Math Core

Multiplying out $(x + y)^2$ is quick, but $(x + y)^{10}$ by hand would take ten rounds of distributing and dozens of terms. The binomial theorem gives every term of $(x + y)^n$ directly, so you can expand a power in one line or jump straight to a single term you need.

Looking for the pattern

Here are the first few powers of $x + y$ :

\begin{aligned} (x+y)^0 &= 1 \\ (x+y)^1 &= x + y \\ (x+y)^2 &= x^2 + 2xy + y^2 \\ (x+y)^3 &= x^3 + 3x^2y + 3xy^2 + y^3 \\ (x+y)^4 &= x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4 \end{aligned}

Three patterns stand out in $(x+y)^n$ :

There are $n + 1$ terms.
The power of $x$ counts down from $n$ to $0$ while the power of $y$ counts up from $0$ to $n$ . In every term the exponents add to $n$ .
The coefficients are symmetric, and they come from Pascal's triangle: each entry is the sum of the two entries above it.

$n$	coefficients
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
6	1, 6, 15, 20, 15, 6, 1

Binomial coefficients

Pascal's triangle is fine for small $n$ , but you'd need ten rows (rows 0 through 9) to reach $n = 9$ . A formula gives any entry directly.

Definition

Binomial coefficient

For integers $0 \le k \le n$ , the binomial coefficient is

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

read " $n$ choose $k$ ." (Recall $n! = n(n-1)\cdots 2 \cdot 1$ and $0! = 1$ .) It is also written ${}_nC_k$ or $C(n, k)$ . The entry in row $n$ , position $k$ of Pascal's triangle (counting from $0$ ) is $\dbinom{n}{k}$ .

To compute one by hand, cancel before multiplying. For example,

\binom{8}{3} = \frac{8!}{3!\,5!} = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = 56.

In general, $\dbinom{n}{k}$ is the product of $k$ numbers counting down from $n$ , divided by $k!$ . Two facts save work: $\dbinom{n}{k} = \dbinom{n}{n-k}$ (the symmetry of the triangle) and $\dbinom{n}{0} = \dbinom{n}{n} = 1$ .

Why the coefficients count choices

When you expand $(x + y)^n = (x + y)(x + y)\cdots(x + y)$ , each term of the result comes from picking either $x$ or $y$ from each of the $n$ factors and multiplying. You get $x^{n-k}y^k$ exactly when you pick $y$ from $k$ of the factors and $x$ from the rest. The number of ways to choose which $k$ factors supply a $y$ is $\dbinom{n}{k}$ , so that's the coefficient. You'll see this "choosing" meaning again in the next lesson on counting.

The addition rule of Pascal's triangle, $\dbinom{n}{k-1} + \dbinom{n}{k} = \dbinom{n+1}{k}$ , is called Pascal's identity. You can prove it by adding the two fractions over a common denominator, and it is exactly the step an induction proof of the binomial theorem needs.

The binomial theorem

For any positive integer $n$ ,

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

The general term is $\dbinom{n}{k} x^{\,n-k} y^{\,k}$ , which is the $(k+1)$ st term.

Worked example: Expanding a binomial

Expand $(2a - 3)^4$ .

Use $x = 2a$ , $y = -3$ , $n = 4$ , and coefficients $1, 4, 6, 4, 1$ :

\begin{aligned} (2a - 3)^4 &= (2a)^4 + 4(2a)^3(-3) + 6(2a)^2(-3)^2 + 4(2a)(-3)^3 + (-3)^4 \\ &= 16a^4 + 4(8a^3)(-3) + 6(4a^2)(9) + 4(2a)(-27) + 81 \\ &= 16a^4 - 96a^3 + 216a^2 - 216a + 81. \end{aligned}

Check: at $a = 1$ , $(2 - 3)^4 = 1$ and $16 - 96 + 216 - 216 + 81 = 1$ . ✓

Common mistake

Keep the whole term in parentheses when you raise it to a power. $(2a)^3 = 8a^3$ , not $2a^3$ , and $(-3)^2 = 9$ . Signs alternate whenever the second term is negative, so check that your signs go $+, -, +, -, \dots$

Finding one term

Usually you don't need the whole expansion. Set up the general term, simplify its power of $x$ , and solve for $k$ .

Worked example: A specific coefficient

Find the coefficient of $x^5$ in $(2x - 3)^8$ .

The general term is $\dbinom{8}{k}(2x)^{8-k}(-3)^k$ . The power of $x$ is $8 - k$ , so $x^5$ needs $k = 3$ :

\binom{8}{3}(2x)^5(-3)^3 = 56 \cdot 32x^5 \cdot (-27) = -48384\,x^5.

The coefficient is $-48384$ .

Worked example: A constant term

Find the constant term of $\left(x^2 + \dfrac{1}{x}\right)^6$ .

The general term is

\binom{6}{k}(x^2)^{6-k}\left(\frac1x\right)^k = \binom{6}{k}x^{12 - 2k}x^{-k} = \binom{6}{k}x^{12 - 3k}.

The constant term has exponent $0$ : $12 - 3k = 0$ gives $k = 4$ . The term is $\dbinom{6}{4} = 15$ .

Two useful consequences

Sums of coefficients. Substituting $x = y = 1$ gives

\sum_{k=0}^{n}\binom{n}{k} = 2^n,

so every row of Pascal's triangle adds to a power of $2$ . More generally, to add all the coefficients of a polynomial, plug in $x = 1$ : the coefficients of $(3x - 1)^5$ add to $(3 - 1)^5 = 32$ .

Approximations. When $y$ is small, the later terms of $(1 + y)^n$ shrink fast, so the first few terms give a good estimate.

Worked example: Estimating a power

Estimate $(1.01)^{10}$ to four decimal places.

Write $1.01 = 1 + 0.01$ :

(1 + 0.01)^{10} = 1 + 10(0.01) + 45(0.01)^2 + 120(0.01)^3 + \cdots

\approx 1 + 0.1 + 0.0045 + 0.00012 = 1.10462.

The next term is $210(0.01)^4 = 0.0000021$ , too small to change the fourth decimal. So $(1.01)^{10} \approx 1.1046$ . A calculator gives $1.104622\ldots$ ✓

Tip

To name "the $r$ th term," use $k = r - 1$ , because the first term has $k = 0$ . The 5th term of $(x + y)^n$ is $\dbinom{n}{4}x^{n-4}y^4$ .