# Binomial Expansion – positive integer powers

## Binomial Expansion Notes

Binomial Expansion is essentially multiplying out brackets. A binomial is two terms added together and this is raised to a power, i.e. $(x+y)^n$. Before learning how to perform a Binomial Expansion, one must understand factorial notation and be familiar with Pascal’s triangle.

### Factorial Notation

When you see an exclamation mark following a number in mathematics it is known as a factorial. For example, 6! is said ‘6 factorial’ and you multiply all of the positive integers less than 6 together:

$6!=6\times 5\times 4 \times 3 \times 2 \times 1=720$

Here are some more examples:

$8!=8\times 7\times 6\times 5\times 4 \times 3 \times 2 \times 1=40,320$

$4!\times 3!=4\times 3\times 2 \times 1\times 3\times 2 \times 1=24\times 6=144$

$9!\div 5!=\frac{9\times 8\times 7\times 6\times 5\times 4 \times 3 \times 2 \times 1}{5\times 4 \times 3 \times 2 \times 1}=9\times 8\times 7\times 6=3,024$

### Pascal's Triangle

Pascal’s triangle is the pyramid of numbers where each row is formed by adding together the two numbers that are directly above it:

The triangle continues on this way, is named after a French mathematician named Blaise Pascal (find out more about Blaise Pascal) and is helpful when performing Binomial Expansions.

Notice that the 5th row, for example, has 6 entries. Like the 0th row, the first entry in any one row is the 0th entry. Consider the first 15 in the 6th row, we call this $^{6}C_{2}$, pronounced ‘6 choose 2’. This can also be written as ${6}\choose{2}$. In general, we write $^{n}C_r$ or ${n}\choose {r}$ and is calculated as

$^{n}C_r=\left(\begin{array}{c}n\\r\end{array}\right)=\frac{n!}{(n-r)!r!}$

This comes from summing all the terms above a given entry and simplifies to a fraction with factorials. $^{n}C_r$ can be thought of as the number of combinations of putting r balls in n buckets. It is also the number of times you get an $x^ry^{n-r}$ term in the expansion of $(x+y)^n$. Hence, this is why Pascal’s triangle is useful in Binomial Expansion. Note that there is a button on your calculator for working out ${n}\choose{r}$ – you don’t necessarily need to calculate the individual factorials. You might also notice that ${{n}\choose{0}}={{n}\choose{n}}=1$ and ${{n}\choose{1}}={{n}\choose{n-1}}=n$ always.

### Binomial Expansion

Suppose now that we wish to expand $(x+y)^n$, i.e. find the Binomial Expansion. In the simple case where n is a relatively small integer value, we expand the expression one bracket at a time. See Examples 1 and 2. Expanding $(x+y)^n$ by hand for larger n becomes a tedious task. The Edexcel Formula Booklet provides the following formula for binomial expansion:

$(a+b)^n=a^n+\left(\begin{array}{c}n\\1\end{array}\right)a^{n-1}b+\left(\begin{array}{c}n\\2\end{array}\right)a^{n-2}b^2+…+\left(\begin{array}{c}n\\r\end{array}\right)a^{n-r}b^r+…+b^n,\hspace{30pt}n\in {\mathbb N}$

where

$\left(\begin{array}{c}n\\r\end{array}\right)=\frac{n!}{(n-r)!r!}$

(see Pascal’s triangle notes for more on this) for when $n\in{\mathbb N}$, i.e for when n is a positive integer. Directly substituting $x$ in place of $a$ and $y$ in place of $b$ results in finding the expansions for larger $n$. Usually only the first few terms are required – see Example 3. You may substitute other expressions or numbers for $a$ and $b$ – see Example 4. Notice that this question asks you for descending powers of $x$ so you may need to swap the variables accordingly. Note that when there are also coefficients inside the brackets, the coefficients in the expansion change dramatically from those given in Pascal’s triangle.

### Relationship to Binomial Probabilities

Consider a binomially distributed random variable with $n$ trials and probability of success $p$ – see Binomial Distribution. If we require r of the trials to be successful (probability $p^r$) we require the remaining $n-r$ trials to be unsuccessful (probability $(1-p)^{n-r}$). The number of combinations in which there can be $r$ successes out of $n$ trials is ${n}\choose{r}$ (see above). Finally, the associated probability is given by

$P(X=r)={{n}\choose{r}}p^r(1-p)^{n-r}$

when $X\sim B(n,p)$ as seen on the Binomial Distribution page. See Statistics Example to see this in practice.

## Examples of Binomial Expansion

Expand $(x+y)^3$.

$\begin{array}{l}&&(x+y)^3\\

&=&(x+y)(x+y)(x+y)\\&=&(x^2+2xy+y^2)(x+y)\\&=&x^3+3x^2y+3xy^2+y^3\end{array}$

Notice that the coefficients of this expansion correspond to the 3rd row of Pascal’s triangle.

Using Example 1, expand $(x+y)^4$.

$\begin{array}{l}&&(x+y)^4\\&=&(x+y)^3(x+y)\\&=&(x^3+3x^2y+3xy^2+y^3)(x+y)\\&=&x^4+4x^3y+6x^2y^2+4xy^3+y^4\end{array}$

using the expansion in Example 1. Notice that the coefficients of this expansion correspond to the 4th row of Pascal’s triangle.

Find the first three terms in the expansion $(x+y)^8$.

There are a number of ways that this can be done. Firstly, we could find the first few entries in the next row of Pascal’s triangle (1, 8, 28, etc) and use those as coefficients:

$(x+y)^8=x^8+8x^7y+28x^6y^2+…$

Alternatively, and recommended since we don’t always have Pascal’s triangle and it could be a row much lower down, calculate the coefficient using the ${n}\choose{r}$ formula:

${{8}\choose{0}} =1$, ${{8}\choose{1}} =8$, ${{8}\choose{2}} =28$.

Find the first three terms, in descending powers of $x$, of the binomial expansion of $(2x+4)^5$.

We do this using the formula above. Perform a direct substitution as follows: $a=2x$, $b=4$ and $n=5$ and take the first three terms.

$(2x+4)^5=(2x)^5+\left(\begin{array}{c}5\\1\end{array}\right)(2x)^{5-1}(4)+\left(\begin{array}{c}5\\2\end{array}\right)(2x)^{5-2}(4)^2..$

Note that

$\left(\begin{array}{c}5\\1\end{array}\right)=\frac{5!}{(5-1)!1!}=\frac{120}{24\times 1}=5$,

and

$\left(\begin{array}{c}5\\2\end{array}\right)=\frac{5!}{(5-2)!2!}=\frac{120}{6\times 2}=10$,

(or seen in Pascal’s triangle) and so the formula becomes

$\begin{array}{l}&&(2x+4)^5\\&=&(2x)^5+5\times(2x)^{4}(4)+10\times(2x)^{3}(4)^2+…\\&=&32x^5+5\times 16x^{4}\times 4+10\times 8x^{3}\times 16+…\\&=&32x^5+320x^4+1280x^3+…\end{array}$

### (Statistics)

Consider the binomially distributed random variable $X\sim B(7,x)$. Find the probability $P(X=3)$ in terms of x. Write your answer as a polynomial in x.

Using the formula:

$P(X=3)={{7}\choose{3}}x^3(1-x)^4$.

We can use Example 2 above to expand $(1-x)^4$:

$\begin{array}{l}P(X=3)&=&35x^3\left(1-4x+6x^2-4x^3+x^4\right)\\&=&35x^3-140x^4+210x^5-140x^4+35x^3\end{array}$

- Given that ${{8}\choose{2}}=\frac{8!}{2!q!}$, write down the value of $q$.
- Given that the coefficient of $x^2$ in the expansion of $\left(p-\frac{x}{8}\right)^8$ is 28, find the value of $p$.
- Using the first three terms of a binomial expansion, estimate the value of $1.995^8$.

- The formula for ‘n choose r’ is given by ${{n}\choose{r}}=\frac{n!}{r!(n-r)!}$. Setting $n=8$ and $r=2$ gives the missing term $n-r=6$ and so $q=6$.
- One can perform the full expansion up to the $x^2$ term or notice that only the coefficient of $x^2$ is required. That is, the coefficient when the term ${{8}\choose{2}}p^6\left(-\frac{x}{8}\right)^2$ is simplified. Notice that $a$ and $b$ are interchangeable in the formula and are chosen so that the power of $x$ is $2$. The coefficient is thus $28\times p^6\times \frac{1}{64}$ and equal to 28 giving $p=2$.
- Find the first 3 terms in the binomial expansion using $p=2\\$:$\begin{array}{l}\left(2-\frac{x}{8}\right)^8&\approx& 2^8-{{8}\choose{1}}\times 2^7\times-\frac{x}{4}+28 x^2\\&=&256-256x+28x^2\end{array}$. Note that we already knew the coefficient of the $x^2$ term. Using this expansion suggests that we should choose x so that $2-\frac{x}{8}=1.995$, that is, $0.005=\frac{x}{8}$ or $x=0.04$. Substituting $x=0.04$ into the expansion gives $\begin{array}{l}(1.995)^8&\approx&256-256\times 0.04+28\times 0.04^2\\&=&245.8048\end{array}$. The actual answer to 7 decimal places, using a calculator, is 250.9245767, so not a great approximation.

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