Binomial Expansion – positive integer powers

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 and ‘n choose r’

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

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

Notice that the 5th row, for example, has 6 entries. The first entry in any one row is known as the 0th entry.  Similarly, the top row with a single 1 is known as the 0th row. This is because it is associated with the expansion of (x+y)^n. Now consider the first number 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:


This comes from summing all the appropriate 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 can expand the expression one bracket at a time. This is demonstrated in the first two Basic Binomial Expansion Examples in Section 3. Expanding (x+y)^n by hand for larger n becomes a tedious task. The Edexcel Formula Booklet provides the following formula for binomial expansion:


for n\in {\mathbb N}, where:


is n choose r as we saw earlier. Note that this expansion is only true for when n\in{\mathbb N}, i.e for when n is a positive integer. ┬áDirectly substituting x in place of a and y in place of b results in finding the expansions of (x+y)^n for larger n. Usually, only the first few terms are required. You may substitute other expressions or numbers for a and b but note that when there are added coefficients, the expanded coefficients change dramatically from those given in Pascal’s triangle. Also beware when the question asks you for descending powers of x – you may need to swap the variables accordingly.

Relationship to Binomial Probabilities

Before moving on to the examples section, take a moment to consider the relationship between Binomial Expansion and Binomial Distribution. Consider a binomially distributed random variable with n trials and probability of success p, that is, X\sim B(n,p). If we require r of the trials to be successful (with probability p^r) then we also require the remaining n-r trials to be unsuccessful (with 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 Pascal’s Triangle and ‘n choose r’ notation in the notes section). It follows that the associated probability is given by:


As we saw earlier in the notes, this looks very much like the general term in the Binomial Expansion. 


1. Expand (x+y)^3.
2. Hence, expand (x+y)^4.


1. \begin{array}{lll}(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}

2. \begin{array}{lll}(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 above. Notice that the coefficients of this expansion correspond to the 4th row of Pascal’s triangle.

1. Find the first three terms in the expansion (x+y)^8.
2. Find the first three terms, in descending powers of x, of the binomial expansion of (2x+4)^5.


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


Alternatively, and recommended since we don’t always have Pascal’s triangle, calculate the coefficient using the {n}\choose{r} formula:

{{8}\choose{0}} =1, \hspace{10pt}{{8}\choose{1}} =8, \hspace{10pt}{{8}\choose{2}} =28.

2. We do this using the Binomial Expansion formula:


Perform a direct substitution with a=2x, b=4 and n=5 and take the first three terms.


Note that

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


\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 expansion becomes

\begin{array}{lll}(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}

1. Given that {{8}\choose{2}}=\frac{8!}{2!q!}, write down the value of q.
2. 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.
3. Using the first three terms of a binomial expansion, estimate the value of 1.995^8.


  1. 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.
  2. 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.
  3. Find the first 3 terms in the binomial expansion using p=2: \left(2-\frac{x}{8}\right)^8\approx 2^8-{{8}\choose{1}}\times 2^7\times-\frac{x}{8}+28 x^2=256-128x+28x^2. 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 1.995^8\approx256-128\times 0.04+28\times 0.04^2=250.9248.  The actual answer to 7 decimal places, using a calculator, is 250.9245767, so a pretty good approximation.


A binomial expansion followed by finding missing coefficients when multiplied by a linear factor.

Binomial Expansion with an unknown coefficient combined with Simultaneous Equations.