The Binomial Theorem

Pascal's triangle, binomial coefficient identities, the binomial theorem and its general term, finding a coefficient or the constant term, and proving identities by substitution, comparing coefficients or counting.

1What you need to know
2Pascal's triangle
3The binomial theorem
4Proving identities
5Common mistakes
6Exam tips
7Links to other topics

1.What you need to know

The binomial theorem is a short focus area, usually worth 1–3 marks: typically finding a coefficient or a constant term, sometimes proving an identity. It follows directly from Permutations and combinations (ME11-PC).

A binomial is an expression with two terms, such as $x + y$ or $2 x - \frac{1}{x}$ .
The coefficients of $(x + y)^{n}$ for small $n$ form Pascal's triangle; the coefficient of $x^{n - r} y^{r}$ is $(r n)$ .
Use $(0 n) = (n n) = 1$ , the symmetry identity and Pascal's identity to simplify expressions.
Derive and apply the binomial theorem for positive integer $n$ .
Find the coefficient of a given power of $x$ , or the constant term.
Prove identities involving binomial coefficients by substituting values, comparing coefficients or a counting argument (no calculus is needed or expected in this focus area).

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2.Pascal's triangle

Expanding $(x + y)^{n}$ for $n = 0, 1, 2, \dots$ gives these coefficients:

$n$	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

Each row starts and ends with 1, is symmetrical, and every inside entry is the sum of the two above it. In binomial notation these facts are

(0 n) (r n) (r n) = (n n) = 1 = (n - r n) = (r - 1 n - 1) + (r n - 1)

The last line is Pascal's identity (for $1 \leq r \leq n - 1$ ). Row $n$ of the triangle lists $(0 n), (1 n), \dots, (n n)$ .

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3.The binomial theorem

When you expand $(x + y)^{n}$ as a product of $n$ brackets $(x + y)$ , each term comes from choosing $x$ or $y$ from each of the $n$ brackets. A term $x^{n - r} y^{r}$ arises from choosing $y$ from exactly $r$ brackets, which can be done in $(r n)$ ways. So

(x + y)^{n} = r = 0 \sum n (r n) x^{n - r} y^{r}

Written out,

(x + y)^{n} = x^{n} + (1 n) x^{n - 1} y + (2 n) x^{n - 2} y^{2} + \dots + y^{n}

There are $n + 1$ terms, the power of $x$ falls by 1 each time, and the powers in every term add to $n$ .

The general term (the term containing $y^{r}$ ) is

T_{r} = (r n) x^{n - r} y^{r}

Be careful: with this labelling $r$ starts at 0, so $T_{0}$ is the first term. Some textbooks call it $T_{r + 1}$ . Either is fine if you are consistent.

When the binomial has coefficients or a negative sign, put the whole term in brackets: for $(2 x - 3)^{n}$ use $x \to 2 x$ and $y \to - 3$ , so the general term is $(r n) (2 x)^{n - r} (- 3)^{r}$ .

Worked example 1: a coefficient

Find the coefficient of $x^{5}$ in the expansion of $(1 + 2 x)^{8}$ .

The general term is

(r 8) 1^{8 - r} (2 x)^{r} = (r 8) 2^{r} x^{r}

For $x^{5}$ we need $r = 5$ :

(5 8) 2^{5} = 56 \times 32 = 1792

The coefficient of $x^{5}$ is $1792$ . (The coefficient does not include the $x^{5}$ itself.)

Worked example 2: the constant term

Find the term independent of $x$ in the expansion of $(x + \frac{2}{x ^{2}})^{9}$ .

Write the general term and collect the powers of $x$ :

T_{r} = (r 9) x^{9 - r} (\frac{2}{x ^{2}})^{r} = (r 9) 2^{r} x^{9 - r} x^{- 2 r} = (r 9) 2^{r} x^{9 - 3 r}

The term is independent of $x$ when $9 - 3 r = 0$ , so $r = 3$ :

T_{3} = (3 9) 2^{3} = 84 \times 8 = 672

The method always works the same way: write the power of $x$ in the general term as a function of $r$ , set it equal to the power you want, and solve. If $r$ is not a whole number between 0 and $n$ , there is no such term (and the coefficient is 0).

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4.Proving identities

There are three standard methods.

1. Substitute a value

Start from $(1 + x)^{n} = \sum_{r = 0}^{n} (r n) x^{r}$ and choose $x$ to make the sum you want.

Worked example 3: sums of coefficients

Prove that $(0 n) + (1 n) + \dots + (n n) = 2^{n}$ , and hence that the sum of the coefficients with even $r$ , $(0 n) + (2 n) + (4 n) + \dots$ , equals $2^{n - 1}$ for $n \geq 1$ .

Substitute $x = 1$ into $(1 + x)^{n} = \sum_{r = 0}^{n} (r n) x^{r}$ :

2^{n} = (0 n) + (1 n) + (2 n) + \dots + (n n)

Now substitute $x = - 1$ . For $n \geq 1$ the left side is $0^{n} = 0$ :

0 = (0 n) - (1 n) + (2 n) - \dots

Add the two equations. The terms with odd $r$ cancel and the terms with even $r$ double:

2^{n} 2^{n - 1} = 2 [(0 n) + (2 n) + \dots] = (0 n) + (2 n) + \dots

which is what we wanted to prove.

2. Compare coefficients

Expand the same expression in two ways and equate the coefficients of one power of $x$ . For example, $(1 + x)^{2 n} = (1 + x)^{n} (1 + x)^{n}$ . The coefficient of $x^{n}$ on the left is $(n 2 n)$ . On the right, an $x^{n}$ term comes from $x^{r}$ in the first bracket times $x^{n - r}$ in the second, so its coefficient is $\sum_{r = 0}^{n} (r n) (n - r n)$ . Using symmetry, this proves

(n 2 n) = r = 0 \sum n (r n)^{2}

3. A counting argument

Count one situation in two ways. For the identity above: choose $n$ people from a group of $n$ boys and $n$ girls. Directly this is $(n 2 n)$ . Alternatively, choose $r$ boys and $n - r$ girls for each $r$ , giving $\sum (r n) (n - r n)$ . Both count the same thing, so they are equal.

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5.Common mistakes

Leaving the coefficient or sign out of a term: $(2 x)^{3} = 8 x^{3}$ , not $2 x^{3}$ , and $(- 3)^{r}$ is negative when $r$ is odd.
Mixing up the term number and $r$ (the 4th term has $r = 3$ when $r$ starts at 0).
Giving the whole term (such as $1792 x^{5}$ ) when the question asks for the coefficient, or the reverse.
Collecting powers wrongly with negative indices: $(x^{- 1})^{r} = x^{- r}$ , so subtract.
Using calculus to prove identities in this focus area when substitution or comparing coefficients is expected.
Forgetting that $0^{n} = 0$ only for $n \geq 1$ ; check the smallest case of an identity.

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6.Exam tips

Always write the general term first, then simplify the power of $x$ . This line usually earns the first mark even if your arithmetic later slips.
State the value of $r$ you are using and why ("for the constant term, $9 - 3 r = 0$ ").
Evaluate the final number unless the question says the answer can be left in $(r n)$ form.
In proofs, state the expansion you start from and the value you substitute (or the coefficient you compare) in words.
Hence in a later part usually means substitute a specific value into the identity you just proved, or compare a specific coefficient.

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7.Links to other topics

Permutations and combinations (ME11-PC): $(r n)$ counts selections, which is why it appears as the coefficient; Pascal's identity has the same combinatorial proof.
Polynomials (ME11-PO): expansions of $(x + a)^{n}$ are polynomials, and comparing coefficients is the same idea used to prove the sums and products of zeros.
Proof by mathematical induction (ME12-PI): binomial coefficient identities can also be proved by induction using Pascal's identity.
The binomial distribution (ME12-BS): the probabilities $(r n) p^{r} (1 - p)^{n - r}$ are the terms of $(q + p)^{n}$ with $q = 1 - p$ , which is why they add to 1.
Advanced, Sequences and Series: sigma notation is used to write the binomial theorem compactly.

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Practise 6 exam-style questions on The Binomial Theorem, each with instant marking and a worked solution.

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The Binomial Theorem

Contents

1.What you need to know

2.Pascal's triangle

3.The binomial theorem

Worked example 1: a coefficient

Worked example 2: the constant term

4.Proving identities

1. Substitute a value

Worked example 3: sums of coefficients

2. Compare coefficients

3. A counting argument

5.Common mistakes

6.Exam tips

7.Links to other topics

Test yourself