Mathematics 9709 · Pure Mathematics 1

The binomial expansion

The binomial expansion is a fast way to multiply out $(a+b)^n$ . You do not have to do it by hand.
Paper 1 asks in two ways. One asks for the first few terms in order of rising powers. The other asks for one term from the middle. You use the same method for both.

Every term of $(a+b)^n$ is $\binom{n}{r}\,a^{\,n-r}\,b^{\,r}$ . There are three parts. The number in front is $\binom{n}{r}$ . The power of $a$ counts down from $n$ . The power of $b$ counts up from $0$ . The number $\binom{n}{r}$ is just a row of Pascal's triangle. You get it from one calculator button, so you do not draw the triangle. To write the whole expansion, use $r=0,1,2,\dots$ . To get one number, pick the single $r$ that gives the power you want.

One term, three pieces

When you multiply $(a+b)^n$ out, every term has the same shape. It is a Pascal number times a power of $a$ times a power of $b$ :

$\binom{n}{r}\,a^{\,n-r}\,b^{\,r}$

The two powers always add to $n$ (here $(n-r)+r=n$ ). This is a quick way to check your work.
$\binom{n}{r}$ (“n choose r”) is the Pascal number. It is the ${}^{n}\mathrm{C}_{r}$ button. Type $n$ , press ${}^{n}\mathrm{C}_{r}$ , then type $r$ .
The counter $r$ goes from $0$ to $n$ . This gives you the whole expansion.

Here is the fast way to answer a question. Pick the one $r$ that gives the power you are asked for. Then work out only that term.

Pascal's triangle. Row n gives the numbers in front for (a + b)ⁿ. Each number is the sum of the two numbers just above it.

The triangle is fine for small $n$ . But in the exam, use the ${}^{n}\mathrm{C}_{r}$ button. It gives $\binom{n}{r}$ right away.
You should also be able to do it by hand. Some questions do not let you use the calculator.
It comes from the factorial $n!$ (“multiply all the way down to $1$ ”), $n!=n(n-1)(n-2)\cdots 3\cdot 2\cdot 1$ :

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

The quick way: write $r$ numbers on top, counting down from $n$ , over $r$ numbers counting down from $r$ :

$\binom{5}{2}=\frac{5\cdot 4}{2\cdot 1}=10,\qquad \binom{8}{4}=\frac{8\cdot 7\cdot 6\cdot 5}{4\cdot 3\cdot 2\cdot 1}=70$

Where the Pascal number comes from

Expanding $(a+b)^n$ means you multiply $n$ copies of $(a+b)$ . To make the term $a^{n-r}b^{r}$ , you pick $b$ from exactly $r$ of those $n$ brackets. You pick $a$ from the rest. The number of ways to pick $r$ brackets out of $n$ is $\binom{n}{r}$ . So you get that many copies of $a^{n-r}b^{r}$ . That is why “n choose r” is also the number in front. It also explains the triangle. Each row adds the two numbers above, because a choice in the next row either takes the new bracket or it does not.

Writing out the first few terms

The most common version asks for the first three or four terms, in ascending powers of $x$ .
“Ascending” means smallest power first. So you write the $r=0$ term, then $r=1$ , then $r=2$ . You stop once you have enough terms. You only need the start, not the whole thing.

Use it like a fill-in template. First write the numbers in front: $\binom{n}{0},\binom{n}{1},\binom{n}{2},\dots$ . Next to each one, the power of $a$ goes down $n,n-1,n-2,\dots$ . At the same time the power of $b$ goes up $0,1,2,\dots$ . Each row is one value of $r$ . The two powers always add to $n$ .

Worked example

Find the first four terms of $(2-3x)^{10}$ in ascending powers of $x$

Here $a=2$ , $b=-3x$ , $n=10$ . Fill the template for $r=0,1,2,3$ . Raise the whole $(-3x)$ each time, including its sign and its number:

$\binom{10}{0}2^{10}+\binom{10}{1}2^{9}(-3x)+\binom{10}{2}2^{8}(-3x)^2+\binom{10}{3}2^{7}(-3x)^3+\dots$

Now work out each row. The numbers in front, $\binom{10}{0},\binom{10}{1},\binom{10}{2},\binom{10}{3}$ , are $1,10,45,120$ (use the button or do it by hand):

$1\cdot 1024+10\cdot 512(-3x)+45\cdot 256\cdot 9x^2+120\cdot 128\cdot(-27x^3)$

Answer

1024-15360x+103680x^2-414720x^3+\dots

The signs flip plus and minus because $b$ is negative. Odd powers of $(-3x)$ are minus. Even powers are plus.

Pulling out a single coefficient

The other version asks for the number in front of one power from the middle. It comes from something like $(x-\tfrac{2}{x})^6$ .
You do not write every term. Instead, find the one $r$ that gives the power you want. Then work out only that term.

Write the general term $\binom{n}{r}a^{n-r}b^{r}$ . Put every $x$ into a single power. Set that power equal to the power you want. Then solve for $r$ . You get one value of $r$ , and one number to work out. “Independent of $x$ ” is the same job, but you set the power to $0$ .

Write the general term and collect all the $x$ powers into a single exponent.
Set that exponent equal to the power you want and solve for $r$ .
Put that one $r$ back in and work out the number.

Worked example

Find the coefficient of $x^2$ in $\left(x - \tfrac{2}{x}\right)^6$

Write the general term. Then split the second power so you can see every $x$ :

$\binom{6}{r}\,x^{6-r}\left(-\tfrac{2}{x}\right)^{r}=\binom{6}{r}(-2)^{r}\,x^{6-r}x^{-r}$

Step 1. Combine the $x$ powers: $x^{6-r}x^{-r}=x^{6-2r}$ , so the term is $\binom{6}{r}(-2)^{r}\,x^{6-2r}$ .

Step 2. We want $x^2$ . So set the power equal to $2$ and solve for $r$ :

$6-2r=2 \;\Rightarrow\; r=2$

Step 3. Substitute $r=2$ : $\binom{6}{2}(-2)^{2}=15\times 4$ .

Answer

\text{coefficient of }x^2 = 60

(9709/12 Jun 2020 Q1a) The answer is just a number. Do not put the $x^2$ in the final line.

For the term independent of $x$ (the constant term), you change only one thing. You set the power to $0$ .
The same three steps work even when there is a number in front of the $x$ and a higher power on the bottom.

Worked example

Find the term independent of $x$ in $\left(3x + \tfrac{2}{x^2}\right)^6$

Here $a=3x$ and $b=\tfrac{2}{x^2}$ . Write the general term. Keep every number with its bracket:

$\binom{6}{r}(3x)^{6-r}\left(\tfrac{2}{x^2}\right)^{r}=\binom{6}{r}3^{\,6-r}\,2^{\,r}\,x^{6-r}x^{-2r}$

Step 1. Collect the $x$ powers: $x^{6-r}x^{-2r}=x^{6-3r}$ .

Step 2. “Independent of $x$ ” means the power is $0$ :

$6-3r=0 \;\Rightarrow\; r=2$

Step 3. Put $r=2$ back in. Do not forget the power on either number:

$\binom{6}{2}\,3^{4}\,2^{2}=15\times 81\times 4$

Answer

\text{term independent of }x = 4860

(9709/12 Mar 2022 Q3a) The mark scheme gives one mark for $6-3r=0$ and one mark for the number $4860$ . So write that power equation down, even if you can see $r=2$ fast.

A coefficient from a product of two brackets

A harder version puts a short bracket in front of the expansion, like $(2+3x^2)(x-\tfrac{2}{x})^6$ .
Now you can make an $x^2$ in more than one way. So you add up all the parts.

Take each piece of the front bracket in turn. Ask: which power from the expansion do I need, so the two multiply to the power I want? Find that one term and multiply. Then add up all the parts.

Worked example

Find the coefficient of $x^2$ in $(2 + 3x^2)\left(x - \tfrac{2}{x}\right)^6$

From the last example, the term in $(x-\tfrac{2}{x})^6$ is $\binom{6}{r}(-2)^{r}\,x^{6-2r}$ . Take each piece of the front bracket in turn.

The $2$ has no $x$ . So it needs the $x^2$ term from the expansion (that is the $60x^2$ we just found):

$2\times 60x^2=120x^2$

The $3x^2$ already has $x^2$ . So it needs the constant term (the $x^0$ term). Set $6-2r=0$ , which gives $r=3$ :

$\binom{6}{3}(-2)^{3}=20\times(-8)=-160$

So this piece gives $3x^2\times(-160)=-480x^2$ .

Add the two parts:

$120+(-480)=-360$

Answer

\text{coefficient of }x^2 = -360

(9709/12 Jun 2020 Q1) The main idea is simple. Match each front piece to the one expansion term that gives the power you want.

Where the marks go

Common mistake

Raise the whole second term. Include its sign and its number. For example

\left(-\tfrac{2}{x}\right)^3=-\tfrac{8}{x^3}

, not

-\tfrac{2}{x^3}

. Forgetting the power on the number is the most common slip here. (9709/12 Mar 2022 Q3)

Common mistake

The question asks for the coefficient. That is a number, not the whole term. Do not leave the

x^2

on your final answer.

Common mistake

In a product of two brackets, check you have found every way to make the power you want. If you miss one part, the answer is wrong, even when each term you did write is right.

Now you try

Find the term independent of $x$ in the expansion of $\left(x+\dfrac{3}{x^2}\right)^{6}$ . (9709/12 Nov 2024 Q4)

Your turn— tap to reveal the worked answer (9709/12 Nov 2024 Q4)

General term, with the powers put together: $\binom{6}{r}x^{6-r}\left(\tfrac{3}{x^2}\right)^{r}=\binom{6}{r}3^{r}\,x^{6-3r}$ .

“Independent of $x$ ” means the power is zero: $6-3r=0\Rightarrow r=2$ .

Put $r=2$ back in: $\binom{6}{2}\,3^{2}=15\times 9$ .

Answer

\text{term independent of }x = 135

Next in this chapter: arithmetic progressions. There you add the same amount each step, instead of using a fixed power.

PreviousSolving trigonometric equations NextArithmetic progressions

The binomial expansion

The binomial expansion is a fast way to multiply out $(a+b)^n$ . You do not have to do it by hand.
Paper 1 asks in two ways. One asks for the first few terms in order of rising powers. The other asks for one term from the middle. You use the same method for both.

One term, three pieces

When you multiply $(a+b)^n$ out, every term has the same shape. It is a Pascal number times a power of $a$ times a power of $b$ :

$\binom{n}{r}\,a^{\,n-r}\,b^{\,r}$

The two powers always add to $n$ (here $(n-r)+r=n$ ). This is a quick way to check your work.
$\binom{n}{r}$ (“n choose r”) is the Pascal number. It is the ${}^{n}\mathrm{C}_{r}$ button. Type $n$ , press ${}^{n}\mathrm{C}_{r}$ , then type $r$ .
The counter $r$ goes from $0$ to $n$ . This gives you the whole expansion.

Here is the fast way to answer a question. Pick the one $r$ that gives the power you are asked for. Then work out only that term.

Pascal's triangle. Row n gives the numbers in front for (a + b)ⁿ. Each number is the sum of the two numbers just above it.

The triangle is fine for small $n$ . But in the exam, use the ${}^{n}\mathrm{C}_{r}$ button. It gives $\binom{n}{r}$ right away.
You should also be able to do it by hand. Some questions do not let you use the calculator.
It comes from the factorial $n!$ (“multiply all the way down to $1$ ”), $n!=n(n-1)(n-2)\cdots 3\cdot 2\cdot 1$ :

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

The quick way: write $r$ numbers on top, counting down from $n$ , over $r$ numbers counting down from $r$ :

$\binom{5}{2}=\frac{5\cdot 4}{2\cdot 1}=10,\qquad \binom{8}{4}=\frac{8\cdot 7\cdot 6\cdot 5}{4\cdot 3\cdot 2\cdot 1}=70$

Where the Pascal number comes from

Writing out the first few terms

The most common version asks for the first three or four terms, in ascending powers of $x$ .
“Ascending” means smallest power first. So you write the $r=0$ term, then $r=1$ , then $r=2$ . You stop once you have enough terms. You only need the start, not the whole thing.

Worked example

Find the first four terms of $(2-3x)^{10}$ in ascending powers of $x$

Here $a=2$ , $b=-3x$ , $n=10$ . Fill the template for $r=0,1,2,3$ . Raise the whole $(-3x)$ each time, including its sign and its number:

$\binom{10}{0}2^{10}+\binom{10}{1}2^{9}(-3x)+\binom{10}{2}2^{8}(-3x)^2+\binom{10}{3}2^{7}(-3x)^3+\dots$

Now work out each row. The numbers in front, $\binom{10}{0},\binom{10}{1},\binom{10}{2},\binom{10}{3}$ , are $1,10,45,120$ (use the button or do it by hand):

$1\cdot 1024+10\cdot 512(-3x)+45\cdot 256\cdot 9x^2+120\cdot 128\cdot(-27x^3)$

Answer

1024-15360x+103680x^2-414720x^3+\dots

The signs flip plus and minus because $b$ is negative. Odd powers of $(-3x)$ are minus. Even powers are plus.

Pulling out a single coefficient

The other version asks for the number in front of one power from the middle. It comes from something like $(x-\tfrac{2}{x})^6$ .
You do not write every term. Instead, find the one $r$ that gives the power you want. Then work out only that term.

Write the general term and collect all the $x$ powers into a single exponent.
Set that exponent equal to the power you want and solve for $r$ .
Put that one $r$ back in and work out the number.

Worked example

Find the coefficient of $x^2$ in $\left(x - \tfrac{2}{x}\right)^6$

Write the general term. Then split the second power so you can see every $x$ :

$\binom{6}{r}\,x^{6-r}\left(-\tfrac{2}{x}\right)^{r}=\binom{6}{r}(-2)^{r}\,x^{6-r}x^{-r}$

Step 1. Combine the $x$ powers: $x^{6-r}x^{-r}=x^{6-2r}$ , so the term is $\binom{6}{r}(-2)^{r}\,x^{6-2r}$ .

Step 2. We want $x^2$ . So set the power equal to $2$ and solve for $r$ :

$6-2r=2 \;\Rightarrow\; r=2$

Step 3. Substitute $r=2$ : $\binom{6}{2}(-2)^{2}=15\times 4$ .

Answer

\text{coefficient of }x^2 = 60

(9709/12 Jun 2020 Q1a) The answer is just a number. Do not put the $x^2$ in the final line.

For the term independent of $x$ (the constant term), you change only one thing. You set the power to $0$ .
The same three steps work even when there is a number in front of the $x$ and a higher power on the bottom.

Worked example

Find the term independent of $x$ in $\left(3x + \tfrac{2}{x^2}\right)^6$

Here $a=3x$ and $b=\tfrac{2}{x^2}$ . Write the general term. Keep every number with its bracket:

$\binom{6}{r}(3x)^{6-r}\left(\tfrac{2}{x^2}\right)^{r}=\binom{6}{r}3^{\,6-r}\,2^{\,r}\,x^{6-r}x^{-2r}$

Step 1. Collect the $x$ powers: $x^{6-r}x^{-2r}=x^{6-3r}$ .

Step 2. “Independent of $x$ ” means the power is $0$ :

$6-3r=0 \;\Rightarrow\; r=2$

Step 3. Put $r=2$ back in. Do not forget the power on either number:

$\binom{6}{2}\,3^{4}\,2^{2}=15\times 81\times 4$

Answer

\text{term independent of }x = 4860

(9709/12 Mar 2022 Q3a) The mark scheme gives one mark for $6-3r=0$ and one mark for the number $4860$ . So write that power equation down, even if you can see $r=2$ fast.

A coefficient from a product of two brackets

A harder version puts a short bracket in front of the expansion, like $(2+3x^2)(x-\tfrac{2}{x})^6$ .
Now you can make an $x^2$ in more than one way. So you add up all the parts.

Take each piece of the front bracket in turn. Ask: which power from the expansion do I need, so the two multiply to the power I want? Find that one term and multiply. Then add up all the parts.

Worked example

Find the coefficient of $x^2$ in $(2 + 3x^2)\left(x - \tfrac{2}{x}\right)^6$

From the last example, the term in $(x-\tfrac{2}{x})^6$ is $\binom{6}{r}(-2)^{r}\,x^{6-2r}$ . Take each piece of the front bracket in turn.

The $2$ has no $x$ . So it needs the $x^2$ term from the expansion (that is the $60x^2$ we just found):

$2\times 60x^2=120x^2$

The $3x^2$ already has $x^2$ . So it needs the constant term (the $x^0$ term). Set $6-2r=0$ , which gives $r=3$ :

$\binom{6}{3}(-2)^{3}=20\times(-8)=-160$

So this piece gives $3x^2\times(-160)=-480x^2$ .

Add the two parts:

$120+(-480)=-360$

Answer

\text{coefficient of }x^2 = -360

(9709/12 Jun 2020 Q1) The main idea is simple. Match each front piece to the one expansion term that gives the power you want.

Where the marks go

Common mistake

Raise the whole second term. Include its sign and its number. For example

\left(-\tfrac{2}{x}\right)^3=-\tfrac{8}{x^3}

, not

-\tfrac{2}{x^3}

. Forgetting the power on the number is the most common slip here. (9709/12 Mar 2022 Q3)

Common mistake

The question asks for the coefficient. That is a number, not the whole term. Do not leave the

x^2

on your final answer.

Common mistake

In a product of two brackets, check you have found every way to make the power you want. If you miss one part, the answer is wrong, even when each term you did write is right.

Now you try

Find the term independent of $x$ in the expansion of $\left(x+\dfrac{3}{x^2}\right)^{6}$ . (9709/12 Nov 2024 Q4)

Your turn— tap to reveal the worked answer (9709/12 Nov 2024 Q4)

General term, with the powers put together: $\binom{6}{r}x^{6-r}\left(\tfrac{3}{x^2}\right)^{r}=\binom{6}{r}3^{r}\,x^{6-3r}$ .

“Independent of $x$ ” means the power is zero: $6-3r=0\Rightarrow r=2$ .

Put $r=2$ back in: $\binom{6}{2}\,3^{2}=15\times 9$ .

Answer

\text{term independent of }x = 135

Next in this chapter: arithmetic progressions. There you add the same amount each step, instead of using a fixed power.