IB Maths AA HLTopic 1 — Number & AlgebraPaper 1 & 2~9 min read
Binomial Theorem
Multiplying out something like (a + b)7 by hand would take ages — and you’d almost certainly slip up somewhere. The binomial theorem is the shortcut. It tells you exactly what every term of the expansion looks like, without writing out a single bracket. Once you’ve got the formula and a feel for which numbers to plug in, expanding any positive-integer power of (a + b) is mechanical. The exam questions almost always ask one of two things: write out the first few terms, or pick out a specific term you care about. Both are easy once you know where the powers go.
📘 What you need to know
The binomial theorem expands (a + b)n for any positive integer n, without multiplying brackets out by hand.
Each term has the form nCr · an−r · br. The powers of a drop while the powers of b rise, and they always add to n.
nCr is the binomial coefficient, given by nCr = n! ÷ [r!(n − r)!]. Your GDC will calculate it directly.
For harder binomials like (1 + 4x)6 or (3 − 2y)5, put each term in brackets first, then apply index laws (e.g. (4x)3 = 64x3).
To find a specific term (like “the term in x5“), use the general termnCr · an−r · br and solve for r — no need to expand the whole thing.
Both the formula and the binomial coefficient are in the formula booklet, so you don’t need to memorise them — just know how to use them.
Watch the signs! When the binomial has a minus, like (a − b)n, treat it as (a + (−b))n. Odd powers of b will be negative.
The binomial theorem formula
Here is the result the whole topic is built around. It says: if you raise (a + b) to the power n, where n is a positive integer, then the expansion has exactly n + 1 terms, and each term follows a neat pattern.
Binomial theorem
(a + b)n = an + nC1an−1b + nC2an−2b2 + … + nCran−rbr + … + bn✓ in the formula booklet
Binomial coefficientnCr = n!r!(n − r)!✓ in the formula booklet
Quick reminder on factorials: 4! means 4 × 3 × 2 × 1 = 24. By convention, 0! = 1. You’ll meet these all the time when calculating nCr by hand.
Anatomy of a general term
The pattern in one line: powers of a count down from n, powers of b count up from 0, and the two always sum to n. If yours don’t, you’ve made a slip.
Both the binomial theorem itself and the formula for nCr sit in the formula booklet — flip to them whenever you start a question. There’s no advantage in trying to memorise the formula; the marks come from getting a, b and n identified correctly and substituting cleanly.
Using the theorem on a clean expansion
Let’s run through (a + b)4 as a warm-up. Substitute n = 4 into the formula:
(a + b)4 = a4 + 4C1a3b + 4C2a2b2 + 4C3ab3 + b4
Now compute the coefficients: 4C1 = 4, 4C2 = 6, 4C3 = 4. Drop them in:
(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
Notice the coefficients 1, 4, 6, 4, 1 are symmetric — they read the same forwards and backwards. That’s a feature of every binomial expansion, and you can use it to halve your work: once you’ve found the first half, mirror it for the second.
🤔 Why is the coefficient row always symmetric?
Because nCr = nCn−r. Choosing r things out of n is the same as choosing which n − r things to leave behind. So 4C1 = 4C3 and 4C0 = 4C4, giving the mirror.
Expanding harder binomials
Real exam questions usually replace a and b with something less friendly — like 1, 3x, or −2y. The trick is to wrap each piece in brackets the moment you substitute, so the index laws apply cleanly.
🧭 Recipe — expanding (1 + 3x)5
Identify a, b, n. Here a = 1, b = 3x, n = 5.
Write the formula skeleton. Substitute a, b, n using brackets: 15 + 5C1(1)4(3x) + 5C2(1)3(3x)2 + …
Apply index laws. (3x)2 = 9x2, (3x)3 = 27x3, etc. The constants compound quickly — be careful.
When the second term is negative — say (2 − y)6 — write it as (2 + (−y))6 and treat b = −y. Now (−y)2 = +y2, (−y)3 = −y3, and so on. The signs in your final answer will alternate: + − + − + − + …
Finding a particular term
If a question asks “find the term in x5” or “find the coefficient of x3“, you don’t want to expand the whole expression — that’s slow and error-prone, especially for big n. Instead, use the general term:
General term in (a + b)nnCr · an−r · br✓ embedded in the formula booklet’s binomial theorem
1
Substitute a, b, n
Plug them into the general term, using brackets.
2
Track the x power only
Combine the xs on the right and set the resulting power equal to the one you want.
3
Solve for r, then sub back
Once you know r, substitute it into the full general term to get the actual coefficient.
Shortcut for the (1 + bx)n case: here a = 1, so an−r = 1. The general term simplifies to nCr · br · xr — and “the term in xk” just means r = k.
Worked examples
WE 1
Expand (a + b)4 in full
Use the binomial theorem to write out the full expansion of (a + b)4.
Step 1: Substitute n = 4 into the formula(a + b)4 = a4 + 4C1 a3b + 4C2 a2b2 + 4C3 ab3 + b4Step 2: Calculate each binomial coefficient4C1 = 4!/(1! · 3!) = 44C2 = 4!/(2! · 2!) = 24/4 = 64C3 = 4 (by symmetry: 4C3 = 4C1)Step 3: Drop the coefficients in(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4check: powers of a drop 4 → 0, powers of b rise 0 → 4, and they always sum to 4 ✓
WE 2
First four terms of (1 + 3x)7
Find the first four terms, in ascending powers of x, in the expansion of (1 + 3x)7.
Step 1: Identify a, b, na = 1, b = 3x, n = 7Step 2: Write the first four terms (r = 0, 1, 2, 3)(1 + 3x)7 = 17 + 7C1(1)6(3x) + 7C2(1)5(3x)2 + 7C3(1)4(3x)3 + …Step 3: Apply index laws and compute coefficients7C1 = 7, 7C2 = 21, 7C3 = 35(3x)2 = 9x2, (3x)3 = 27x3= 1 + 7(3x) + 21(9x2) + 35(27x3) + …= 1 + 21x + 189x2 + 945x3 + …(1 + 3x)7 ≈ 1 + 21x + 189x2 + 945x3“ascending powers” means smallest power first, so start from the constant — easy mark to lose by going the other way
WE 3
Coefficient of x3 in (2 − x)8
Find the coefficient of x3 in the expansion of (2 − x)8.
Step 1: Identify a, b, n (careful with the minus!)a = 2, b = −x, n = 8Step 2: Use the general term nCr an−r brgeneral term = 8Cr (2)8−r (−x)r= 8Cr · 28−r · (−1)r · xrStep 3: Match the power of x to 3 → r = 38C3 = 8!/(3! · 5!) = 5628−3 = 25 = 32(−1)3 = −1Step 4: Multiplycoefficient = 56 × 32 × (−1) = −1792coefficient of x3 is −1792odd power of (−x) → negative coefficient. If you got +1792, you forgot the sign on b
WE 4
Find the term in x11 in (3x + x2)7
Find the term in x11 in the expansion of (3x + x2)7.
Step 1: Identify a, b, na = 3x, b = x2, n = 7Step 2: Write the general termgeneral term = 7Cr (3x)7−r (x2)rStep 3: Track the x-power onlyx7−r · x2r = x7−r+2r = x7+rneed 7 + r = 11, so r = 4Step 4: Substitute r = 4 to get the full term7C4 (3x)3 (x2)4 (from 7C4 = 35)= 35 · 27x3 · x8= 35 · 27 · x11 = 945x11term in x11 is 945x11when both terms have x in them, the trick is to combine the powers and set them equal to your target
WE 5
Coefficient of x4 in (1 + 2x)6
Find the coefficient of x4 in the expansion of (1 + 2x)6.
Step 1: Identify a, b, na = 1, b = 2x, n = 6Step 2: Use the simplified general term (since a = 1)general term = 6Cr (1)6−r (2x)r = 6Cr · 2r · xrStep 3: For x4, use r = 46C4 = 6!/(4! · 2!) = 1524 = 16coefficient = 15 × 16 = 240coefficient of x4 is 240when a = 1, the (1)n−r part vanishes — life is much easier. Spot this whenever you can.
💡 Top tips
Always wrap substitutions in brackets. Writing 3x2 when you mean (3x)2 is the single most common error in this topic. Brackets first, then expand.
Check the powers add to n. Every term should have a?·b? with the two indices summing to n. If they don’t, find your slip before going further.
Use the GDC for big nCr values on Paper 2. For something like 15C6, factorial arithmetic by hand wastes time.
For “term in xk” questions, don’t expand the whole binomial. Use the general term and solve for r.
“Term” vs “coefficient” matters. The term in x3 might be 240x3, while the coefficient of x3 is just 240. Read the question carefully.
Use symmetry to halve your work.nCr = nCn−r, so once you’ve found 8C0, 8C1, 8C2, 8C3, 8C4, the rest are mirrors.
For minus signs, treat as (−b). Then odd r gives a negative term, even r gives a positive one. Expect alternating signs.
⚠ Common mistakes
Forgetting to raise the constant. In (3x)4, you must do 34·x4 = 81x4, not 3x4. The bracket binds the power to the whole thing.
Losing the minus sign. When the binomial is (a − b)n, every odd-r term is negative. Forgetting this flips the sign of half your answer.
Confusing “term” with “coefficient”. The “term in x5” includes the x5; the “coefficient of x5” is just the number in front.
Reading “ascending” as “descending”. Ascending powers means smallest first (constant, then x, then x2, …). Always check before writing.
Using the wrong r in the general term. The first term corresponds to r = 0, not r = 1. So nC0 = 1 gives the very first term.
Setting up the wrong equation for “term in xk“. Track only the powers of x, not the constants — and remember to combine all the x-powers (from a and b alike) before solving.
Stopping after finding r. Once you know r, you still need to plug it back in and compute the actual coefficient — finding r alone is only halfway.
The binomial theorem is one of those topics where you’ll feel slow at first and then suddenly very fast. The whole game is recognising what a, b, and n are, bracketing them carefully, and either expanding the first few terms or hunting down a single one with the general term. Practise both styles — three or four questions of each — and you’ll have it nailed before the next note on Pascal’s triangle.
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