Finding Specific Terms (AA HL)
You rarely need to write out an entire binomial expansion. Exam questions almost always ask for one piece of it - the coefficient of \(x^5\), the term in \(x^3y^2\), or the term that has no \(x\) in it at all - and there's a direct route to each that skips every term you don't need. It's part of the broader Binomial Theorem topic.
12 questions on this sub-topic.
The general term
Covered under IB syllabus reference SL1.9: the binomial theorem for \((a+b)^n\), \(n \in \mathbb{N}\), using \(\binom{n}{r}\) found by formula or technology - the general term below is just one piece of that expansion picked out and studied on its own.
General term
\(T_{r+1}=\binom{n}{r}a^{n-r}b^r\)
The \((r+1)\)th term of \((a+b)^n\). This is not printed separately in the formula booklet - it comes straight out of the summation form of the binomial theorem, which is booklet material.
Finding a specific coefficient
Set the powers of \(a\) and \(b\) in the general term equal to what the question asks for, solve for \(r\), then substitute back to get the coefficient.
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
Find the coefficient of \(x^3\) in the expansion of \((2+x)^6.\)
Worked solution
\(\binom{6}{r}2^{6-r}x^{r}.\) M1
Choose \(r=3\): A1
\(\binom{6}{3}2^{3} = 20\times 8\) M1
\(= 160.\) A1
Find the term independent of \(x\) in the expansion of \(\left(2x - \dfrac{3}{x^2}\right)^{9}.\)
Worked solution
\(\binom9k (2x)^{9-k}\left(-\dfrac{3}{x^2}\right)^k\) M1 \(= \binom9k 2^{9-k}(-3)^k x^{9-k-2k}.\) A1
\(9 - 3k = 0 \Rightarrow k = 3.\) M1
\(\binom93 2^{6}(-3)^3 = 84\cdot 64\cdot(-27).\) M1 \(= -145152.\) A1
Find the term independent of \(x\) in \(\left(3x^2-\dfrac{1}{2x}\right)^9.\)
Worked solution
\(\binom{9}{r}(3x^2)^{9-r}\left(-\tfrac{1}{2x}\right)^r\) M1 \(= \binom{9}{r}3^{9-r}(-\tfrac12)^r x^{18-2r-r}.\) A1
\(18-3r=0 \Rightarrow r=6.\) M1
\(\binom{9}{6}3^{3}(-\tfrac12)^6 = 84\cdot 27\cdot\tfrac{1}{64}\) A1 \(= \dfrac{2268}{64} = \dfrac{567}{16}.\) A1
Common mistakes
- Forgetting to raise the whole second term to the power \(r\). In \(\left(2x-\tfrac{3}{x^2}\right)^9\), a term like \(-\tfrac{3}{x^2}\) must be raised to \(k\) as a whole, sign included - dropping the negative sign or only raising part of the fraction gives the wrong coefficient.
- Solving for the wrong power of \(x\). "The term independent of \(x\)" means the power of \(x\) equals zero, not one - set the exponent expression equal to \(0\) before solving for \(r\).
- Writing out the full expansion to find one term. This wastes time and invites arithmetic slips - go straight to the general term, solve for the value of \(r\) you need, then substitute only that single value.
Ready to practise properly?
12 specific-term questions, marked instantly like the real exam.
Quick answers
How do I find a specific term without expanding the whole bracket?
Write down the general term \(T_{r+1}=\binom{n}{r}a^{n-r}b^r\), work out which value of \(r\) gives the power you want, then substitute that one value of \(r\) back in.
How do I find the term independent of \(x\)?
Write the general term, collect the power of \(x\) into a single expression in \(r\), set that power equal to \(0\), solve for \(r\), then substitute back to get the constant term.