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The binomial expansion lets you expand \((a+bx)^n\), for a positive integer n, without multiplying out brackets. The expansion stops after n + 1 terms, and you use it to find terms, coefficients and unknown constants. It comes up on Paper 1.
Key facts
\(^nC_r = \binom{n}{r} = \dfrac{n!}{r!(n-r)!}\)
\((a+b)^n = a^n + \binom{n}{1}a^{n-1}b\) \(+ \binom{n}{2}a^{n-2}b^2 + \dots + b^n\), for positive integer n
General term: \(\binom{n}{r}a^{n-r}b^r\)
Notes
Positive integer powers
diagram
\(n! = n \times (n-1) \times \dots \times 2 \times 1\), and \(0! = 1\)
\(^nC_r = \binom{n}{r} = \dfrac{n!}{r!(n-r)!}\). These are the numbers in row n of Pascal's triangle; your calculator has an nCr button.
Each number is the sum of the two above it (3 + 3 = 6). Row n gives the coefficients of (a + b)ⁿ.
Find a coefficient in a product, or an unknown constant
5 marks (2 + 3)Grade C
Only work out the terms you need.
For a product such as \((3+x)(2-3x)^6\), find every pair of terms that multiply to give the power you want.
For an unknown constant, write the coefficient in terms of it, set it equal to the given value and solve.
Keep both roots unless the question rules one out (e.g. 'k is positive').
Example. (a) Given that \((2-3x)^6 = 64 - 576x + 2160x^2 + \dots\), find the coefficient of \(x^2\) in the expansion of \((3+x)(2-3x)^6\). (b) The coefficient of \(x^2\) in the expansion of \((1+kx)^8\) is 252. Find the possible values of the constant k.
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(a) \(3 \times 2160 + 1 \times (-576)\)
M1
\(= 5904\)
A1
(b) Term in \(x^2\): \(\binom{8}{2}(kx)^2 = 28k^2x^2\)
M1
\(28k^2 = 252 \Rightarrow k^2 = 9\)
M1
\(k = 3\) or \(k = -3\)
A1
Don’t lose marks
Leaving out the bracket round the second term: writing \(-3x^2\) instead of \((-3x)^2 = 9x^2\)
More tips
Memory tricks
Pascal's triangle is quicker than nCr for small n: row 4 is 1, 4, 6, 4, 1.
Both binomial expansions are in the formulae booklet: check the formula there rather than relying on memory.
Exam technique
'Ascending powers of x' means the constant term first, then x, then \(x^2\), and so on.
Show the unsimplified terms (binomial coefficients and brackets) before simplifying, so you earn the method mark even if you slip.
Give coefficients as exact integers or fractions, not decimals, unless you are told otherwise.
When a part says 'hence', use the expansion you have just found.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade E
Expand \((a+b)^n\) for positive integer nUse Pascal's triangle or \(^nC_r\) to write down all n + 1 terms, e.g. \((1+2x)^4 = 1 + 8x + 24x^2 + 32x^3 + 16x^4\)
Grade D
Find a given term or coefficientThe term in \(x^r\) in \((a+bx)^n\) is \(\binom{n}{r}a^{n-r}(bx)^r\), so you can find it without writing out the whole expansion.
Grade C
Find an unknown constant from a coefficientWrite the coefficient in terms of the constant, set it equal to the given value and solve, e.g. \(28k^2 = 252\)
Grade A
Use an expansion to approximate a valueChoose a value of x inside the valid range that turns the expression into the number you want, then substitute it.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy6 marks
(a) Find the first four terms, in ascending powers of \(x\), of the binomial expansion of \((3 + 2x)^{6}\), giving each term in its simplest form.[4]
(b) Find the coefficient of \(x^{4}\) in the expansion of \((3 + 2x)^{6}\).[2]
M1 for \((-2) \times 90 + 3 \times 15\) using their coefficients
A1 for \(-135\)
Worked solution: The \(x^2\) terms come from \((-2) \times 90x^2\) and \(3x \times 15x\). Coefficient \(= -180 + 45 = -135\).
Question 3Hard7 marks
In the binomial expansion of \((1 + kx)^n\), in ascending powers of \(x\), where \(n\) is a positive integer and \(k\) is a constant, the coefficient of \(x\) is \(40\) and the coefficient of \(x^2\) is \(700\).
(a) Find the value of \(n\) and the value of \(k\).[5]
(b) Find the coefficient of \(x^3\).[2]
Show the answer and mark scheme
(a)Answer: \(n = 8, \ k = 5\)
B1 for \(nk = 40\)
B1 for \(\frac{n(n - 1)}{2}k^2 = 700\)
M1 for eliminating one unknown, e.g. dividing to obtain \((n - 1)k = 35\)
A1 for \(k = 5\)
A1 for \(n = 8\)
Worked solution: \(nk = 40\) (1) and \(\frac{n(n - 1)k^2}{2} = 700\) (2). (2) ÷ (1): \(\frac{(n - 1)k}{2} = \frac{35}{2}\), so \(nk - k = 35\). Using (1): \(40 - k = 35\), so \(k = 5\) and \(n = 40 \div 5 = 8\).
(b)Answer: \(7000\)
M1 for \(\frac{n(n - 1)(n - 2)}{3!}k^3\) with their \(n\) and \(k\)