# Binomial Expansion

A binomial expansion is a method used to allow us to expand and simplify algebraic expressions in the form $$(x+y)^n$$ into a sum of terms of the form $$ax^by^c$$If $$n$$ is an integer, $$b$$ and $$c$$ also will be integers, and $$b + c = n$$.

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We can expand expressions in the form $$(x+y)^n$$ by multiplying out every single bracket, but this might be very long and tedious for high values of $$n$$ such as in $$(x+y)^{20}$$ for example. This is where using the Binomial Theorem comes in useful.

## The binomial theorem

The binomial theorem allows us to expand an expression of the form $$(x+y)^n$$ into a sum. A general formula for a binomial expression is:

$(x+y)^n = \binom{n}{0}x^ny^0 + \binom{n}{1}x^{n-1}y^1 + \binom{n}{2}x^{n-2}y^2 + \dots + \binom{n}{n-1}x^1y^{n-1} + \binom{n}{n}x^0y^n.$

Which can be simplified to:

\begin{align} (x+y)^n &= \sum\limits_{k=0}^n \binom{n}{k} x^{n-k}y^k \\ &= \sum\limits_{k=0}^n \binom{n}{k} x^ky^{n-k} . \end{align}

Where both $$n$$ and $$k$$ are integers. This is also known as the binomial formula. The notation

$\binom{n}{k}$

can be referred to as '$$n$$ choose $$k$$' and gives a number called the binomial coefficient which is the number of different combinations of ordering $$k$$ objects out of a total of $$n$$ objects. The equation for the binomial coefficient ($$n$$ choose $$k$$ or $$^nC_r$$ on a calculator) is given by:

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

Where '!' means factorial. Factorial means the product of an integer with all the integers below it. For example for $$5$$ choose $$3$$, we would have:

\begin{align} \binom{5}{3} &= \frac{5!}{3!(5-3)!} \\ &= \frac{5\cdot 4\cdot 3 \cdot 2 \cdot 1}{(3\cdot 2\cdot 1)(2\cdot 1)} \\ &= 10. \end{align}

## How do you do a binomial expansion?

To understand how to do a binomial expansion, we will look at an example. Let's say we want to expand $$(x+y)^4$$. In this case, $$n = 4$$ and $$k$$ will vary between $$0$$ and $$4$$. Using the formula for the binomial expansion, we can write:

$(x+y)^4 = \binom{4}{0}x^4y^0 + \binom{4}{1}x^3y^1 + \binom{4}{2}x^2y^2 + \binom{4}{3}x^1y^3+\binom{4}{4}x^0y^4.$

We can now use the equation for the binomial coefficient to find all the constant terms in this expression. For the first term we have:

\begin{align} \binom{4}{0} &= \frac{4!}{0!(4-0)!} \\ &= \frac{4 \cdot 3\cdot 2\cdot 1}{1\cdot (4 \cdot 3\cdot 2\cdot 1 )} \\ &= 1. \end{align}

Repeating this for all five coefficients, we end up with binomial coefficients of $$1$$, $$4$$, $$6$$, $$4$$, $$1$$ in order. Therefore, our expression for the binomial expansion simplifies to:

$x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4.$

Note that $$y$$ could also be replaced by any number.

## Binomial expansion formula

To summarise the above explanation, the expansion formula can be written as:

$(x+y)^n = \sum _{k=0}^{n} \binom{n}{k}x^{n-k}y^k = \sum _{k=0}^{n} \binom{n}{k}x^{k}y^{n-k}$

Where $$\binom{n}{k}$$ is the binomial coefficient of each term.

## Binomial expansion for fractional and negative powers

Sometimes you will encounter algebraic expressions where n is not a positive integer but a negative integer or a fraction. Let's consider the expression $$\sqrt{1-2x}$$ which can also be written as

$(1- 2x)^\dfrac{1}{2}$ where $$x < 0.5$$. In this case, it becomes hard to find the formula to find the binomial coefficients,

because we can't find the factorials for a negative or rational number. However, if we look at an example for a positive integer, we can find a more general expression that we can then also apply to negative and fractional numbers. For example for

$\binom{6}{3}$

we have

\begin{align} \binom{6}{3}&= \frac{6!}{3!(6-3)!} \\ &= \frac{6\cdot 5\cdot 4}{3!} \\ &= \frac{6(6-1)(6-2)}{3!}. \end{align}

From this we observe that

$\binom{n}{k} = \frac{n(n-1)(n-2)(n-3)\dots (n-k+1)}{k!}$

and therefore the more general expression for the binomial theorem is the infinite formula

$(a+b)^n = \frac{a^n}{0!} + \frac{na^{n-1}b}{1!} + \frac{n(n-1)a^{n-2}b^2}{2!} + \frac{n(n-1)(n-2)a^{n-3}b^3}{3!} + \dots$

Let's look at $$\sqrt{1-2x}$$. In this case $$a = -2x$$, $$b = 1$$ and $$n =1/2$$. Substituting this we get:

\begin{align} \frac{(-2x)^\frac{1}{2}}{0!} &+ \frac{\left(-\frac{1}{2}\right) (-2x)^{-\frac{1}{2}}\cdot 1 }{1!} \\ &\quad + \frac{\left(-\frac{1}{2}\right) \left(-\frac{1}{2}\right) (-2x)^{-\frac{3}{2}}\cdot 1^2 }{2!} \\ &\quad + \frac{\left(-\frac{1}{2}\right) \left(-\frac{1}{2}\right) \left(-\frac{3}{2}\right) (-2x)^{-\frac{5}{2}}\cdot 1^3 }{3!} + \dots \end{align}

Using Mac Laurin's expansion we can say that the above expression converges to

$\sqrt{1-2x} = 1 - x - \frac{x^2}{2} - \frac{x^3}{2}.$

## Binomial expansion questions

We have collected a few questions with step-by-step solutions to help you understand how the binomial theorem and binomial expansion can be applied or asked about in an exam.

Exercise 1

Expand $$(x + 2)^4$$ using the binomial theorem.

Solution:

Using the binomial theorem, we have:

$$(x + 2)^4 = \binom{4}{0}x^4(2)^0 + \binom {4}{1}x^3(2)^1 + \binom{4}{2}x^2(2)^2 + \binom{4}{3}x(2)^3 + \binom{4}{4}(2)^4$$

Evaluating the coefficients, we get:

$$(x + 2)^4 = x^4 + 8x^3 + 24x^2 + 32x + 16$$

Therefore, $$(x + 2)^4$$ expands to $$x^4 + 8x^3 + 24x^2 + 32x + 16$$.

Exercise 2

Find the coefficient of $$x^3$$ in the expansion of $$(2x + 1)^5$$.

Solution:

Using the binomial theorem, the expansion of $$(2x + 1)^5$$ is:

$$(2x + 1)^5 = \binom{5}{0}(2x)^0(1)^5 + \binom{5}{1}(2x)^1(1)^4 + \binom{5}{2}(2x)^2(1)^3 + \binom{5}{3}(2x)^3(1)^2 + \binom{5}{4}(2x)^4(1)^1 + \binom{5}{5}(2x)^5(1)^0$$

To find the coefficient of $$x^3$$, we need to look at the term with $$(2x)^3$$:

$$\binom{5}{3}(2x)^3(1)^2 = 10(2x)^3$$

Evaluating the term, we get:

$$10(2x)^3 = 80x^3$$

Therefore, the coefficient of $$x^3$$ in the expansion of $$(2x + 1)^5$$ is 80.

Exercise 3

Find the first three terms in the expansion of $$(1 - 3x)^6$$.

Solution:

Using the binomial theorem, the expansion of $$(1 - 3x)^6$$ is:

$$(1 - 3x)^6 = \binom{6}{0}(1)^6(-3x)^0 + \binom{6}{1}(1)^5(-3x)^1 + \binom{6}{2}(1)^4(-3x)^2 + ...$$

To find the first three terms, we need to evaluate the terms with $$(1)^6, (1)^5, \text{and} \space (1)^4$$:

$$\binom {6}{0}(1)^6(-3x)^0 = 1$$

$$\binom{6}{1}(1)^5(-3x)^1 = -18x$$

$$\binom{6}{2}(1)^4(-3x)^2 = 162x^2$$

Therefore, the first three terms in the expansion of $$(1 - 3x)^6$$ are $$1, -18x, \text{and } 162x^2$$.

## Binomial Expansion - Key takeaways

• A binomial expansion helps us to simplify algebraic expressions into a sum
• The formula for the binomial expansion is:

$(x+y)^n = \binom{n}{0}x^ny^0 + \binom{n}{1}x^{n-1}y^1 + \binom{n}{2}x^{n-2}y^2 + \dots + \binom{n}{n-1}x^1y^{n-1} + \binom{n}{n}x^0y^n$

• The binomial coefficients or constant terms in this expression are found using:$\binom{n}{k} = \frac{n!}{k!(n-k)!}$
• $(1+a)^n = 1 + na+ \frac{n(n-1)}{2!}a^2 + \frac{n(n-1)(n-2)}{3!}a^3 + \dots$

#### Flashcards in Binomial Expansion 40

###### Learn with 40 Binomial Expansion flashcards in the free StudySmarter app

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How do you find the constant term in a binomial expansion?

The constant term is found by using the formula

n choose k=n!/k!(n-k)!

What is binomial expansion?

A binomial expansion is a method that allows us to simplify complex algebraic expressions into a sum.

How do you do a binomial expansion?

You can use the binomial expansion formula

(x+y)^n=(nC0)x^n y^0+(nC1)x^/n-1)y^1+(nC2)x^(n-2)y^2+...+(nCn-1)x^1y^(n-1)+(nCn)x^0y^n

## Test your knowledge with multiple choice flashcards

Which of these is the correct expansion of $$(x^2-y)^2$$?

Given that the coefficient of $$x^2$$ in the expansion of $$(1+ax)^7$$ is $$525$$, find the possible values of $$a$$.

Which of these is equivalent to $$n!$$?

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