To find one term of (a + b)ⁿ, use the general term: nCr × aⁿ⁻ʳ × bʳ. Choose r so that the power of x is the one you want, then multiply just that one term. This saves many lines when n is large, for example in a question about x⁴ in a power of 7.
If full expansions still feel unsteady, revisit expanding a positive integer power first.
What does the general term say?
The term with bʳ is the (r + 1)th term, because r starts at 0. It equals nCr × aⁿ⁻ʳ × bʳ, where nCr is the binomial coefficient, available as the nCr key on your calculator.
The powers of a and b always add up to n. So if you know the power of one part, the other follows, and r is easy to find.
A method that avoids wrong-r errors
- Write the general term with the brackets kept around each part.
- Decide which part carries x and set up an equation for its power. If (3x) has power n − r and you need x⁴, then n − r = 4.
- Solve for r and check that 0 ≤ r ≤ n.
- Substitute r into nCr, both powers and the numbers, and simplify.
- Answer what was asked: the term, or only the coefficient.
Worked example
Find the coefficient of x⁴ in the expansion of (3x + 2)⁷.
Step 1, general term: nCr (3x)⁷⁻ʳ (2)ʳ = 7Cr (3x)⁷⁻ʳ 2ʳ.
Step 2, find r: the power of x is 7 − r. We need 7 − r = 4, so r = 3.
Step 3, substitute: 7C3 × (3x)⁴ × 2³ = 35 × 81x⁴ × 8.
Step 4, simplify: 81 × 8 = 648, and 35 × 648 = 22 680.
The term is 22 680x⁴, so the coefficient of x⁴ is 22 680.
The mistake to watch for
A common slip is to take r equal to the power you want, instead of n minus that power.
Mistaken working: r = 4, so 7C4 × (3x)³ × 2⁴ = 35 × 27x³ × 16 = 15 120x³
The student set r = 4 because the question said x⁴. That gives a term in x³, not x⁴, so the power of x is wrong and so is the coefficient.
The correction is to ask which part carries x and what its power is: here (3x) has power 7 − r, not r. Always test your r by asking “does this give x⁴?” before multiplying numbers.
Check yourself
Try these, then open each answer.
1. Find the coefficient of x³ in (1 + 2x)⁶.
Show answer
General term: 6Cr × 1⁶⁻ʳ × (2x)ʳ. For x³, r = 3. So 6C3 × 2³ = 20 × 8 = 160.
2. Find the coefficient of x² in (x + 4)⁵.
Show answer
General term: 5Cr × x⁵⁻ʳ × 4ʳ. For x², 5 − r = 2, so r = 3. Then 5C3 × 4³ = 10 × 64 = 640.
3. Find the third term in the expansion of (x + 3)⁶, in descending powers of x.
Show answer
The third term has r = 2. So 6C2 × x⁴ × 3² = 15 × 9 × x⁴ = 135x⁴.
Where this leads next
Next, see how a known coefficient can give you an unknown, in comparing coefficients to determine a constant. You can also return to the module overview to see the whole route.
If you can expand a bracket but hesitate when a question asks for a single term, a teacher in online one-to-one Additional Mathematics tuition can practise setting up r with you.