Finding Specific Terms (AA SL)
Expanding the whole of \((a+b)^n\) just to reach one particular term wastes time you don't have in an exam. The general term formula lets you jump straight to the term you need - a coefficient, a term in \(x^3\), or a constant term - without writing out everything before it. It's part of the broader Binomial Theorem topic.
22 questions on this sub-topic.
The formula
This sits under IB syllabus reference SL1.9, the same binomial theorem point covered on the parent topic page: expansion of \((a+b)^n\), \(n\in\mathbb{N}\), using Pascal's triangle and \(nC_r\).
A specific term
\[T_{r+1}=\binom{n}{r}a^{n-r}b^r\]
Set the power of the target variable equal to the power you want, solve for \(r\), then substitute.
Follows directly from the booklet formulaReading the index
\(T_{r+1}\), not \(T_r\)
The term you get when you substitute \(r\) is labelled \(T_{r+1}\), because the first term of the expansion (\(r=0\)) is \(T_1\), not \(T_0\).
Need the full syllabus wording and formula-booklet reference table? See Binomial Theorem.
Worked examples
Find the term in \(x^3\) in the expansion of \((2 + x)^6\).
Worked solution
For \((a+b)^n\) the term in position \(r+1\) is \(\binom{n}{r}a^{n-r}b^r\); choosing \(r\) controls the power of \(b\). Here \(a=2,\ b=x,\ n=6\), so \(T_{r+1}=\binom{6}{r}2^{6-r}x^{r}.\) M1
We need \(x^3\), so the exponent on \(x\) tells us \(r=3\). A1
\(\binom{6}{3}=20\) and \(2^{3}=8\), so \(T_4=20\times8\times x^3=160x^3.\) A1
Consider \(\left(2x + \dfrac{1}{x}\right)^6\).
(a) Write the general term.
(b) Find the constant term (the term independent of \(x\)).
Worked solution
(a) The powers of \(x\) come from both factors, so track them together: \(T_{r+1}=\binom{6}{r}(2x)^{6-r}\left(\tfrac1x\right)^{r}\) A1
which combines to \(=\binom{6}{r}2^{6-r}x^{6-2r}.\) A1
(b) The constant term has \(x^0\), so set the exponent to zero: \(6-2r=0\Rightarrow r=3\). M1
Substituting, \(T_4=\binom{6}{3}2^{3}=20\times8=160.\) A1
Common mistakes
- Solving for \(r\) but forgetting to substitute back. Finding \(r=3\) answers "which term?", not "what is the coefficient?" - the value of \(r\) still needs to go back into \(\binom{n}{r}a^{n-r}b^r\) to get the actual numerical answer.
- Losing track of the power on the coefficient of \(x\). When the term itself contains a coefficient, like \(2x\) or \(\tfrac1x\), that whole bracket is raised to the power, not just the \(x\) - dropping the \(2^{6-r}\) or similar factor is a common slip.
- Confusing \(T_{r+1}\) with \(T_r\). The term obtained by substituting a given \(r\) is labelled \(T_{r+1}\), since \(r=0\) gives the first term \(T_1\) - stating "the 4th term" when you mean \(r=4\) is an easy indexing error.
Ready to practise properly?
21 questions on finding specific terms, marked instantly like the real exam.
Quick answers
How do I find a specific term in a binomial expansion without expanding everything?
Use the general term \(T_{r+1}=\binom{n}{r}a^{n-r}b^r\). Set the power in the term you want equal to the power of \(b\), solve for \(r\), then substitute \(r\) back in to get the actual coefficient or term.
What is the general term formula for a binomial expansion?
\(T_{r+1}=\binom{n}{r}a^{n-r}b^r\), where \(r\) runs from \(0\) to \(n\). It isn't given separately in the formula booklet, but it follows directly from the full expansion formula that is. See the parent topic's GDC guidance for checking values of \(nC_r\).