IB Maths AA HLTopic 1 — Number & AlgebraPaper 1 & 2HL only~10 min read
Extension of The Binomial Theorem
So far we’ve only expanded (a + b)n when n was a positive whole number — like 5, 8, or 12. Easy enough. But what if n were a fraction, like 1⁄2, or negative, like −3? Things like √(1 + x) or 1/(1 − x)2 can also be expanded as a sum of powers of x — but the expansion goes on forever, and it only works inside a certain range of x. That’s the extension of the binomial theorem, and it’s HL-only territory. Master the recipe and you’ll have a clean way of approximating roots, reciprocals, and other awkward expressions to whatever accuracy you need.
📘 What you need to know
The extension lets you expand (a + b)n when n is any rational number — including fractions and negatives — not just positive integers.
The expansion has infinitely many terms (so you take “the first few” rather than the whole thing).
It only converges when |b| < |a|. If you ignore this, your expansion is meaningless for that value.
The cleanest version is the (1 + x)n form, valid when |x| < 1.
If the expression isn’t already in (1 + something) form — e.g. √(4 + x) or 1/(3 − 2x) — you must factor out the constant first to get it there.
For positive integer n, the extension formula gives exactly the same answer as the regular binomial theorem (it’s a more general version of the same result).
The formula is in the formula booklet. You don’t need to memorise it — just know how to apply it cleanly.
The extension formula
Here’s the formula in its book form. It looks busier than the regular binomial theorem, but it’s the same idea — a coefficient, multiplied by powers of two things — just with the coefficients written out as products of n‘s instead of nCr.
Extension of the binomial theorem (n ∈ ℚ)
(a + b)n = an
[ 1 + n·(ba) + n(n − 1)2!·(ba)2 + n(n − 1)(n − 2)3!·(ba)3 + … ]
✓ in the formula booklet
And the simpler version — the one you’ll actually use most of the time — is what you get when a = 1:
Look at the coefficient pattern. Each new coefficient is the previous one times “next descending integer” times “next x” divided by “next factorial“. Once you’ve written 1, nx, n(n−1)x2/2!, the next one almost writes itself: n(n−1)(n−2)x3/3!. Spot the pattern, and you don’t need to look at the formula again for the rest of the question.
Why validity matters (|x| < 1)
Here’s something genuinely new. When n was a positive integer, the expansion always stopped after a finite number of terms — so it didn’t matter what x was. But for fractional or negative n, the expansion goes on forever, and it only adds up to the right answer when x is small enough.
Specifically, (1 + x)n only makes sense as an infinite expansion when |x| < 1. Outside this range, the terms get bigger instead of smaller, and the sum doesn’t converge.
🤔 Why does it only work when |x| < 1?
Because each term contains a power of x — x, then x2, then x3 — and we want those to shrink as we go further. If |x| < 1, then each new xk is smaller than the last, and the sum settles down to a finite answer. If |x| ≥ 1, the powers grow (or stay equal) and the sum blows up.
The general rule for (a + b)n: the expansion is valid when |b| < |a|, i.e. the second term is smaller in size than the first. After you factor out an, this becomes the |b/a| < 1 condition.
Two scenarios you’ll meet
Almost every exam question on the extension lands in one of two camps. Knowing which camp you’re in is half the work.
Case A — already in (1 + …) form
(1 + x)n, (1 − 2x)−3, …
Just substitute straight into the (1 + x)n formula. Be careful with signs when the inside is “1 − something”. Validity: |x| < 1 (or whatever the inside variable is).
Case B — needs factoring first
√(4 + x), 1⁄(3 − 2x), …
Pull the constant out as a power. E.g. (4 + x)1/2 = 41/2(1 + x/4)1/2 = 2(1 + x/4)1/2. Then expand the bracket using Case A.
Which approach to use
The three-step recipe
Almost every “expand using the extension” question follows the same three steps. Memorise them once and you’re set.
1
Get to (1 + …)n form
If the base isn’t already 1, factor out the constant. E.g. (4 + x)1/2 = 2(1 + x/4)1/2.
2
Substitute and simplify
Plug in the inside expression as x in the (1+x)n formula. Compute each coefficient one at a time, then simplify the powers.
3
State validity
Mention the range of x for which the expansion converges. If you factored out, the condition becomes |inside-thing| < 1.
A note on Step 3. Some questions explicitly ask “for what values of x is this valid?” — others don’t. Even when not asked, write the validity range anyway. It costs you nothing and protects against a sneaky follow-up later in the question.
Worked examples
WE 1
Expand (1 + x)−2 up to and including the x3 term
Find the first four terms in the expansion of (1 + x)−2, and state the values of x for which the expansion is valid.
Step 1: Already in (1 + x)n form, with n = −2use: (1 + x)n = 1 + nx + n(n−1)/2! · x2 + n(n−1)(n−2)/3! · x3 + …Step 2: Compute each coefficient with n = −2term in x: nx = −2xterm in x2: (−2)(−3)/2! = 6/2 = 3 → 3x2term in x3: (−2)(−3)(−4)/3! = −24/6 = −4 → −4x3Step 3: Put it together + state validity(1 + x)−2 ≈ 1 − 2x + 3x2 − 4x3, valid for |x| < 1notice the pattern: 1, −2, 3, −4 — coefficients alternate sign and grow by 1 each time
WE 2
Expand (1 − 2x)1/2 up to x3
Find the first four terms, in ascending powers of x, in the expansion of √(1 − 2x). State the validity range.
Step 1: Rewrite — already (1 + …)n with the “…” being −2x and n = 1/2let y = −2x, so we want (1 + y)1/2Step 2: Apply the formula with n = 1/2(1+y)1/2 = 1 + (1/2)y + (1/2)(−1/2)/2! · y2 + (1/2)(−1/2)(−3/2)/3! · y3 + …= 1 + y/2 − y2/8 + y3/16 − …Step 3: Substitute y = −2xy/2 = −x−y2/8 = −(4x2)/8 = −x2/2y3/16 = (−8x3)/16 = −x3/2√(1 − 2x) ≈ 1 − x − x2/2 − x3/2, valid for |2x| < 1, i.e. |x| < 1/2when the inside is “1 − 2x”, validity becomes |2x| < 1 — divide both sides by 2
WE 3
Expand √(4 + x) up to x3
Find the first four terms in the expansion of √(4 + x) in ascending powers of x, and state the validity range.
Step 1: Factor out the 4 (Case B!)(4 + x)1/2 = [4(1 + x/4)]1/2 = 41/2 · (1 + x/4)1/2 = 2(1 + x/4)1/2Step 2: Use the (1 + y)1/2 result with y = x/4(1 + y)1/2 ≈ 1 + y/2 − y2/8 + y3/16y = x/4 → y/2 = x/8y2/8 = (x2/16)/8 = x2/128y3/16 = (x3/64)/16 = x3/1024Step 3: Multiply each term by 22 · [1 + x/8 − x2/128 + x3/1024] = 2 + x/4 − x2/64 + x3/512√(4 + x) ≈ 2 + x/4 − x2/64 + x3/512, valid for |x/4| < 1, i.e. |x| < 4forgetting to multiply through by the 2 at the end is the most common slip in this style of question
WE 4
Coefficient of x4 in (1 − 3x)−3
Find the coefficient of x4 in the expansion of (1 − 3x)−3.
Step 1: This is (1 + y)n with y = −3x and n = −3general coefficient of yr: n(n−1)(n−2)…(n−r+1)/r!Step 2: For x4, we need the y4 term (since y4 = (−3x)4 = 81x4)coefficient of y4 = (−3)(−4)(−5)(−6)/4!= 360 / 24 = 15Step 3: Multiply by (−3)4 = 81 to get the coefficient of x415 × 81 = 1215coefficient of x4 = 1215, valid for |3x| < 1, i.e. |x| < 1/3all four (−n)’s in the numerator are negative, so an even power of y gives a positive result
WE 5
Use the expansion to estimate √1.04
Use the first three terms of the expansion of (1 + x)1/2 to estimate √1.04 to four decimal places.
Step 1: Recall the first three terms of (1 + x)1/2(1 + x)1/2 ≈ 1 + x/2 − x2/8Step 2: Pick x to make 1 + x = 1.04x = 0.04 (check: |x| = 0.04 < 1 ✓)Step 3: Substitute≈ 1 + 0.04/2 − (0.04)2/8= 1 + 0.02 − 0.0016/8= 1 + 0.02 − 0.0002 = 1.0198√1.04 ≈ 1.0198actual value is 1.0198039…, so three terms gets us correct to 4 d.p. — this is the practical power of the extension
💡 Top tips
Always factor out first if the base isn’t 1. The (1 + x)n form is much cleaner. (3 − x)−1 becomes (1/3)(1 − x/3)−1 — easy from there.
Write coefficients one at a time. Don’t try to do it in your head — work out nx, then n(n−1)x2/2!, then n(n−1)(n−2)x3/3! step by step. Sign errors love rushed work.
State validity even when not asked. “Valid for |x| < …” at the end is often a free mark.
If the inside is “1 − ax”, treat x as “−ax”. The minus signs will cascade through the powers — keep track.
Don’t forget the multiplier when factoring out. If you turned √(4 + x) into 2(1 + x/4)1/2, the final answer must be multiplied by that 2.
Use the pattern, not the formula, after the first two terms. Each new coefficient is the previous one × (next descending integer in n) / (next factorial). Saves looking up the booklet.
For approximation problems, choose x small. The smaller |x|, the faster the series converges and the fewer terms you need for accuracy.
⚠ Common mistakes
Ignoring the sign when the inside is “1 − ax”. If you treat (1 − 2x)n as if it were (1 + 2x)n, every odd power has the wrong sign.
Forgetting to factor out first when the base is not 1. Substituting straight into the (1 + x)n formula won’t work for √(4 + x).
Forgetting to multiply by the constant after factoring. If √(4 + x) = 2 · (something), the final answer needs the 2 carried through every term.
Using the extension when n is a positive integer. Technically valid, but slower — for positive-integer n, just use the regular binomial theorem.
Stating the wrong validity range. If the variable inside is 2x or x/4, the validity is on that, not on plain x. So |2x| < 1 means |x| < 1/2, not |x| < 1.
Mistreating factorials in the denominators. The x2 term has 2! = 2 in the denominator, the x3 term has 3! = 6 — easy to mix up. Write them out.
Stopping at too few terms. “Up to and including x3” means four terms total (constant, x, x2, x3). Re-read the question before you stop.
The extension is a topic where the formula does most of the heavy lifting — your job is mainly to factor cleanly, substitute carefully, and watch the signs. Once you’ve worked through three or four full questions, the recipe becomes second nature: get to (1 + something) form, plug into the formula, simplify, state validity. Notice that this technique connects beautifully to the Taylor series you’ll meet in calculus — the extension is essentially the Taylor series of (1 + x)n around x = 0. Same idea, different topic. Maths really is one big web of connected results.
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