1.04PureStretch

Binomial Expansion

Q: When is a binomial expansion finite?

Only when n is a non-negative integer. Then there are exactly n + 1 terms.

Q: Why does the rational-power expansion need |x| < 1?

The series is infinite, and it only converges — adds up to a finite value equal to (1+x)^n — when the powers of x shrink, which needs |x| < 1.

Q: Should I use Pascal’s triangle or the formula?

Pascal’s triangle is quick for small n. For n above about 6, or for a single coefficient, use nr.

Q: How is this connected to probability?

nr counts the ways to get r successes in n trials, which is why it appears in the binomial distribution.

Expanding $(2 + x)^{2}$ by hand is quick. Expanding $(2 + x)^{9}$ by hand is miserable. The $\textbf{binomial theorem}$ writes down any term of $(a + b)^{n}$ directly, and a second version extends the idea to powers like $-1$ and $\tfrac12$ , where the expansion never stops.

45 min Sequences and Series

The big picture

Every term in $(a + b)^{n}$ comes from choosing either $a$ or $b$ from each of the $n$ brackets. The coefficient of $a^{n-r}b^{r}$ is simply the number of ways to choose which $r$ brackets supply the $b$ — that is $\binom{n}{r}$ , the same number that counts selections in probability and appears in Pascal’s triangle. So the theorem is really a counting argument. For positive integer $n$ the expansion is finite; for negative or fractional $n$ the coefficients never reach zero, the series is infinite, and it only converges for small enough $x$ . Those infinite expansions are how calculators approximate roots, and they reappear in the binomial distribution in statistics.

What you'll be able to do

Use Pascal’s triangle and $\binom{n}{r} = \dfrac{n!}{r!(n-r)!}$ for binomial coefficients
Expand $(a + bx)^{n}$ for positive integer $n$ and find specific terms
Expand $(1 + x)^{n}$ for rational $n$ and state the range of validity
Use binomial expansions for approximations

Binomial coefficients

The coefficients in $(a + b)^{n}$ form row $n$ of $\textbf{Pascal’s triangle}$ , where each entry is the sum of the two above it. Row 4 is $1, 4, 6, 4, 1$ .

For larger $n$ , use the formula $\binom{n}{r} = {}^{n}\mathrm{C}_{r} = \dfrac{n!}{r!(n-r)!}$ , where $n! = n \times (n-1) \times \dots \times 1$ and $0! = 1$ .

The coefficients are symmetric: $\binom{n}{r} = \binom{n}{n-r}$ , because choosing $r$ brackets to supply $b$ is the same as choosing $n - r$ to supply $a$ .

\binom{n}{r} = \dfrac{n!}{r!\,(n-r)!}

Number of ways to choose r from n.

Why this works

Calculating $\binom{10}{3}$ by the formula, most of $10!$ cancels with $7!$ : $\dfrac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$ . Always cancel before multiplying.

Positive integer powers

For positive integer $n$ : $(a + b)^{n} = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^{r}$ . There are $n + 1$ terms; the powers of $a$ fall from $n$ to $0$ as the powers of $b$ rise from $0$ to $n$ .

With $(a + bx)^{n}$ , the " $b$ " in the theorem is the whole of $bx$ . Put it in brackets so that its coefficient is raised to the power too.

The general term is $\binom{n}{r} a^{n-r}(bx)^{r}$ , which lets you find one term without expanding everything.

(a + b)^{n} = a^{n} + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^{2} + \dots + b^{n}

The binomial theorem, n a positive integer.

Example 1Find the first three terms, in ascending powers of

x

, of

(2 - 3x)^{6}

r = 0

2^{6} = 64

r = 1

\binom{6}{1} 2^{5}(-3x) = 6 \times 32 \times (-3x) = -576x

r = 2

\binom{6}{2} 2^{4}(-3x)^{2} = 15 \times 16 \times 9x^{2} = 2160x^{2}

Answer

64 - 576x + 2160x^{2}

Example 2Find the coefficient of

x^{4}

\left(x + \dfrac{2}{x}\right)^{8}

1General term:

\binom{8}{r} x^{8-r}\left(\dfrac{2}{x}\right)^{r} = \binom{8}{r} 2^{r} x^{8-2r}

2Need

8 - 2r = 4

, so

r = 2

3Coefficient:

\binom{8}{2} \times 2^{2} = 28 \times 4

Answer

112

Tip — Negative signs inside the bracket are the biggest source of lost marks. Write $(-3x)^{2}$ and $(-3x)^{3}$ explicitly so the sign of each term is automatic.

Rational powers

When rational $n$ is negative or is not an integer, the binomial expansion of $(1+x)^n$ is an infinite series: the coefficients do not terminate. For a non-negative integer $n$ , the expansion terminates and is valid for every real $x$ .

For the non-terminating expansions in this course, use the standard validity range $|x|<1$ . Do not assume convergence at either endpoint without a separate justification; the usual exam validity statement uses the strict inequality.

For $(a + bx)^{n}$ , first take out a factor of $a^{n}$ : $(a + bx)^{n} = a^{n}\left(1 + \dfrac{b}{a}x\right)^{n}$ , which is valid for $\left|\dfrac{bx}{a}\right| < 1$ , that is $|x| < \left|\dfrac{a}{b}\right|$ .

(1 + x)^{n} = 1 + nx + \dfrac{n(n-1)}{2!}x^{2} + \dfrac{n(n-1)(n-2)}{3!}x^{3} + \dots, \quad |x| < 1

The general binomial expansion.

Example 1Expand

\sqrt{4 + x}

up to and including the term in

x^{2}

, stating the range of validity.

\sqrt{4 + x} = 2\left(1 + \dfrac{x}{4}\right)^{1/2}

2With

n = \tfrac12

and

\dfrac{x}{4}

in place of

x

1 + \tfrac12\left(\dfrac{x}{4}\right) + \dfrac{\tfrac12\left(-\tfrac12\right)}{2}\left(\dfrac{x}{4}\right)^{2}

= 1 + \dfrac{x}{8} - \dfrac{x^{2}}{128}

4Multiply by 2. Valid when

\left|\dfrac{x}{4}\right| < 1

Answer

2 + \dfrac{x}{4} - \dfrac{x^{2}}{64}

, valid for

|x| < 4

Why this works

For a positive integer $n$ , the factor $(n - n)$ eventually appears in every coefficient and all later terms vanish. For any other $n$ that factor never arrives — which is exactly why the series goes on forever.

Approximations

For small $x$ , the higher powers are tiny, so the first few terms of an expansion give a good approximation.

To approximate a number, choose $x$ so that the bracket matches it, and check that the value lies inside the range of validity.

Example 1Use the expansion of

\sqrt{4 + x}

above to estimate

\sqrt{4.2}

1Set

x = 0.2

, which satisfies

|x| < 4

2 + \dfrac{0.2}{4} - \dfrac{0.04}{64} = 2 + 0.05 - 0.000625

Answer

2.049375

(the true value is

2.049390\dots

)

Tip — The smaller $x$ is, the better the approximation. Say why an approximation is good: " $x$ is small, so the omitted terms in $x^{3}$ and beyond are negligible."

Think like an examiner

Keep $(bx)^{r}$ bracketed, including negative signs.
Use the general term to find a single coefficient instead of expanding fully.
For rational powers, take out $a^{n}$ before expanding and state the range of validity.
When approximating, show that the chosen $x$ lies within the range of validity.

Common misconceptions

Writing the

x^{2}

term of

(1 + 2x)^{5}

10 \times 2x^{2} = 20x^{2}

Raise the whole term:

(2x)^{2} = 4x^{2}

, so the term is

40x^{2}

Expanding

(4 + x)^{1/2}

4^{1/2}(1 + x)^{1/2}

Factor out correctly:

2\left(1 + \tfrac{x}{4}\right)^{1/2}

Stating validity as

|x| < 1

for

(1 + 3x)^{-1}

Need

|3x| < 1

, so

|x| < \tfrac13

Forgetting that

0! = 1

\binom{n}{0} = \binom{n}{n} = 1

Binomial expansion

\binom{n}{r} = \dfrac{n!}{r!(n-r)!}

Binomial coefficient.

\binom{n}{r} a^{n-r}(bx)^{r}

General term.

(1+x)^{n} = 1 + nx + \tfrac{n(n-1)}{2!}x^{2} + \dots

Rational n,

|x| < 1

Stretch yourself

In the expansion of $(1 + kx)^{n}$ , where $n$ is a positive integer, the coefficient of $x$ is $12$ and the coefficient of $x^{2}$ is $60$ . Find $n$ and $k$ .

Hint — Write both coefficients in terms of $n$ and $k$ , then divide one equation by the square of the other to eliminate $k$ .

Reveal solution

SolutionThe coefficient of

x

nk = 12

, and of

x^{2}

\dfrac{n(n-1)}{2}k^{2} = 60

. From the first,

k = \dfrac{12}{n}

, so

\dfrac{n(n-1)}{2} \times \dfrac{144}{n^{2}} = 60

, which simplifies to

\dfrac{72(n-1)}{n} = 60

. Then

72n - 72 = 60n

, so

n = 6

and

k = 2

. Check:

(1 + 2x)^{6}

has

x

-coefficient

6 \times 2 = 12

✓ and

x^{2}

-coefficient

15 \times 4 = 60

✓.

\blacksquare

Questions students ask

When is a binomial expansion finite?

Only when

n

is a non-negative integer. Then there are exactly

n + 1

terms.

Why does the rational-power expansion need

|x| < 1

The series is infinite, and it only converges — adds up to a finite value equal to

(1+x)^{n}

— when the powers of

x

shrink, which needs

|x| < 1

Should I use Pascal’s triangle or the formula?

Pascal’s triangle is quick for small

n

. For

n

above about 6, or for a single coefficient, use

\binom{n}{r}

How is this connected to probability?

\binom{n}{r}

counts the ways to get

r

successes in

n

trials, which is why it appears in the binomial distribution.

Key takeaways

Coefficients are $\binom{n}{r}$ , from Pascal’s triangle or $\dfrac{n!}{r!(n-r)!}$ .
For positive integer $n$ , $(a + b)^{n}$ has $n + 1$ terms with general term $\binom{n}{r}a^{n-r}b^{r}$ .
For rational $n$ , $(1 + x)^{n}$ is an infinite series valid for $|x| < 1$ .
For $(a + bx)^{n}$ with rational $n$ , factor out $a^{n}$ ; validity is $|x| < |a/b|$ .
The first few terms give approximations when $x$ is small.