The Binomial Theorem for Rational Powers

You already know how to expand $(1 + x)^5$: it’s a polynomial with six terms. But what about $(1 + x)^{-1}$ or $\sqrt{1 + x} = (1 + x)^{1/2}$? The binomial theorem still works for negative and fractional powers, with one big change: the expansion never stops. It becomes an infinite series, and it’s only valid for small enough $x$. These series give quick, accurate approximations, and they’re a first look at the power series you’ll meet in calculus.

Key ideas

The pattern for any power n

For a positive integer $n$, the binomial expansion of $(1 + x)^n$ can be written

$$(1 + x)^n = 1 + nx + \frac{n(n - 1)}{2!}x^2 + \frac{n(n - 1)(n - 2)}{3!}x^3 + \dots$$

When $n$ is a positive integer, a factor $(n - n)$ eventually appears, every later coefficient is $0$, and the expansion stops after $n + 1$ terms. These coefficients are the entries $\dbinom{n}{r}$ of Pascal’s triangle.

The same formula works when $n$ is any rational number ($n \in \mathbb{Q}$), such as $n = -2$ or $n = \tfrac{1}{2}$. Now no factor is ever zero, so the series goes on forever:

$$(1 + x)^n = 1 + nx + \frac{n(n - 1)}{2!}x^2 + \frac{n(n - 1)(n - 2)}{3!}x^3 + \dots, \qquad |x| \lt 1$$

The coefficient of $x^r$ is

$$\frac{n(n - 1)(n - 2)\cdots(n - r + 1)}{r!}$$

with $r$ factors on top. Your GDC’s nCr function usually only accepts whole numbers, so work these coefficients out by hand.

When the series is valid

For a negative or fractional $n$, the series only gives the right value (converges) when

$$|x| \lt 1, \quad \text{that is,} \quad -1 \lt x \lt 1$$

Outside that interval the terms grow instead of shrinking, and adding more of them takes you further from the true value. The graph shows this for $(1 + x)^{-1} = 1 - x + x^2 - x^3 + \dots$

Inside $-1 \lt x \lt 1$ the partial sums close in on $y = \dfrac{1}{1 + x}$. Outside it, they fly away.

You’ll recognize this particular series: it’s the infinite geometric series with first term $1$ and ratio $-x$, which also needs $|x| \lt 1$.

Brackets of the form (a + bx)

The formula needs a $1$ at the front of the bracket. For $(a + bx)^n$, factor out $a$ first:

$$(a + bx)^n = \left(a\left(1 + \frac{bx}{a}\right)\right)^n = a^n\left(1 + \frac{bx}{a}\right)^n$$

Then expand with $\dfrac{bx}{a}$ in place of $x$. Two things change:

• Don’t forget the factor $a^n$ at the front (for example $4^{1/2} = 2$ or $2^{-3} = \dfrac{1}{8}$).
• The series is valid when $\left|\dfrac{bx}{a}\right| \lt 1$, that is, $|x| \lt \left|\dfrac{a}{b}\right|$.

For $(1 + bx)^n$ the condition is $|bx| \lt 1$, so $|x| \lt \dfrac{1}{|b|}$. For example, $(1 - 4x)^{1/2}$ is valid for $|x| \lt \dfrac{1}{4}$.

Approximations

If $x$ is small, the powers $x^2, x^3, \dots$ get small very fast, so the first few terms give a good approximation. To approximate a number like $\sqrt{1.02}$:

1. Write it as a binomial: $\sqrt{1.02} = (1 + 0.02)^{1/2}$.
2. Check the value of $x$ is inside the interval of validity: $|0.02| \lt 1$ ✓.
3. Substitute into the first few terms:

$$(1 + x)^{1/2} \approx 1 + \frac{1}{2}x - \frac{1}{8}x^2 \quad\Rightarrow\quad \sqrt{1.02} \approx 1 + 0.01 - 0.00005 = 1.00995$$

The true value is $1.009950\ldots$, so three terms already give five correct decimal places. The smaller $x$ is, the better the approximation.

A connection to calculus

These series are examples of power series: infinite polynomials that represent a function on an interval. In calculus you’ll build many more, such as the series for $e^x$ and $\ln(1 + x)$; see representing functions as power series.

Worked examples

Example 1: A negative power

Expand $(1 + x)^{-2}$ up to and including the term in $x^3$, and state the values of $x$ for which the expansion is valid.

Solution. Use the formula with $n = -2$:

$$\begin{aligned} (1 + x)^{-2} &= 1 + (-2)x + \frac{(-2)(-3)}{2!}x^2 + \frac{(-2)(-3)(-4)}{3!}x^3 + \dots \\ &= 1 - 2x + \frac{6}{2}x^2 - \frac{24}{6}x^3 + \dots \\ &= 1 - 2x + 3x^2 - 4x^3 + \dots \end{aligned}$$

The expansion is valid for $|x| \lt 1$.

Check with a small value, $x = 0.1$: $(1.1)^{-2} = 0.826\ldots$, and $1 - 0.2 + 0.03 - 0.004 = 0.826$ ✓.

Example 2: A fractional power and an approximation

(a) Expand $\sqrt{1 - 4x}$ up to and including the term in $x^3$, and state the interval of validity.

(b) Use your expansion with $x = 0.01$ to approximate $\sqrt{0.96}$.

Solution.

(a) $\sqrt{1 - 4x} = (1 + (-4x))^{1/2}$, so use $n = \tfrac{1}{2}$ with $-4x$ in place of $x$. Keep $-4x$ in brackets, because it gets squared and cubed:

$$\begin{aligned} (1 - 4x)^{1/2} &= 1 + \tfrac{1}{2}(-4x) + \frac{\tfrac{1}{2}\left(-\tfrac{1}{2}\right)}{2!}(-4x)^2 + \frac{\tfrac{1}{2}\left(-\tfrac{1}{2}\right)\left(-\tfrac{3}{2}\right)}{3!}(-4x)^3 + \dots \\ &= 1 - 2x + \left(-\tfrac{1}{8}\right)(16x^2) + \left(\tfrac{1}{16}\right)(-64x^3) + \dots \\ &= 1 - 2x - 2x^2 - 4x^3 + \dots \end{aligned}$$

It’s valid when $|-4x| \lt 1$, that is, $|x| \lt \dfrac{1}{4}$.

(b) With $x = 0.01$, $1 - 4x = 0.96$, and $0.01$ is inside the interval of validity:

$$\sqrt{0.96} \approx 1 - 2(0.01) - 2(0.01)^2 - 4(0.01)^3 = 1 - 0.02 - 0.0002 - 0.000004 = 0.979796$$

A calculator gives $\sqrt{0.96} = 0.9797959\ldots$, so the approximation is correct to $6$ decimal places.

Example 3: Factoring out a first

Expand $(2 + x)^{-3}$ up to and including the term in $x^2$. State the interval of validity.

Solution. Factor out $2$ so the bracket starts with $1$:

$$(2 + x)^{-3} = 2^{-3}\left(1 + \frac{x}{2}\right)^{-3} = \frac{1}{8}\left(1 + \frac{x}{2}\right)^{-3}$$

Expand with $n = -3$ and $\dfrac{x}{2}$ in place of $x$:

$$\begin{aligned} \left(1 + \frac{x}{2}\right)^{-3} &= 1 + (-3)\left(\frac{x}{2}\right) + \frac{(-3)(-4)}{2!}\left(\frac{x}{2}\right)^2 + \dots \\ &= 1 - \frac{3}{2}x + 6\left(\frac{x^2}{4}\right) + \dots \\ &= 1 - \frac{3}{2}x + \frac{3}{2}x^2 + \dots \end{aligned}$$

Multiply by $\dfrac{1}{8}$:

$$(2 + x)^{-3} = \frac{1}{8} - \frac{3}{16}x + \frac{3}{16}x^2 + \dots$$

It’s valid when $\left|\dfrac{x}{2}\right| \lt 1$, that is, $|x| \lt 2$.

Example 4: A coefficient in a product

Find the coefficient of $x^2$ in the expansion of $\dfrac{3 - x}{\sqrt{1 + 2x}}$, and state when the expansion is valid.

Solution. Write the expression as a product: $(3 - x)(1 + 2x)^{-1/2}$. Expand the second bracket with $n = -\tfrac{1}{2}$ and $2x$ in place of $x$, up to $x^2$:

$$\begin{aligned} (1 + 2x)^{-1/2} &= 1 + \left(-\tfrac{1}{2}\right)(2x) + \frac{\left(-\tfrac{1}{2}\right)\left(-\tfrac{3}{2}\right)}{2!}(2x)^2 + \dots \\ &= 1 - x + \tfrac{3}{8}(4x^2) + \dots \\ &= 1 - x + \tfrac{3}{2}x^2 + \dots \end{aligned}$$

Now multiply by $3 - x$, collecting only the terms that give $x^2$:

$$(3 - x)\left(1 - x + \tfrac{3}{2}x^2 + \dots\right): \qquad 3 \times \tfrac{3}{2}x^2 + (-x) \times (-x) = \tfrac{9}{2}x^2 + x^2 = \tfrac{11}{2}x^2$$

The coefficient of $x^2$ is $\dfrac{11}{2}$. The expansion is valid for $|2x| \lt 1$, that is, $|x| \lt \dfrac{1}{2}$.

Common mistakes

Leaving out the interval of validity. For negative or fractional powers, the expansion is only true for some values of $x$. Always state the condition, such as $|x| \lt 1$ or $|x| \lt \dfrac{1}{4}$.

Not factoring out a, or forgetting to raise it to the power n. $(4 - x)^{1/2}$ is not $4\left(1 - \dfrac{x}{4}\right)^{1/2}$. It’s $4^{1/2}\left(1 - \dfrac{x}{4}\right)^{1/2} = 2\left(1 - \dfrac{x}{4}\right)^{1/2}$.

Not raising the whole term to the power. In $(1 - 4x)^{1/2}$, the $x^2$ term uses $(-4x)^2 = 16x^2$, not $-4x^2$. Keep the whole term in brackets, including its sign.

Sign slips in the coefficients. With $n = -\tfrac{1}{2}$, the factors are $-\tfrac{1}{2}$, $-\tfrac{3}{2}$, $-\tfrac{5}{2}$, and so on: each one is $1$ less than the one before. Write each factor out rather than doing it in your head.

Approximating with an x outside the interval. $\sqrt{3} = (1 + 2)^{1/2}$ looks tempting, but $x = 2$ is outside $|x| \lt 1$, so the series doesn’t converge to $\sqrt{3}$ at all. Choose a form where $x$ is small.

Using the calculator’s nCr button. For fractional or negative $n$, nCr usually gives an error. Work out $\dfrac{n(n - 1)\cdots(n - r + 1)}{r!}$ by hand.

Practice

1. (Warm-up) Expand $(1 + x)^{-3}$ up to and including the term in $x^3$.

Solution

$$\begin{aligned} (1 + x)^{-3} &= 1 + (-3)x + \frac{(-3)(-4)}{2!}x^2 + \frac{(-3)(-4)(-5)}{3!}x^3 + \dots \\ &= 1 - 3x + 6x^2 - 10x^3 + \dots \end{aligned}$$

valid for $|x| \lt 1$.

2. (Warm-up) Expand $\sqrt[3]{1 + x}$ up to and including the term in $x^2$.

Solution

Use $n = \tfrac{1}{3}$:

$$\begin{aligned} (1 + x)^{1/3} &= 1 + \tfrac{1}{3}x + \frac{\tfrac{1}{3}\left(-\tfrac{2}{3}\right)}{2!}x^2 + \dots \\ &= 1 + \tfrac{1}{3}x - \tfrac{1}{9}x^2 + \dots \end{aligned}$$

valid for $|x| \lt 1$.

3. (Warm-up) State the values of $x$ for which each expansion is valid.

• (a) $(1 - 5x)^{-2}$
• (b) $(4 + 3x)^{1/2}$

Solution

(a) Valid when $|-5x| \lt 1$, so $|x| \lt \dfrac{1}{5}$.

(b) $(4 + 3x)^{1/2} = 2\left(1 + \dfrac{3x}{4}\right)^{1/2}$, valid when $\left|\dfrac{3x}{4}\right| \lt 1$, so $|x| \lt \dfrac{4}{3}$.

4. (Core) Expand $\dfrac{1}{\sqrt{1 - 2x}}$ up to and including the term in $x^3$, and state the interval of validity.

Solution

$\dfrac{1}{\sqrt{1 - 2x}} = (1 + (-2x))^{-1/2}$. Use $n = -\tfrac{1}{2}$:

$$\begin{aligned} (1 - 2x)^{-1/2} &= 1 + \left(-\tfrac{1}{2}\right)(-2x) + \frac{\left(-\tfrac{1}{2}\right)\left(-\tfrac{3}{2}\right)}{2!}(-2x)^2 + \frac{\left(-\tfrac{1}{2}\right)\left(-\tfrac{3}{2}\right)\left(-\tfrac{5}{2}\right)}{3!}(-2x)^3 + \dots \\ &= 1 + x + \tfrac{3}{8}(4x^2) + \left(-\tfrac{5}{16}\right)(-8x^3) + \dots \\ &= 1 + x + \tfrac{3}{2}x^2 + \tfrac{5}{2}x^3 + \dots \end{aligned}$$

Valid for $|-2x| \lt 1$, that is, $|x| \lt \dfrac{1}{2}$.

5. (Core)

• (a) Expand $\sqrt{4 - x}$ up to and including the term in $x^2$, and state the interval of validity.
• (b) Use your expansion to approximate $\sqrt{3.96}$.

Solution

(a) Factor out $4$:

$$\sqrt{4 - x} = 4^{1/2}\left(1 - \frac{x}{4}\right)^{1/2} = 2\left(1 - \frac{x}{4}\right)^{1/2}$$$$\begin{aligned} \left(1 - \frac{x}{4}\right)^{1/2} &= 1 + \tfrac{1}{2}\left(-\frac{x}{4}\right) + \frac{\tfrac{1}{2}\left(-\tfrac{1}{2}\right)}{2!}\left(-\frac{x}{4}\right)^2 + \dots \\ &= 1 - \frac{x}{8} - \frac{x^2}{128} + \dots \end{aligned}$$

So

$$\sqrt{4 - x} = 2 - \frac{x}{4} - \frac{x^2}{64} + \dots, \qquad |x| \lt 4$$

(b) $3.96 = 4 - 0.04$, so use $x = 0.04$ (inside $|x| \lt 4$):

$$\sqrt{3.96} \approx 2 - 0.01 - \frac{0.0016}{64} = 2 - 0.01 - 0.000025 = 1.989975$$

(A calculator gives $1.9899748\ldots$)

6. (Core) Use the first three terms of the expansion of $(1 + x)^{1/2}$ to approximate $\sqrt{1.02}$. Hence approximate $\sqrt{102}$.

Solution

$(1 + x)^{1/2} = 1 + \tfrac{1}{2}x - \tfrac{1}{8}x^2 + \dots$ for $|x| \lt 1$. With $x = 0.02$:

$$\sqrt{1.02} \approx 1 + 0.01 - \tfrac{1}{8}(0.0004) = 1 + 0.01 - 0.00005 = 1.00995$$

Since $102 = 100 \times 1.02$,

$$\sqrt{102} = 10\sqrt{1.02} \approx 10.0995$$

(A calculator gives $10.099505\ldots$)

7. (Core) Find the term in $x^3$ in the expansion of $(2 - x)^{-2}$.

Solution

$$(2 - x)^{-2} = 2^{-2}\left(1 - \frac{x}{2}\right)^{-2} = \frac{1}{4}\left(1 - \frac{x}{2}\right)^{-2}$$

The $x^3$ term inside the bracket uses $n = -2$, $r = 3$:

$$\frac{(-2)(-3)(-4)}{3!}\left(-\frac{x}{2}\right)^3 = (-4)\left(-\frac{x^3}{8}\right) = \frac{x^3}{2}$$

Multiply by $\dfrac{1}{4}$: the term in $x^3$ is $\dfrac{1}{8}x^3$.

8. (Challenge) The expansion of $(1 + ax)^n$ begins $1 - 6x + 27x^2 + \dots$, where $a$ and $n$ are constants.

• (a) Find $a$ and $n$.
• (b) Find the coefficient of $x^3$.
• (c) State the interval of validity.

Solution

(a) Compare coefficients:

$$na = -6 \qquad\text{and}\qquad \frac{n(n - 1)}{2}a^2 = 27$$

From the first, $a^2 = \dfrac{36}{n^2}$. Substitute into the second:

$$\begin{aligned} \frac{n(n - 1)}{2} \cdot \frac{36}{n^2} &= 27 \\ \frac{18(n - 1)}{n} &= 27 \\ 18n - 18 &= 27n \\ n &= -2 \end{aligned}$$

Then $a = \dfrac{-6}{-2} = 3$.

(b) With $n = -2$ and $3x$ in place of $x$:

$$\frac{(-2)(-3)(-4)}{3!}(3x)^3 = (-4)(27x^3) = -108x^3$$

The coefficient of $x^3$ is $-108$.

(c) Valid for $|3x| \lt 1$, that is, $|x| \lt \dfrac{1}{3}$.

9. (Challenge)

• (a) Show that, for small $x$, $\sqrt{\dfrac{1 + x}{1 - x}} \approx 1 + x + \dfrac{1}{2}x^2$.
• (b) By choosing $x = \dfrac{1}{9}$, use part (a) to find a fraction that approximates $\sqrt{5}$. How close is it?

Solution

(a) Write it as a product of two binomials, both valid for $|x| \lt 1$:

$$\sqrt{\frac{1 + x}{1 - x}} = (1 + x)^{1/2}(1 - x)^{-1/2}$$$$(1 + x)^{1/2} = 1 + \tfrac{1}{2}x - \tfrac{1}{8}x^2 + \dots$$$$(1 - x)^{-1/2} = 1 + \left(-\tfrac{1}{2}\right)(-x) + \frac{\left(-\tfrac{1}{2}\right)\left(-\tfrac{3}{2}\right)}{2!}(-x)^2 + \dots = 1 + \tfrac{1}{2}x + \tfrac{3}{8}x^2 + \dots$$

Multiply, keeping terms up to $x^2$:

$$\begin{aligned} &1 + \left(\tfrac{1}{2} + \tfrac{1}{2}\right)x + \left(\tfrac{3}{8} + \tfrac{1}{4} - \tfrac{1}{8}\right)x^2 \\ &= 1 + x + \tfrac{1}{2}x^2 \end{aligned}$$

(b) With $x = \dfrac{1}{9}$:

$$\sqrt{\frac{1 + \frac{1}{9}}{1 - \frac{1}{9}}} = \sqrt{\frac{10/9}{8/9}} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}$$

So

$$\frac{\sqrt{5}}{2} \approx 1 + \frac{1}{9} + \frac{1}{2}\cdot\frac{1}{81} = \frac{162 + 18 + 1}{162} = \frac{181}{162}$$

and $\sqrt{5} \approx \dfrac{181}{81} = 2.2346\ldots$ The true value is $\sqrt{5} = 2.2361\ldots$, so the approximation is out by only about $0.0015$.