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 $binomial theorem$ writes down any term of $(a + b)^{n}$ directly, and a second version extends the idea to powers like $- 1$ and $\frac{1}{2}$ , 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 $(r n)$ , 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 $(r n) = \frac{n !}{r ! ( n - r )!}$ for binomial coefficients
Expand $(a + b x)^{n}$ 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 $Pascal’s triangle$ , 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 $(r n) =^{n} C_{r} = \frac{n !}{r ! ( n - r )!}$ , where $n! = n \times (n - 1) \times \dots \times 1$ and $0! = 1$ .

The coefficients are symmetric: $(r n) = (n - r n)$ , because choosing $r$ brackets to supply $b$ is the same as choosing $n - r$ to supply $a$ .

(r n) = \frac{n !}{r ! ( n - r )!}

Number of ways to choose r from n.

Why this works

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

Positive integer powers

For positive integer $n$ : $(a + b)^{n} = \sum_{r = 0}^{n} (r n) a^{n - r} b^{r}$ . 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 $(a + b x)^{n}$ , the " $b$ " in the theorem is the whole of $b x$ . Put it in brackets so that its coefficient is raised to the power too.

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

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

The binomial theorem, n a positive integer.

r = 0

2^{6} = 64

r = 1

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

r = 2

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

Answer

64 - 576 x + 2160 x^{2}

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

Rational powers

For any rational $n$ — negative or fractional included — there is a binomial expansion for $(1 + x)^{n}$ , but it is an $infinite series$ because the coefficients never become zero.

It is valid only for $∣ x ∣ < 1$ . Outside that range the terms do not shrink and the series does not converge to the function.

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

(1 + x)^{n} = 1 + n x + \frac{n ( n - 1 )}{2 !} x^{2} + \frac{n ( n - 1 ) ( n - 2 )}{3 !} x^{3} + \dots, ∣ x ∣ < 1

The general binomial expansion.

4 + x = 2 (1 + \frac{x}{4})^{1/2}

2With

n = \frac{1}{2}

and

\frac{x}{4}

in place of

x

1 + \frac{1}{2} (\frac{x}{4}) + \frac{\frac{1}{2} ( - \frac{1}{2} )}{2} (\frac{x}{4})^{2}

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

4Multiply by 2. Valid when

\frac{x}{4} < 1

Answer

2 + \frac{x}{4} - \frac{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.

1Set

x = 0.2

, which satisfies

∣ x ∣ < 4

2 + \frac{0.2}{4} - \frac{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 $(b x)^{r}$ 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 + 2 x)^{5}

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

Raise the whole term:

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

, so the term is

40 x^{2}

Expanding

(4 + x)^{1/2}

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

Factor out correctly:

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

Stating validity as

∣ x ∣ < 1

for

(1 + 3 x)^{- 1}

Need

∣3 x ∣ < 1

, so

∣ x ∣ < \frac{1}{3}

Forgetting that

0! = 1

(0 n) = (n n) = 1

Binomial expansion

(r n) = \frac{n !}{r ! ( n - r )!}

Binomial coefficient.

(r n) a^{n - r} (b x)^{r}

General term.

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

Rational n,

∣ x ∣ < 1

Stretch yourself

In the expansion of $(1 + k x)^{n}$ , 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 $k$ .

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

(r n)

How is this connected to probability?

(r n)

counts the ways to get

r

successes in

n

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

Key takeaways

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

How this fits the course

Build on

Expanding Brackets

Probability Distributions

Leads to

Trigonometric Ratios

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Binomial Expansion

Binomial coefficients

Positive integer powers

Rational powers

Approximations

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Binomial Expansion

Binomial coefficients

Positive integer powers

Rational powers

Approximations

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