Binomial Expansions with Unknowns (AA HL)

Most binomial questions give you every number up front, but this group hides one - an unknown power \(n\), an unknown coefficient inside the bracket, or a condition linking two terms - and asks you to recover it algebraically. The expansion itself works exactly as normal; the extra step is turning the given condition into an equation and solving it. It's part of the broader Binomial Theorem topic.

30 questions on this sub-topic.

Practise binomial expansions with unknowns → Try exam-style questions

The formulas you need

Covered under IB syllabus reference SL1.9: the binomial theorem for expansion of \((a+b)^n\) with \(n \in \mathbb{N}\), using Pascal's triangle or \(\binom{n}{r}\), found by formula or by technology. Both formulas below are in the formula booklet.

Binomial theorem

\((a+b)^n=\displaystyle\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r\)

Positive integer \(n\). When \(n\) is itself the unknown, leave it as a letter in this formula and build an equation from whatever condition the question gives you.

Binomial coefficient

\(\binom{n}{r}=\dfrac{n!}{r!(n-r)!}\)

Use this to compare or equate two coefficients symbolically - the ratio \(\binom{n}{r+1}/\binom{n}{r}\) often simplifies an "equal coefficients" condition down to a single linear equation.

Need the full syllabus wording and formula-booklet reference table? See Binomial Theorem. GDC methods for these expansions are covered in the parent topic's GDC section.

Worked examples

1
Medium
No calc
[4 marks]

Find the coefficient of \(x^4\) in \((1+x+x^2)(1+x)^6.\)

Worked solution

\(x^4:15,\ x^3:20,\ x^2:15.\) M1 A1
(each factor of \(1, x, x^2\) shifts the power): \(1\cdot 15 + 1\cdot 20 + 1\cdot 15\) M1 \(= 50.\) A1

M1 Relevant coefficients A1 \(15, 20, 15\) M1 Collect contributions A1 Coefficient \(=50\)
2
Hard
No calc
[4 marks]

In the expansion of \((1 + x)^{n}\), the coefficients of \(x^4\) and \(x^5\) are equal.

Find \(n.\)

Worked solution

\(\binom n4 = \binom n5.\) M1
\(\dfrac{\binom n5}{\binom n4} = \dfrac{n-4}{5}.\) M1 A1
\(\dfrac{n-4}{5} = 1 \Rightarrow n - 4 = 5 \Rightarrow n = 9.\) A1

M1 Equate coefficients M1 Form the ratio A1 Ratio \(=\tfrac{n-4}{5}\) A1 \(n=9\)
3
Easy
No calc
[4 marks]

Expand \((2-x)^4\) fully.

Worked solution

\(\binom40 2^4 - \binom41 2^3 x + \binom42 2^2 x^2 - \binom43 2x^3 + \binom44 x^4.\) M1
\(16 - 32x + 24x^2 - 8x^3 + x^4.\) A1 A1 A1

M1 Expansion with coefficients A1 \(16-32x\) A1 \(24x^2-8x^3\) A1 \(+x^4\), full expansion

Common mistakes

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Quick answers

How do I expand a binomial like \((1+x)^n\)?

Use the binomial theorem \((a+b)^n=\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r\), reading the coefficients \(\binom{n}{r}\) off Pascal's triangle or from the formula \(\dfrac{n!}{r!(n-r)!}\).

What if the power \(n\) is unknown in the question?

Keep \(n\) as a letter in the general term, turn whatever condition the question gives (two coefficients equal, one coefficient stated) into an equation, and solve that equation for \(n\).

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