Binomial theorem questions are some of the most predictable marks on an IB Maths AA SL paper. There are only a handful of question types, and one method handles all of them: write down the general term, choose the power you need, and solve. This post shows that method, then works through each question type the way an examiner would mark it.
What the syllabus expects
At SL you need to expand $(a+b)^n$ for a positive integer $n$ and use $\binom{n}{r}$, either from Pascal's triangle or with technology. The full expansion is:
$$(a+b)^n = a^n + \binom{n}{1}a^{n-1}b + \dots + \binom{n}{r}a^{n-r}b^r + \dots + b^n, \qquad \binom{n}{r}=\frac{n!}{r!\,(n-r)!}$$
Check your formula booklet for how it sets out the binomial expansion. Whichever way it is printed, the term with $b^r$ in it is the part that does all the work:
The general term
The term containing $b^r$ is $\binom{n}{r}a^{n-r}b^r$. Every "find the coefficient", "find the constant term" and "find $a$" question starts by writing this down with the question's own $a$, $b$ and $n$.
The five question types
- First few terms: "Find the first three terms in ascending powers of $x$."
- One coefficient: "Find the coefficient of $x^3$."
- The constant term (the term independent of $x$), usually with an $x$ in a denominator.
- An unknown constant inside the bracket, found from a given coefficient, often with an extra bracket multiplying the expansion.
- An unknown power $n$, found from a relationship between coefficients.
Types 1 and 2 are the quicker marks. Types 3 to 5 are harder, because you have to set up an equation before you can solve anything.
Type 1: the first few terms
Example 1 · Paper 1 · 4 marks
Find the first three terms, in ascending powers of $x$, in the expansion of $(2-x)^5$.
Solution
Here $a=2$, $b=-x$ and $n=5$. Write each term with its binomial coefficient:
$$\binom{5}{0}2^5 + \binom{5}{1}2^4(-x) + \binom{5}{2}2^3(-x)^2$$
The coefficients are $1, 5, 10$ (Pascal's triangle), so the terms are $32 - 80x + 80x^2$.
Marks: (M1) for the correct structure of at least two terms, (A1)(A1)(A1) for each correct term.
Where the marks go
The sign. With $b=-x$, the term $\binom{5}{1}2^4(-x)$ is negative. Put brackets round $(-x)$ every time and the sign looks after itself.
Type 2: one coefficient
Do not expand the whole bracket. Write the general term, collect the powers of $x$, and set the power equal to the one you want.
Example 2 · Paper 2 · 5 marks
Find the coefficient of $x^3$ in the expansion of $\left(3x-\dfrac{2}{x}\right)^7$.
Solution
General term: $\binom{7}{r}(3x)^{7-r}\left(-\dfrac{2}{x}\right)^r$.
The power of $x$ is $(7-r) - r = 7-2r$. Setting $7-2r=3$ gives $r=2$.
The term is $\binom{7}{2}(3x)^5\left(-\dfrac{2}{x}\right)^2 = 21 \times 243x^5 \times \dfrac{4}{x^2} = 20412x^3$.
The coefficient is $20412$.
Marks: (M1) for a valid general term, (M1) for an equation in $r$ from the powers of $x$, (A1) $r=2$, (M1) substituting, (A1) $20412$.
Two habits protect these marks: put a bracket round $(3x)$ before raising it to a power, so the $3$ is raised to the power too, and write the power of $x$ as an expression in $r$ before you solve.
Type 3: the term independent of $x$
"Independent of $x$" means the power of $x$ is zero. The method is the same as Type 2, with the target power set to $0$.
Example 3 · Paper 1 · 5 marks
Find the term independent of $x$ in the expansion of $\left(x^2+\dfrac{1}{2x}\right)^6$.
Solution
General term: $\binom{6}{r}(x^2)^{6-r}\left(\dfrac{1}{2x}\right)^r$.
The power of $x$ is $2(6-r) - r = 12-3r$. Setting $12-3r=0$ gives $r=4$.
The term is $\binom{6}{4}(x^2)^2\left(\dfrac{1}{2x}\right)^4 = 15 \times x^4 \times \dfrac{1}{16x^4} = \dfrac{15}{16}$.
The question asks for the term, so the answer is the number $\frac{15}{16}$. If a question asks for the term containing $x^3$, the answer includes the $x^3$.
Type 4: an unknown constant, with an extra bracket
When the expansion is multiplied by another bracket, an $x^2$ term can arise in more than one way. List every way before you add.
Example 4 · Paper 1 · 6 marks
The coefficient of $x^2$ in the expansion of $(1+2x)(1+ax)^6$ is $27$. Find the possible values of $a$.
Solution
The first three terms of $(1+ax)^6$ are $1 + 6ax + 15a^2x^2$.
An $x^2$ term comes from $1 \times 15a^2x^2$ and from $2x \times 6ax$, so the coefficient of $x^2$ is $15a^2 + 12a$.
Setting $15a^2+12a = 27$ gives $5a^2+4a-9=0$, so $(5a+9)(a-1)=0$.
So $a=1$ or $a=-\dfrac{9}{5}$.
Missing the $2x \times 6ax$ cross term is the most common way to lose most of these marks: the equation becomes $15a^2=27$ and every mark after the first is gone. Keep both answers unless the question restricts $a$.
Type 5: finding the power $n$
When $n$ is unknown you cannot use Pascal's triangle, so write $\binom{n}{r}$ algebraically:
$$\binom{n}{1}=n, \qquad \binom{n}{2}=\frac{n(n-1)}{2}, \qquad \binom{n}{3}=\frac{n(n-1)(n-2)}{6}$$
Example 5 · Paper 1 · 5 marks
In the expansion of $(1+2x)^n$, where $n$ is a positive integer, the coefficient of $x^2$ is ten times the coefficient of $x$. Find $n$.
Solution
Coefficient of $x$: $\binom{n}{1}\times 2 = 2n$.
Coefficient of $x^2$: $\binom{n}{2}\times 2^2 = \dfrac{n(n-1)}{2}\times 4 = 2n(n-1)$.
So $2n(n-1) = 10 \times 2n$, which gives $2n^2 - 22n = 0$, so $2n(n-11)=0$.
Since $n$ is a positive integer, $n=11$.
Check: in $(1+2x)^{11}$ the coefficient of $x$ is $22$ and of $x^2$ is $220$.
Factorise rather than dividing both sides by $n$. Dividing hides the solution $n=0$; a complete answer finds it and rejects it with a reason: $n$ must be a positive integer.
A checklist for the exam
- Write the general term $\binom{n}{r}a^{n-r}b^r$ with the question's own $a$, $b$ and $n$, with brackets round both.
- Write the power of $x$ as an expression in $r$, then solve for $r$.
- On Paper 2, use nCr on your GDC for the numbers, but still write the term out: the method marks are for the working, not the button.
- For a product of brackets, list every pair of terms that gives the power you want.
- Answer the question asked: a coefficient is a number, a term includes its power of $x$.
Every example here follows the same pattern as the worked examples and 37 practice questions on the AA SL binomial theorem questions page. The SL AA formulas page collects the course formulas in one place; check them against your own copy of the official formula booklet.
Practise this topic
FAQ
Is the binomial theorem on Paper 1 or Paper 2 in IB Maths AA SL?
It can appear on either. On Paper 1 you work out $\binom{n}{r}$ by hand or from Pascal's triangle; on Paper 2 your GDC's nCr function does it for you, but you still need to show the general term.
Is the binomial expansion formula in the IB formula booklet?
The IB subject guide lists the expansion of $(a+b)^n$, $n \in \mathbb{N}$, as SL content. Check your own copy of the formula booklet to see how it is set out there. Either way, learn the general term $\binom{n}{r}a^{n-r}b^r$ by heart, because it is the line every question uses.
Do AA SL students need negative or fractional powers of $n$?
No. At SL, $n$ is a positive integer. Expansions with negative or fractional $n$ are AA HL content.
37 exam-style AA SL binomial theorem questions, easy to hard, each with a step-by-step mark scheme.
Try the binomial theorem questions →