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TMUA Binomial Expansion

The binomial expansion gives (a+b)n(a+b)^n for a positive integer nn without multiplying out. A question may ask for just one coefficient or term, not the whole expansion.

Paper 1 and Paper 2 topic. Specification codes: MM2.4. Last checked 12 November 2026.

The general term

(a+b)n=∑k=0n(nk)an−kbk(a+b)^n=\sum_{k=0}^{n}\binom nk a^{n-k}b^k, where (nk)=n!k!(n−k)!\binom nk=\dfrac{n!}{k!(n-k)!}. The term with bkb^k is (nk)an−kbk\binom nk a^{n-k}b^k.

Finding one term

Write the power of xx in the general term, set it equal to the one you want, and solve for kk. Remember that aa and bb include their own numbers and signs.

Products of brackets

For an expression like (1+x)(1+2x)4(1+x)(1+2x)^4, expand the power to the order you need and collect the matching terms.

Checks and shortcuts

A quick check is to substitute an easy value such as x=1x=1 into the original expression and into your expansion. For (2+x)5(2+x)^5 the coefficients must add up to 35=2433^5=243.

Pascal's triangle gives (nk)\binom nk for small nn without factorials: row 5 reads 1, 5, 10, 10, 5, 1. Use the symmetry (nk)=(nn−k)\binom nk=\binom n{n-k} to save effort, and be ready to simplify factorial expressions such as 7!5!=42\dfrac{7!}{5!}=42.

Because there is no calculator, the numbers stay small. If a coefficient looks messy, recheck which term you have picked.

Worked examples

Example 1

Find the coefficient of x3x^3 in (2+x)5(2+x)^5.

Show the worked solution

The x3x^3 term is (53)⋅22⋅x3=10⋅4 x3=40x3\binom53\cdot2^2\cdot x^3=10\cdot4\,x^3=40x^3.

Answer: 40

Example 2

Find the constant term in (x+2x)6\left(x+\dfrac2x\right)^6.

Show the worked solution

The general term is (6k)x6−k(2x)k=(6k)2kx6−2k\binom6k x^{6-k}\left(\tfrac2x\right)^k=\binom6k2^kx^{6-2k}. The power is zero when k=3k=3, giving 20×8=16020\times8=160.

Answer: 160

Example 3

Find the coefficient of x2x^2 in (1+x)(1+2x)4(1+x)(1+2x)^4.

Show the worked solution

(1+2x)4=1+8x+24x2+…(1+2x)^4=1+8x+24x^2+\dots. The x2x^2 terms of the product come from 1⋅24x21\cdot24x^2 and x⋅8xx\cdot8x, so 24+8=3224+8=32.

Answer: 32

Common mistakes

  • Forgetting to raise the numerical part of bb to the power kk, for example (2x)3=8x3(2x)^3=8x^3, not 2x32x^3.
  • Missing a sign when bb is negative.
  • Taking only one route to the required power in a product of brackets.
  • Using (nk)\binom nk with the wrong kk for the power of xx asked for.

Test yourself on this topic with our free practice questions.

Quick answers

How do you find a specific term in a binomial expansion?

Use (nk)an−kbk\binom nk a^{n-k}b^k, choose kk so the power of xx matches, and simplify.

What does nCr mean?

(nr)\binom nr is the number of ways to choose rr items from nn, equal to n!r!(n−r)!\dfrac{n!}{r!(n-r)!}.

How do you find the constant term in a binomial expansion?

Set the total power of xx in the general term equal to zero and solve for kk.

Practise it under timed conditions

A free timed sample paper with a worked solution for every question. No sign-up. For two full papers and a weak-spot drill, see Full Access.

Free sample paper

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