, 3, 1 are rows of Pascal’s triangle, and this is no accident. Every coefficient in the expansion of (a + b)^n is a combination count \binom{n}{r}, for a reason that is about choosing. The binomial theorem states this, and with it any power of a binomial can be written down directly, and any single term found without the others.

Definition

A binomial is a sum of two terms, such as a + b, x - 2 (which is x + (-2)) or 2x^2 + 5y.

The binomial theorem says that for any numbers a and b and any whole number n (that is, n = 0, 1, 2, \dots),

\begin{aligned} (a + b)^n &= \binom{n}{0}a^n + \binom{n}{1}a^{n - 1}b \\ &\quad + \binom{n}{2}a^{n - 2}b^2 + \cdots \\ &\quad + \binom{n}{n - 1}ab^{n - 1} + \binom{n}{n}b^n \end{aligned}

In sigma notation, which writes a sum by giving its typical term and the range of the counter r,

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

The numbers \binom{n}{r} = \dfrac{n!}{r!\,(n - r)!} are the combination counts. In this role they are called binomial coefficients. The term \binom{n}{r}a^{n - r}b^r is the general term. Because the counter starts at r = 0, the general term is the (r + 1)-th term of the expansion counting from the left.

Intuition

Write (a + b)^3 as three brackets, (a + b)(a + b)(a + b), and multiply them out with the distributive property. Each product in the full expansion is made by picking one letter from each bracket, a or b, and multiplying the three picks. There are 2 \times 2 \times 2 = 8 ways to pick, so before like terms are collected there are 8 products.

Figure 1Each term of the expanded product comes from picking a or b in every bracket. The number of picks with exactly r letters b is \binom{3}{r}, which is why the coefficients are 1, 3, 3, 1.

A pick such as b, a, a (from brackets one, two and three) gives baa, which is a^2b because the order of a product of numbers does not matter. So every pick with exactly one b gives a^2b, and there are as many of those as there are ways to choose which one of the three brackets supplies the b: \binom{3}{1} = 3. Likewise \binom{3}{2} = 3 picks have two b’s and give ab^2, and one pick each gives a^3 or b^3. Collecting:

(a + b)^3 = 1a^3 + 3a^2b + 3ab^2 + 1b^3

Nothing about the number 3 mattered. In (a + b)^n, a term a^{n - r}b^r arises once for each way to choose the r brackets, out of n, that supply a b, and that is \binom{n}{r} ways.

Results

The binomial theorem. For any numbers a, b and any whole number n,

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

The general term. The term with b^r, which is the (r + 1)-th term, is

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

Shape of the expansion. (a + b)^n has n + 1 terms. The power of a falls from n to 0 while the power of b rises from 0 to n, and in every term the two powers add to n. The coefficients are row n of Pascal’s triangle, so they read the same from either end: \binom{n}{r} = \binom{n}{n - r}.

A difference. Writing a - b as a + (-b),

(a - b)^n = \sum_{r=0}^{n} (-1)^r \binom{n}{r} a^{n - r} b^r

so the signs alternate, starting with +.

Sums of a row. Putting a = b = 1, and then a = 1, b = -1 (for n \ge 1):

\binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{n} = 2^n

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

Why it is true

By counting. Write (a + b)^n as a product of n brackets, each a + b. By the distributive property, applied over and over, the product is the sum of every product formed by taking one term, a or b, from each bracket. Each such choice is a list of n letters, so there are 2^n products.

Take a choice that picks b from exactly r brackets and a from the other n - r. By the commutative and associative properties of multiplication, its factors can be reordered and grouped as a^{n - r}b^r. So each choice gives some a^{n - r}b^r, and the choice is fixed by naming which r of the n brackets supply the b’s. The number of ways to name them is the number of r-element subsets of a set of n brackets, which is \binom{n}{r}. Collecting like terms therefore gives a^{n - r}b^r the coefficient \binom{n}{r}, for each r from 0 to n.

By Pascal’s rule. The same result follows from building each power from the one before. If row n - 1 gives the coefficients of (a + b)^{n - 1}, multiply by a + b:

(a + b)^n = a\,(a + b)^{n - 1} + b\,(a + b)^{n - 1}

A term a^{n - r}b^r in the result comes from two places: a times the term a^{n - 1 - r}b^r, whose coefficient is \binom{n - 1}{r}, and b times the term a^{n - r}b^{r - 1}, whose coefficient is \binom{n - 1}{r - 1}. So the new coefficient is \binom{n - 1}{r - 1} + \binom{n - 1}{r}, which is \binom{n}{r} by Pascal’s rule; the two end coefficients stay 1. Starting from (a + b)^1 = a + b, each power in turn has the coefficients of the next row.

The row sums come from substituting numbers: the theorem holds for every a and b, so with a = b = 1 the left side is 2^n and each term is \binom{n}{r}, and with a = 1, b = -1 the left side is 0^n = 0 when n \ge 1.

Worked examples

Example 1

A power read off the triangle

Expand (x + 2)^5.

Solution

Here a = x, b = 2 and n = 5. The coefficients are row 5 of Pascal’s triangle, 1, 5, 10, 10, 5, 1. The power of x falls from 5 to 0 while the power of 2 rises from 0 to 5:

\begin{aligned} (x + 2)^5 &= x^5 + 5x^4(2) + 10x^3(2^2) \\ &\quad + 10x^2(2^3) + 5x(2^4) + 2^5 \end{aligned}

Working out each number,

\begin{aligned} (x + 2)^5 &= x^5 + 10x^4 + 40x^3 \\ &\quad + 80x^2 + 80x + 32 \end{aligned}

Check. At x = 1 the left side is 3^5 = 243, and the right side is 1 + 10 + 40 + 80 + 80 + 32 = 243.

Example 2

Numbers inside both terms

Expand (2x - 3)^5, and find the coefficient of x^3.

Solution

Write the binomial as a sum, 2x + (-3), so a = 2x and b = -3. The whole of a is raised to each power, the 2 included, and the sign travels with b. Using row 5 again, the term with b^r is \binom{5}{r}(2x)^{5 - r}(-3)^r:

r	\binom{5}{r}	(2x)^{5 - r}	(-3)^r	term
0	1	32x^5	1	32x^5
1	5	16x^4	-3	-240x^4
2	10	8x^3	9	720x^3
3	10	4x^2	-27	-1080x^2
4	5	2x	81	810x
5	1	1	-243	-243

\begin{aligned} (2x - 3)^5 &= 32x^5 - 240x^4 + 720x^3 \\ &\quad - 1080x^2 + 810x - 243 \end{aligned}

The coefficient of x^3 is \binom{5}{2} \cdot 2^3 \cdot (-3)^2 = 10 \cdot 8 \cdot 9 = 720. It is not the binomial coefficient 10 alone: the numbers inside a and b are raised to their powers and multiplied in too. The signs alternate because odd powers of -3 are negative.

Check. At x = 1 the left side is (-1)^5 = -1, and 32 - 240 + 720 - 1080 + 810 - 243 = -1.

Example 3

One term from a long expansion

Find the coefficient of x^6 in (x - 2)^{10}, and say which term of the expansion it is.

Solution

The expansion has 11 terms, and only one is needed. With a = x, b = -2 and n = 10, the general term is

\binom{10}{r}x^{10 - r}(-2)^r

The power of x is 10 - r, so x^6 needs r = 4. Then

\binom{10}{4}(-2)^4 = 210 \times 16 = 3360

using \binom{10}{4} = \dfrac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210. The term is 3360x^6, and since r = 4 it is the fifth term, counting x^{10} as the first.

Example 4

A term with no x

Find the term of \left(x^2 - \dfrac{2}{x}\right)^6 that does not contain x. Does the expansion have a term in x?

Solution

With a = x^2 and b = -\dfrac{2}{x}, the general term is

\begin{aligned} &\binom{6}{r}(x^2)^{6 - r}\left(-\frac{2}{x}\right)^r \\ &\quad = \binom{6}{r}(-2)^r x^{12 - 3r} \end{aligned}

The power of x comes from (x^2)^{6 - r} = x^{12 - 2r}, by the power of a power rule, multiplied by x^{-r} from the r factors of \dfrac{1}{x}, so it is 12 - 2r - r = 12 - 3r.

No x means 12 - 3r = 0, so r = 4. The term is

\binom{6}{4}(-2)^4 = 15 \times 16 = 240

A term in x would need 12 - 3r = 1, that is r = \dfrac{11}{3}. The counter r only takes whole values 0 to 6, so there is no such term: the coefficient of x is 0. The powers that do occur are x^{12}, x^9, x^6, x^3, x^0, x^{-3}, x^{-6}, stepping down by 3.

Example 5

A decimal power without a calculator

Use the binomial theorem to find 1.01^{10} correct to five decimal places.

Solution

Write 1.01 = 1 + 0.01. With a = 1, every power of a is 1, so

\begin{aligned} 1.01^{10} &= 1 + \binom{10}{1}(0.01) \\ &\quad + \binom{10}{2}(0.01)^2 \\ &\quad + \binom{10}{3}(0.01)^3 + \cdots \end{aligned}

The first four terms are 1, 10 \times 0.01 = 0.1, 45 \times 0.0001 = 0.0045 and 120 \times 0.000001 = 0.00012, which add to 1.10462.

Each later term is far smaller. The next is 210 \times 0.01^4 = 0.0000021, and the one after that is 252 \times 0.01^5 = 0.0000000252. All the remaining terms are positive and together come to about 0.0000021, too little to change the fifth decimal place. So

1.01^{10} \approx 1.10462

The full value is 1.1046221\ldots. This is why a small rate of growth repeated a few times is close to “the rate times the number of times” plus a small correction: the first two terms are 1 + 10 \times 0.01, and the third, 0.0045, is the correction.

Example 6

A sum read backwards

Find the value of

\binom{6}{0} + 2\binom{6}{1} + 2^2\binom{6}{2} + \cdots + 2^6\binom{6}{6}

and say what each term counts among the 6-letter words made from the letters A, B, C.

Solution

The sum is \displaystyle\sum_{r=0}^{6}\binom{6}{r}1^{6 - r}2^r, which is the binomial theorem with a = 1, b = 2 and n = 6, read from right to left. So it equals

(1 + 2)^6 = 3^6 = 729

Check by adding the terms: 1 + 12 + 60 + 160 + 240 + 192 + 64 = 729.

What the terms count: a 6-letter word from A, B, C has 3 choices in each place, so there are 3^6 = 729 words. Sort them by how many places hold a letter other than A. If r places do, those places can be chosen in \binom{6}{r} ways, and each of them holds B or C, in 2^r ways, while the other places hold A. So the term 2^r\binom{6}{r} counts the words with exactly r letters that are not A; for example, 160 words have exactly three. The algebra and the count are the same argument: the bracket (1 + 2) is “an A, or one of two other letters”.

Boundaries

The power must be a whole number. The theorem is about (a + b)^n for n = 0, 1, 2, \dots, because it counts choices from n brackets. For a negative or fractional power, such as (1 + x)^{-1} or (1 + x)^{1/2}, there are no brackets to count, and the expansion, when it exists, never stops: that is the binomial series, which also needs x to be small.

Two terms only. The theorem expands a sum of exactly two terms. A trinomial such as (a + b + c)^n needs more than one combination count per term; it can be handled by grouping it as \big(a + (b + c)\big)^n and expanding twice.

The coefficient of a term is not always \binom{n}{r}. When a or b carries a number or a sign, those are raised to their powers too: in (2x - 3)^5 the coefficient of x^3 is 720, not 10. The binomial coefficient counts the picks; the rest of the coefficient is what each pick multiplies to.

Term number and r are off by one. The term with b^r is the (r + 1)-th term, because the first term has r = 0. And the counting of terms starts from the a^n end, so it depends on which term of the binomial is called a.

The order of the factors must not matter. The proof reorders baa into a^2b, which is allowed for numbers and for letters standing for numbers. For objects whose products depend on order, such as Matrices, (A + B)^2 is A^2 + AB + BA + B^2, and the theorem fails unless AB = BA.

The case n = 0 and powers of zero. (a + b)^0 = 1 and the theorem gives \binom{0}{0} = 1, as long as a^0 and b^0 are read as 1 even when a or b is 0. For that reason the alternating row sum is 0 only for n \ge 1: at n = 0 it is \binom{0}{0} = 1.

Where this goes next

The binomial theorem is where counting meets algebra, and both sides gain. On the counting side, an identity between combination counts can be proved by expanding the same polynomial two ways, or, in the other direction, an algebraic identity can be proved by counting one set in two ways, which is the method of combinatorial proof. On the algebra side, it makes polynomial identities with high powers manageable.

In probability, (p + q)^n with p + q = 1 expands into the chances of 0, 1, \dots, n successes in n independent tries, each term \binom{n}{r}p^r q^{n - r} being one of them: that is the binomial distribution, and it takes its name from this theorem. Allowing the power to be any real number gives the binomial series, an infinite sum that underlies many approximations like the one in Example 5.

Exercises

Expanding

Expand (x + 1)^6.
Expand (w - 4)^4.
Expand (3x + 2)^3.
Expand (2a + b)^5.
Expand \left(x - \dfrac{1}{x}\right)^4.

Finding one term

How many terms does (a + b)^{12} have, and what is the coefficient of a^5b^7?
Find the coefficient of x^3 in (x + 1)^{11}.
Find the coefficient of x^4 in (x - 2)^7.
Find the fourth term of (2x + y)^8.
Find the coefficient of x^2 in (1 - 2x)^9.
Find the middle term of (x + 2y)^6.
Find the term that does not contain x in \left(x + \dfrac{1}{x}\right)^8.
Find the term in x^5 in the expansion of \left(x^2 + \dfrac{3}{x}\right)^7.

Using the theorem

Write 101 as 100 + 1 and use the binomial theorem to find 101^4 without a calculator.
Use the first three terms of the expansion of (1 - 0.01)^8 to estimate 0.99^8.
In the expansion of (1 + kx)^8 the coefficient of x^2 is 112. Find every possible value of k.
Write \displaystyle (\sqrt{2} + 1)^4 in the form \displaystyle p + q\sqrt{2}, with p and q whole numbers.
Find \binom{10}{0} - \binom{10}{1} + \binom{10}{2} - \cdots - \binom{10}{9} + \binom{10}{10}.

Reasoning

Explain, without the factorial formula, why in (a + b)^n the coefficient of a^{n - r}b^r equals the coefficient of a^rb^{n - r}.
Explain why (a + b)^n has n + 1 terms once like terms are collected, and why in each term the powers of a and b add to n.
Explain why \binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \cdots + (-1)^n\binom{n}{n} = 0 for every n \ge 1, and why the equation is false for n = 0.
By finding the coefficient of x^n on both sides of (1 + x)^{2n} = (1 + x)^n(1 + x)^n, explain why \binom{2n}{n} = \binom{n}{0}^2 + \binom{n}{1}^2 + \cdots + \binom{n}{n}^2. Check it for n = 4.
The powers 11^2 = 121 and 11^3 = 1331 spell out rows 2 and 3 of Pascal’s triangle, but 11^5 = 161{,}051 does not spell out row 5. Explain both facts with the binomial theorem.

Answers

x^6 + 6x^5 + 15x^4 + 20x^3 + 15x^2 + 6x + 1. With b = 1 every power of b is 1, so the coefficients are row 6, 1, 6, 15, 20, 15, 6, 1, as they stand.
w^4 - 16w^3 + 96w^2 - 256w + 256. With b = -4 and row 4: 4 \times (-4) = -16, 6 \times 16 = 96, 4 \times (-64) = -256, and (-4)^4 = 256.
27x^3 + 54x^2 + 36x + 8. With a = 3x, b = 2 and row 3: (3x)^3 = 27x^3, 3(3x)^2(2) = 54x^2, 3(3x)(2^2) = 36x, and 2^3 = 8.
32a^5 + 80a^4b + 80a^3b^2 + 40a^2b^3 + 10ab^4 + b^5. The term with b^r is \binom{5}{r}2^{5 - r}a^{5 - r}b^r: 32, 5 \times 16 = 80, 10 \times 8 = 80, 10 \times 4 = 40, 5 \times 2 = 10, 1.
x^4 - 4x^2 + 6 - \dfrac{4}{x^2} + \dfrac{1}{x^4}. The general term is \binom{4}{r}x^{4 - r}\left(-\dfrac{1}{x}\right)^r = (-1)^r\binom{4}{r}x^{4 - 2r}, for r = 0 to 4.
13 terms, and the coefficient is 792. There are n + 1 = 13 terms; a^5b^7 has r = 7, and \binom{12}{7} = \binom{12}{5} = \dfrac{12 \times 11 \times 10 \times 9 \times 8}{120} = 792.
165. The term x^3 has r = 8 (the power of 1), so the coefficient is \binom{11}{8} = \binom{11}{3} = \dfrac{11 \times 10 \times 9}{6} = 165.
-280. The general term is \binom{7}{r}x^{7 - r}(-2)^r, and x^4 needs r = 3: 35 \times (-8) = -280.
1792x^5y^3. The fourth term has r = 3: \binom{8}{3}(2x)^5y^3 = 56 \times 32\,x^5y^3.
144. Here a = 1 and b = -2x, so the term in x^2 has r = 2: \binom{9}{2}(-2x)^2 = 36 \times 4x^2 = 144x^2.
160x^3y^3. There are 7 terms, so the middle one is the fourth, with r = 3: \binom{6}{3}x^3(2y)^3 = 20 \times 8\,x^3y^3.
70. The general term is \binom{8}{r}x^{8 - r}x^{-r} = \binom{8}{r}x^{8 - 2r}, and 8 - 2r = 0 gives r = 4, so the term is \binom{8}{4} = 70.
945x^5. The general term is \binom{7}{r}(x^2)^{7 - r}\left(\dfrac{3}{x}\right)^r = \binom{7}{r}3^r x^{14 - 3r}. 14 - 3r = 5 gives r = 3, and 35 \times 27 = 945.
104{,}060{,}401. With row 4: 100^4 + 4 \times 100^3 + 6 \times 100^2 + 4 \times 100 + 1, which is 100{,}000{,}000 + 4{,}000{,}000 + 60{,}000 + 400 + 1. Each part sits in its own pair of digits, so the row 1, 4, 6, 4, 1 can be read in the answer.
0.9228. The first three terms are 1, 8 \times (-0.01) = -0.08 and 28 \times 0.0001 = 0.0028, and 1 - 0.08 + 0.0028 = 0.9228. (The exact value is 0.92274\ldots.)
k = 2 or k = -2. The term in x^2 is \binom{8}{2}(kx)^2 = 28k^2x^2, so 28k^2 = 112, k^2 = 4.
\displaystyle 17 + 12\sqrt{2}. The terms are \displaystyle (\sqrt{2})^4 = 4, \displaystyle 4(\sqrt{2})^3 = 8\sqrt{2}, \displaystyle 6(\sqrt{2})^2 = 12, \displaystyle 4\sqrt{2} and 1. Collecting, 4 + 12 + 1 = 17 and \displaystyle 8\sqrt{2} + 4\sqrt{2} = 12\sqrt{2}.
0. It is \sum_{r=0}^{10}\binom{10}{r}1^{10 - r}(-1)^r = (1 - 1)^{10} = 0.
The coefficient of a^{n - r}b^r counts the ways to choose which r of the n brackets supply a b. The coefficient of a^rb^{n - r} counts the ways to choose which r brackets supply an a. Swapping the letters in every bracket matches each choice of the first kind with exactly one of the second, so the two counts are equal.
Each product in the full expansion takes exactly one letter from each of the n brackets, so it has n factors in all; if r of them are b, the other n - r are a, and the powers add to n. The number of b’s can be any of 0, 1, \dots, n, and that number fixes the term a^{n - r}b^r, so collecting like terms leaves n + 1 different terms, each with a coefficient \binom{n}{r} of at least 1, so none of them vanishes.
Put a = 1 and b = -1 in the theorem. The right side becomes the alternating sum, and the left side is (1 - 1)^n = 0^n, which is 0 for n \ge 1. For n = 0 the left side is 0^0, which the theorem reads as 1, and the sum is the single term \binom{0}{0} = 1, so the equation would say 1 = 0.
On the left, the coefficient of x^n in (1 + x)^{2n} is \binom{2n}{n}. On the right, an x^n is made by taking x^k from the first factor and x^{n - k} from the second, for each k from 0 to n, with coefficient \binom{n}{k}\binom{n}{n - k} = \binom{n}{k}^2 by symmetry. Adding over k gives the sum of squares, and two equal polynomials have equal coefficients. For n = 4: 1 + 16 + 36 + 16 + 1 = 70 = \binom{8}{4}.
11^n = (10 + 1)^n = \sum_{r=0}^{n}\binom{n}{r}10^{n - r}, so 11^n is row n written as the digits of a number in base ten, with each entry in its own place. That works when every entry is at most 9, as in rows 2 and 3: 121 and 1331. Row 5 is 1, 5, 10, 10, 5, 1, and an entry 10 does not fit in one digit, so it carries into the next place: 100{,}000 + 50{,}000 + 10{,}000 + 1{,}000 + 50 + 1, which is 161{,}051.