The binomial expansion - Edexcel A-Level Maths (Paper 2) Exam Questions

Practise 18 exam-style questions on The binomial expansion for Edexcel A-Level Maths (Paper 2). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

18 questions
2–11 marks each
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18 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

3 marks

In the binomial expansion of
\[\left(kx^2 - \dfrac{1}{x}\right)^7, \quad x ≠ 0\]
where \(k\) is a positive constant, the coefficient of \(\dfrac{1}{x}\) is \(-84\).

Find the value of \(k\).

[Binomial series: \((a + b)^n = a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \dots + \binom{n}{r}a^{n-r}b^r + \dots + b^n \quad (n \in \mathbb{N})\), where \(\binom{n}{r} = {}^n\mathrm{C}_r = \dfrac{n!}{r!(n - r)!}\)]

Model answer

General term: \(\binom{7}{r}(kx^2)^{7-r}\left(-\dfrac{1}{x}\right)^r = \binom{7}{r}k^{7-r}(-1)^r x^{14 - 2r - r} = \binom{7}{r}k^{7-r}(-1)^r x^{14 - 3r}\)
Term in \(\dfrac{1}{x} = x^{-1}\): \(14 - 3r = -1\), so \(r = 5\)
The coefficient is \(\binom{7}{5}k^2(-1)^5 = -21k^2\)
\(-21k^2 = -84\), so \(k^2 = 4\)
\(k > 0\), so \(k = 2\)

Explanation

Each term of the expansion is \(\binom{7}{r}(kx^2)^{7-r}\left(-\dfrac{1}{x}\right)^r\), a multiple of \(x^{14 - 3r}\), so the term in \(\dfrac{1}{x}\) is the one with \(r = 5\). The power 5 is odd, so the term is negative, \(-21k^2\), and \(-21k^2 = -84\) gives \(k^2 = 4\), so \(k = 2\) as \(k\) is positive. Common errors are choosing the wrong value of \(r\), losing the minus sign and leaving the answer as \(k = \pm 2\).

Mark scheme

1 mark
M1 Writes the general term \(\binom{7}{r}(kx^2)^{7-r}\left(-\dfrac{1}{x}\right)^r\) and finds its power of \(x\), \(14 - 3r\), or otherwise identifies the term in \(\dfrac{1}{x}\) as the one with \(r = 5\), i.e. \(\binom{7}{5}(kx^2)^2\left(-\dfrac{1}{x}\right)^5\); or a correct final answer of \(k = 2\) with no working
An answer of \(k = 2\) with no working scores 3 marks. Writing out the terms of the expansion until the term in \(\dfrac{1}{x}\) is reached is a valid method. Using the term with \(r = 4\) or \(r = 6\) is M0.
1 mark
A1 \(-21k^2 = -84\) oe
Accept \(21k^2 = 84\) or \(k^2 = 4\). A sign error giving \(21k^2 = -84\) (no real solution) is A0.
1 mark
A1 \(k = 2\)
Depends on the M mark. cao; \(k = \pm 2\) as the final answer is A0, since \(k\) is positive; isw if \(k = 2\) is clearly chosen.

Question 2

5 marks

A student is asked to find the binomial expansion of \((2 - 3x)^{-3}\), in ascending powers of \(x\), up to and including the term in \(x^2\). The student's working is shown below.

Line 1: \((2 - 3x)^{-3} = 2\left(1 - \dfrac{3x}{2}\right)^{-3}\)
Line 2: \(= 2\left[1 + (-3)\left(-\dfrac{3x}{2}\right) + \dfrac{(-3)(-4)}{2!}\left(-\dfrac{3x^2}{2}\right) + \dots\right]\)
Line 3: \(= 2\left[1 + \dfrac{9x}{2} - 9x^2 + \dots\right]\)
Line 4: \(= 2 + 9x - 18x^2 + \dots\)

Identify the two errors in the student's working.

Find the correct expansion, up to and including the term in \(x^2\), and state the range of values of \(x\) for which it is valid.

[Binomial series: \((1 + x)^n = 1 + nx + \dfrac{n(n - 1)}{1 \times 2}x^2 + \dots + \dfrac{n(n - 1)\dots(n - r + 1)}{1 \times 2 \times \dots \times r}x^r + \dots \quad (\lvert x \rvert < 1,\ n \in \mathbb{R})\)]

Model answer

Error 1: in Line 1 the factor should be \(2^{-3} = \dfrac{1}{8}\), not 2, because \((2 - 3x)^{-3} = 2^{-3}\left(1 - \dfrac{3x}{2}\right)^{-3}\).
Error 2: in Line 2 the third term should use \(\left(-\dfrac{3x}{2}\right)^2 = \dfrac{9x^2}{4}\), not \(-\dfrac{3x^2}{2}\).
Correct working:
\(\left(1 - \dfrac{3x}{2}\right)^{-3} = 1 + (-3)\left(-\dfrac{3x}{2}\right) + \dfrac{(-3)(-4)}{2}\left(-\dfrac{3x}{2}\right)^2 + \dots = 1 + \dfrac{9x}{2} + \dfrac{27x^2}{2} + \dots\)
\((2 - 3x)^{-3} = \dfrac{1}{8}\left(1 + \dfrac{9x}{2} + \dfrac{27x^2}{2} + \dots\right) = \dfrac{1}{8} + \dfrac{9}{16}x + \dfrac{27}{16}x^2 + \dots\)
Valid for \(\left\lvert \dfrac{3x}{2} \right\rvert < 1\), i.e. \(\lvert x \rvert < \dfrac{2}{3}\)

Explanation

When the 2 is taken out of \((2 - 3x)^{-3}\) it must be raised to the power \(-3\), giving \(\dfrac{1}{8}\), and in each term the whole of \(-\dfrac{3x}{2}\) must be raised to the power. With both corrections the expansion is \(\dfrac{1}{8} + \dfrac{9}{16}x + \dfrac{27}{16}x^2\), valid for \(\lvert x \rvert < \dfrac{2}{3}\). Examiners' reports describe both errors: not raising the factor taken out to the power, and losing the brackets round the term in \(x\).

Mark scheme

1 mark
B1 Error in Line 1: the factor taken out should be \(2^{-3} = \dfrac{1}{8}\), not 2; or a correct final answer of \(\dfrac{1}{8} + \dfrac{9}{16}x + \dfrac{27}{16}x^2\), \(\lvert x \rvert < \dfrac{2}{3}\) with no working
An answer of \(\dfrac{1}{8} + \dfrac{9}{16}x + \dfrac{27}{16}x^2\) and \(\lvert x \rvert < \dfrac{2}{3}\) with no working scores 3 marks (the M1, the A1 and the final B1); the two errors must be identified for the first two marks. Accept "the 2 should be raised to the power \(-3\)" or "it should be \(\frac{1}{8}\)".
1 mark
B1 Error in Line 2: the third term should contain \(\left(-\dfrac{3x}{2}\right)^2 = \dfrac{9x^2}{4}\), not \(-\dfrac{3x^2}{2}\) (the whole of \(-\dfrac{3x}{2}\) must be squared)
Accept "they did not square the \(-\frac{3x}{2}\)" or "the bracket should be squared, giving \(+\frac{9x^2}{4}\)". Do not accept vague statements such as "the third term is wrong" with no explanation.
1 mark
M1 A correct method: \(\left(1 - \dfrac{3x}{2}\right)^{-3}\) expanded with the correct structure for the \(x^2\) term, \(\dfrac{(-3)(-4)}{2!}\left(-\dfrac{3x}{2}\right)^2\), and multiplied by \(2^{-3}\)
Independent of the B marks. The factor must be \(2^{-3}\) (or \(\frac{1}{8}\)) for this mark.
1 mark
A1 \(\dfrac{1}{8} + \dfrac{9}{16}x + \dfrac{27}{16}x^2\)
cao; exact equivalents such as \(0.125 + 0.5625x + 1.6875x^2\) are accepted. Ignore any term in \(x^3\). isw after a correct expansion.
1 mark
B1 \(\lvert x \rvert < \dfrac{2}{3}\)
Accept \(-\dfrac{2}{3} < x < \dfrac{2}{3}\). Do not accept \(\lvert x \rvert < 1\), \(\lvert x \rvert < 2\) or \(\lvert x \rvert < \dfrac{3}{2}\).

Question 3

7 marks

Use the binomial expansion to write
\[(27 - 9x)^{\frac{2}{3}} = p + qx + rx^2 + sx^3 + \dots\]
where \(p\), \(q\), \(r\) and \(s\) are rational constants to be found.

Give the values of \(x\) for which this expansion is valid.

Hence, or otherwise, find the first three terms, in ascending powers of \(x\), of the expansion of \((27 - 9x)^{\frac{5}{3}}\), giving each coefficient in its simplest form.

Model answer

\((27 - 9x)^{\frac{2}{3}} = 27^{\frac{2}{3}}\left(1 - \dfrac{x}{3}\right)^{\frac{2}{3}} = 9\left(1 - \dfrac{x}{3}\right)^{\frac{2}{3}}\)
\(= 9\left[1 + \dfrac{2}{3}\left(-\dfrac{x}{3}\right) + \dfrac{\frac{2}{3}\left(-\frac{1}{3}\right)}{2}\left(-\dfrac{x}{3}\right)^2 + \dfrac{\frac{2}{3}\left(-\frac{1}{3}\right)\left(-\frac{4}{3}\right)}{6}\left(-\dfrac{x}{3}\right)^3 + \dots\right]\)
\(= 9\left[1 - \dfrac{2x}{9} - \dfrac{1}{9} \times \dfrac{x^2}{9} + \dfrac{4}{81} \times \left(-\dfrac{x^3}{27}\right) + \dots\right]\)
\(= 9\left[1 - \dfrac{2x}{9} - \dfrac{x^2}{81} - \dfrac{4x^3}{2187} + \dots\right]\)
\(= 9 - 2x - \dfrac{1}{9}x^2 - \dfrac{4}{243}x^3 + \dots\)
So \(p = 9\), \(q = -2\), \(r = -\dfrac{1}{9}\), \(s = -\dfrac{4}{243}\)
Valid for \(\left\lvert -\dfrac{x}{3} \right\rvert < 1\), i.e. \(\lvert x \rvert < 3\)
\((27 - 9x)^{\frac{5}{3}} = (27 - 9x)(27 - 9x)^{\frac{2}{3}} = (27 - 9x)\left(9 - 2x - \dfrac{1}{9}x^2 + \dots\right)\)
\(= 243 - 54x - 3x^2 - 81x + 18x^2 + \dots = 243 - 135x + 15x^2 + \dots\)

Explanation

Take out \(27^{\frac{2}{3}} = 9\) so that the bracket is \(\left(1 - \dfrac{x}{3}\right)^{\frac{2}{3}}\), expand it with \(n = \dfrac{2}{3}\), and multiply every term by 9; the expansion is valid when \(\left\lvert \dfrac{x}{3} \right\rvert < 1\), that is \(\lvert x \rvert < 3\). Since \((27 - 9x)^{\frac{5}{3}} = (27 - 9x)(27 - 9x)^{\frac{2}{3}}\), multiplying the expansion by \(27 - 9x\) gives \(243 - 135x + 15x^2\) without a second binomial expansion. Common errors are taking out 27 instead of \(27^{\frac{2}{3}}\), sign errors from \(\left(-\dfrac{x}{3}\right)^3\), and giving the range as \(\lvert x \rvert < 1\).

Mark scheme

1 mark
B1 Takes out the factor \(27^{\frac{2}{3}} = 9\): \((27 - 9x)^{\frac{2}{3}} = 9\left(1 - \dfrac{x}{3}\right)^{\frac{2}{3}}\); or correct final answers \(p = 9\), \(q = -2\), \(r = -\dfrac{1}{9}\), \(s = -\dfrac{4}{243}\), \(\lvert x \rvert < 3\) and \(243 - 135x + 15x^2\) with no working
An answer of \(p = 9\), \(q = -2\), \(r = -\dfrac{1}{9}\), \(s = -\dfrac{4}{243}\), \(\lvert x \rvert < 3\) and \(243 - 135x + 15x^2\) with no working scores 7 marks. Taking out 27 instead of \(27^{\frac{2}{3}}\) is B0, but the M mark can still be earned.
1 mark
M1 Expands \(\left(1 + kx\right)^{\frac{2}{3}}\), \(k ≠ 1\), with the correct structure for the \(x^2\) or \(x^3\) term, e.g. \(\dfrac{\frac{2}{3}\left(-\frac{1}{3}\right)}{2!}\left(-\dfrac{x}{3}\right)^2\) or \(\dfrac{\frac{2}{3}\left(-\frac{1}{3}\right)\left(-\frac{4}{3}\right)}{3!}\left(-\dfrac{x}{3}\right)^3\)
Independent of the B mark. Condone missing brackets on \(-\dfrac{x}{3}\) for this mark only.
1 mark
A1 \(p = 9\), \(q = -2\) and \(r = -\dfrac{1}{9}\) (i.e. \(9 - 2x - \dfrac{1}{9}x^2\))
All three needed, following the correct factor 9. The values may be seen in the expansion.
1 mark
A1 \(s = -\dfrac{4}{243}\)
Depends on the M mark. Ignore any further terms; isw once a correct expansion is seen. A rounded decimal such as \(s = -0.0165\) or \(r = -0.111\) is A0.
1 mark
B1 \(\lvert x \rvert < 3\)
Accept \(-3 < x < 3\). Do not accept \(\lvert x \rvert < 1\), \(\lvert x \rvert < \dfrac{1}{3}\), \(\lvert x \rvert \leqslant 3\) or \(x < 3\).
1 mark
M1 Uses \((27 - 9x)^{\frac{5}{3}} = (27 - 9x)(27 - 9x)^{\frac{2}{3}}\) and multiplies their expansion by \(27 - 9x\), collecting the terms as far as \(x^2\); or expands \(243\left(1 - \dfrac{x}{3}\right)^{\frac{5}{3}}\) directly with the correct structure for the \(x^2\) term
Independent of the earlier marks. In the product method both products giving \(x^2\) (\(27 \times rx^2\) and \(-9x \times qx\)) must be used. Squaring their expansion, or multiplying it by 27 only, is M0.
1 mark
A1 \(243 - 135x + 15x^2\)
cao. From \(243 + (-54 - 81)x + (-3 + 18)x^2\). Ignore any extra terms; isw.

Question 4

8 marks

\[\mathrm{f}(x) = \dfrac{\sqrt{1 + ax}}{1 + bx}\]
where \(a\) and \(b\) are non-zero constants.

In the binomial expansion of \(\mathrm{f}(x)\), in ascending powers of \(x\), the coefficient of \(x\) is zero and the coefficient of \(x^2\) is \(-2\).

Find the possible pairs of values of \(a\) and \(b\).

State, with a reason, the range of values of \(x\) for which the expansion is valid.

Model answer

\((1 + ax)^{\frac{1}{2}} = 1 + \dfrac{1}{2}(ax) + \dfrac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2}(ax)^2 + \dots = 1 + \dfrac{a}{2}x - \dfrac{a^2}{8}x^2 + \dots\)
\((1 + bx)^{-1} = 1 - bx + b^2x^2 + \dots\)
\(\mathrm{f}(x) = \left(1 + \dfrac{a}{2}x - \dfrac{a^2}{8}x^2\right)\left(1 - bx + b^2x^2\right) + \dots\)
\(= 1 + \left(\dfrac{a}{2} - b\right)x + \left(b^2 - \dfrac{ab}{2} - \dfrac{a^2}{8}\right)x^2 + \dots\)
Coefficient of \(x\): \(\dfrac{a}{2} - b = 0\), so \(a = 2b\)
Coefficient of \(x^2\): \(b^2 - \dfrac{(2b)b}{2} - \dfrac{(2b)^2}{8} = b^2 - b^2 - \dfrac{b^2}{2} = -\dfrac{b^2}{2} = -2\)
So \(b^2 = 4\): \(b = 2, a = 4\) or \(b = -2, a = -4\)
In both cases \(\lvert a \rvert = 4\) and \(\lvert b \rvert = 2\). The expansion of \((1 + ax)^{\frac{1}{2}}\) needs \(\lvert 4x \rvert < 1\) and that of \((1 + bx)^{-1}\) needs \(\lvert 2x \rvert < 1\); both must hold, so the expansion is valid for \(\lvert x \rvert < \dfrac{1}{4}\).

Explanation

Expand the numerator with \(n = \dfrac{1}{2}\) and the denominator as \((1 + bx)^{-1}\), multiply, and equate the coefficients of \(x\) and \(x^2\) to 0 and \(-2\). The first equation gives \(a = 2b\), and substituting into the second gives \(b^2 = 4\), so there are two pairs of values. The product is valid only where both expansions are valid, which is \(\lvert x \rvert < \dfrac{1}{4}\); a common error is to give the wider range \(\lvert x \rvert < \dfrac{1}{2}\).

Mark scheme

1 mark
M1 Expands \((1 + ax)^{\frac{1}{2}}\) with the correct structure for the \(x^2\) term, \(\dfrac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2!}(ax)^2\); or correct final answers of \(a = 4, b = 2\) and \(a = -4, b = -2\), with \(\lvert x \rvert < \dfrac{1}{4}\), and no working (special case: 2 marks)
Answers of \(a = 4, b = 2\) and \(a = -4, b = -2\) and \(\lvert x \rvert < \dfrac{1}{4}\) with no working score 2 marks (special case: the A1 for both pairs and the B1 for the range). Condone \(ax^2\) for \((ax)^2\) for this mark only.
1 mark
A1 \((1 + ax)^{\frac{1}{2}} = 1 + \dfrac{1}{2}ax - \dfrac{1}{8}a^2x^2 + \dots\)
1 mark
B1 \((1 + bx)^{-1} = 1 - bx + b^2x^2 + \dots\)
Independent of the other marks.
1 mark
M1 Multiplies the two expansions and collects the coefficients of \(x\) and \(x^2\)
At least two of the three products giving \(x^2\) must be correct for their expansions.
1 mark
A1 \(\dfrac{a}{2} - b = 0\) and \(b^2 - \dfrac{ab}{2} - \dfrac{a^2}{8} = -2\) oe
Both equations needed for this mark.
1 mark
M1 (dM1) Solves simultaneously: substitutes \(a = 2b\) (or \(b = \frac{a}{2}\)) to obtain an equation in one variable, e.g. \(b^2 - b^2 - \dfrac{b^2}{2} = -2\)
Depends on the second M mark (not the point just before).
1 mark
A1 \(a = 4, b = 2\) and \(a = -4, b = -2\)
Both pairs needed, correctly paired. \(a = \pm 4, b = \pm 2\) is acceptable only if the pairing is made clear.
1 mark
B1 \(\lvert x \rvert < \dfrac{1}{4}\), because both expansions must be valid: \(\lvert ax \rvert < 1\) needs \(\lvert x \rvert < \dfrac{1}{4}\) and \(\lvert bx \rvert < 1\) needs \(\lvert x \rvert < \dfrac{1}{2}\), and the narrower range is needed
The range and a reason that refers to both expansions (or to the more restrictive one) are needed. \(\lvert x \rvert < \dfrac{1}{2}\) or \(\lvert x \rvert < 1\) is B0.

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