The binomial expansion - Edexcel A-Level Maths (Paper 1) Exam Questions

Practise 19 exam-style questions on The binomial expansion for Edexcel A-Level Maths (Paper 1). 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.

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3–9 marks each
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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

By substituting a suitable value of \(x\) into the binomial expansion of \((1 + x)^n\), where \(n\) is a positive integer, explain why
\[\sum_{r=0}^{n}\binom{n}{r} = 2^n\]
Hence find the value of \(n\) for which
\[\binom{n}{1} + \binom{n}{2} + \binom{n}{3} + \dots + \binom{n}{n - 1} = 1022\]

[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

\((1 + x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \dots + \binom{n}{n}x^n\)
Put \(x = 1\): \(2^n = \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \dots + \binom{n}{n}\), so \(\sum_{r=0}^{n}\binom{n}{r} = 2^n\)
The given sum leaves out \(\binom{n}{0} = 1\) and \(\binom{n}{n} = 1\), so
\(2^n - 2 = 1022\)
\(2^n = 1024\)
\(n = 10\)

Explanation

Every term of the expansion of \((1 + x)^n\) is a binomial coefficient times a power of \(x\), so putting \(x = 1\) turns the expansion into the sum of the coefficients, which equals \((1 + 1)^n = 2^n\). The given sum leaves out the first and last coefficients, both equal to 1, so \(2^n - 2 = 1022\) and \(n = 10\). A common error is to remove only one of the end terms.

Mark scheme

1 mark
B1 Substitutes \(x = 1\) into \((1 + x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \dots + \binom{n}{n}x^n\) to give \(2^n = \binom{n}{0} + \binom{n}{1} + \dots + \binom{n}{n}\). This mark is for the explanation, so no working scores B0; a correct final answer of \(n = 10\) with no working scores the M1 and A1
An answer of \(n = 10\) with no working scores 2 marks: the B1 is only for the explanation. The expansion with \(x = 1\) must be seen, e.g. \((1 + 1)^n = 1 + n + \binom{n}{2} + \dots + 1\); just "put \(x = 1\)" without the expansion, or a check for one value of \(n\), is B0.
1 mark
M1 Uses \(\binom{n}{0} = \binom{n}{n} = 1\) to form \(2^n - 2 = 1022\) oe
Accept \(2^n = 1024\). Removing only one end term (\(2^n - 1 = 1022\)) or using \(2^n = 1022\) is M0.
1 mark
A1 \(n = 10\)
cao. \(n\) may be found from \(2^n = 1024\) by inspection or by logarithms.

Question 2

5 marks

Find the binomial expansion of
\[\dfrac{3 - x}{(1 + 2x)^2}\]
in ascending powers of \(x\), up to and including the term in \(x^3\). Give each coefficient as an integer.

State the range of values of \(x\) for which the expansion 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

\(\dfrac{3 - x}{(1 + 2x)^2} = (3 - x)(1 + 2x)^{-2}\)
\((1 + 2x)^{-2} = 1 + (-2)(2x) + \dfrac{(-2)(-3)}{2!}(2x)^2 + \dfrac{(-2)(-3)(-4)}{3!}(2x)^3 + \dots\)
\(= 1 - 4x + 12x^2 - 32x^3 + \dots\)
\((3 - x)(1 - 4x + 12x^2 - 32x^3) = 3 - 12x + 36x^2 - 96x^3 - x + 4x^2 - 12x^3 + \dots\)
\(= 3 - 13x + 40x^2 - 108x^3 + \dots\)
Valid for \(\lvert 2x \rvert < 1\), i.e. \(\lvert x \rvert < \dfrac{1}{2}\)

Explanation

Write the fraction as \((3 - x)(1 + 2x)^{-2}\) and use the expansion of \((1 + x)^n\) with \(n = -2\) and \(2x\) in place of \(x\), then multiply by \(3 - x\), keeping only the terms up to \(x^3\). The expansion is valid when \(\lvert 2x \rvert < 1\). Common errors are writing \((2x)^2\) as \(2x^2\) and sign slips with the alternating coefficients.

Mark scheme

1 mark
M1 Writes \((1 + 2x)^{-2}\) and expands it with the correct structure for the \(x^2\) or \(x^3\) term, e.g. \(\dfrac{(-2)(-3)}{2!}(2x)^2\) or \(\dfrac{(-2)(-3)(-4)}{3!}(2x)^3\); or a correct final answer of \(3 - 13x + 40x^2 - 108x^3\) with no working
An answer of \(3 - 13x + 40x^2 - 108x^3\) with no working scores 4 marks (the B1 for the range is marked separately). Condone missing brackets on \(2x\) for this mark. Alternative method: partial fractions \(-\dfrac{1}{2(1 + 2x)} + \dfrac{7}{2(1 + 2x)^2}\) and expanding each term scores M1 for a correct structure and the marks below in the same way.
1 mark
A1 \((1 + 2x)^{-2} = 1 - 4x + 12x^2 - 32x^3\)
1 mark
M1 Multiplies their expansion by \((3 - x)\) and collects the terms up to \(x^3\)
Depends on the first M mark. At least three of the four terms must come from correct products.
1 mark
A1 \(3 - 13x + 40x^2 - 108x^3\)
cao; isw once a correct expansion is seen. Ignore any terms in \(x^4\).
1 mark
B1 \(\lvert x \rvert < \dfrac{1}{2}\) oe, e.g. \(-\dfrac{1}{2} < x < \dfrac{1}{2}\)
Independent of the other marks. Do not accept \(x < \dfrac{1}{2}\), \(\lvert x \rvert \leqslant \dfrac{1}{2}\) or \(\lvert x \rvert < 2\).

Question 3

6 marks

In the binomial expansion of \((3 + 2x)^n\), where \(n\) is an integer with \(n \geqslant 4\), the coefficient of \(x^3\) is equal to the coefficient of \(x^4\).

Using algebra, find the value of \(n\), and hence find the value of this coefficient.

Model answer

Coefficient of \(x^3\): \(\binom{n}{3}3^{n-3}2^3 = 8\binom{n}{3}3^{n-3}\)
Coefficient of \(x^4\): \(\binom{n}{4}3^{n-4}2^4 = 16\binom{n}{4}3^{n-4}\)
\(8\binom{n}{3}3^{n-3} = 16\binom{n}{4}3^{n-4}\)
Dividing by \(8 \times 3^{n-4}\): \(3\binom{n}{3} = 2\binom{n}{4}\)
\(\binom{n}{4} = \dfrac{n!}{4!(n - 4)!} = \dfrac{n!}{3!(n - 3)!} \times \dfrac{n - 3}{4} = \binom{n}{3} \times \dfrac{n - 3}{4}\)
So \(3 = \dfrac{2(n - 3)}{4}\), \(n - 3 = 6\), \(n = 9\)
Coefficient \(= \binom{9}{3} \times 3^6 \times 2^3 = 84 \times 729 \times 8 = 489888\)

Explanation

Write both coefficients with the correct powers of 3 and 2, set them equal and cancel: \(\dfrac{3^{n-3}}{3^{n-4}} = 3\) and \(\binom{n}{4} \div \binom{n}{3} = \dfrac{n - 3}{4}\). This gives a linear equation with solution \(n = 9\), and then either coefficient is \(84 \times 729 \times 8 = 489888\). A common error is to leave out the powers of 3, which gives \(n = 5\).

Mark scheme

1 mark
M1 Writes the coefficient of \(x^3\) or of \(x^4\) with the correct structure: \(\binom{n}{3}3^{n-3}2^3\) or \(\binom{n}{4}3^{n-4}2^4\). An answer with no working scores 0 marks
An answer with no working scores 0 marks: the question asks for algebra, so \(n = 9\) found by trying values scores none of the first four marks (the last two marks may still be earned by substituting \(n = 9\)). Accept \({}^n\mathrm{C}_3\) notation and factorial forms.
1 mark
A1 Both coefficients correct: \(8\binom{n}{3}3^{n-3}\) and \(16\binom{n}{4}3^{n-4}\) oe
1 mark
M1 (dM1) Sets them equal and simplifies the binomial coefficients and the powers of 3 and 2 to a linear equation in \(n\), e.g. \(\dfrac{n - 3}{4} = \dfrac{3}{2}\)
Depends on the first M mark. Uses \(\binom{n}{4} = \binom{n}{3} \times \dfrac{n - 3}{4}\) (or cancels the factorials correctly) and \(3^{n-3} = 3 \times 3^{n-4}\). A common error is to leave out the powers of 3, which gives \(n = 5\).
1 mark
A1 \(n = 9\)
cao
1 mark
M1 Substitutes their \(n\) into either coefficient, e.g. \(\binom{9}{3} \times 3^6 \times 2^3\)
Independent of the previous marks, but their \(n\) must be an integer of at least 4.
1 mark
A1 489888
cao. Accept 489 888.

Question 4

8 marks

Find the binomial expansion of \((1 - x)^{-2}\), \(\lvert x \rvert < 1\), in ascending powers of \(x\), up to and including the term in \(x^3\), simplifying each coefficient.

Show that, for every positive integer \(r\), the coefficient of \(x^r\) in this expansion is \(r + 1\).

Hence find the exact value of
\[\sum_{r=0}^{\infty} \dfrac{r + 1}{4^r}\]
Find also the exact value of
\[\sum_{r=1}^{\infty} \dfrac{r}{4^r}\]

Model answer

\((1 - x)^{-2} = 1 + (-2)(-x) + \dfrac{(-2)(-3)}{2!}(-x)^2 + \dfrac{(-2)(-3)(-4)}{3!}(-x)^3 + \dots\)
\(= 1 + 2x + 3x^2 + 4x^3 + \dots\)
The term in \(x^r\) is \(\dfrac{(-2)(-3)(-4)\dots(-(r + 1))}{1 \times 2 \times \dots \times r}(-x)^r\)
The numerator has \(r\) negative factors, so it is \((-1)^r \times 2 \times 3 \times \dots \times (r + 1) = (-1)^r(r + 1)!\)
So the term is \(\dfrac{(-1)^r(r + 1)!}{r!} \times (-1)^r x^r = (r + 1)x^r\), since \((-1)^{2r} = 1\) and \(\dfrac{(r + 1)!}{r!} = r + 1\).
So \((1 - x)^{-2} = \displaystyle\sum_{r=0}^{\infty}(r + 1)x^r\) for \(\lvert x \rvert < 1\).
\(x = \dfrac{1}{4}\) is in the range of validity: \(\displaystyle\sum_{r=0}^{\infty}\dfrac{r + 1}{4^r} = \left(1 - \dfrac{1}{4}\right)^{-2} = \left(\dfrac{3}{4}\right)^{-2} = \dfrac{16}{9}\)
\(\displaystyle\sum_{r=0}^{\infty}\dfrac{1}{4^r} = \dfrac{1}{1 - \frac{1}{4}} = \dfrac{4}{3}\)
\(\displaystyle\sum_{r=1}^{\infty}\dfrac{r}{4^r} = \sum_{r=0}^{\infty}\dfrac{r}{4^r} = \sum_{r=0}^{\infty}\dfrac{r + 1}{4^r} - \sum_{r=0}^{\infty}\dfrac{1}{4^r} = \dfrac{16}{9} - \dfrac{4}{3} = \dfrac{4}{9}\)

Explanation

With \(n = -2\) and \(-x\) in place of \(x\), the \(r\) negative factors in the numerator and the \((-1)^r\) from \((-x)^r\) cancel in sign, leaving \(\dfrac{(r + 1)!}{r!} = r + 1\). Since \(x = \dfrac{1}{4}\) lies in the range of validity, the first sum is \(\left(\dfrac{3}{4}\right)^{-2} = \dfrac{16}{9}\), and subtracting the geometric series \(\displaystyle\sum_{r=0}^{\infty}\dfrac{1}{4^r} = \dfrac{4}{3}\) gives \(\dfrac{4}{9}\). Common errors are sign slips from \((-x)^r\) and subtracting a geometric series that starts at \(r = 1\).

Mark scheme

1 mark
M1 Uses the binomial series with \(n = -2\) and \(-x\) in place of \(x\), with the correct structure for the \(x^2\) or \(x^3\) term, e.g. \(\dfrac{(-2)(-3)}{2!}(-x)^2\); or correct final answers of \(1 + 2x + 3x^2 + 4x^3\), \(\dfrac{16}{9}\) and \(\dfrac{4}{9}\) with no working (special case: 4 marks)
Answers of \(1 + 2x + 3x^2 + 4x^3\), \(\dfrac{16}{9}\) and \(\dfrac{4}{9}\) with no working score 4 marks (special case: M1 A1 for the expansion, since a correct expansion with no working scores both marks, and the A1 for each sum); the show-that needs working, and the method marks for the sums need the method seen. Condone missing brackets on \(-x\) if recovered.
1 mark
A1 \(1 + 2x + 3x^2 + 4x^3\)
Ignore any extra terms. A common error is \(1 - 2x + 3x^2 - 4x^3\) (using \(x\) instead of \(-x\)): A0.
1 mark
M1 Writes the general term with \(n = -2\): \(\dfrac{(-2)(-3)\dots(-2 - r + 1)}{1 \times 2 \times \dots \times r}(-x)^r\), with the last factor of the numerator \(-(r + 1)\)
The numerator must have \(r\) factors ending in \(-(r + 1)\), or be written as \((-1)^r \times 2 \times 3 \times \dots \times (r + 1)\).
1 mark
A1* Simplifies to \(\dfrac{(r + 1)!}{r!} = r + 1\), dealing with the signs: \((-1)^r \times (-1)^r = 1\)
cso. Checking a few values of \(r\) only is A0*: it does not show the result for every \(r\).
1 mark
M1 Substitutes \(x = \dfrac{1}{4}\) (within \(\lvert x \rvert < 1\)) into \(\displaystyle\sum_{r=0}^{\infty}(r + 1)x^r = (1 - x)^{-2}\)
Must use \((1 - x)^{-2}\), not the four-term expansion (which gives an approximation, 1.75, only).
1 mark
A1 \(\left(\dfrac{3}{4}\right)^{-2} = \dfrac{16}{9}\)
Exact. 1.78 on its own is A0.
1 mark
M1 A valid method for \(\displaystyle\sum_{r=1}^{\infty}\dfrac{r}{4^r}\): subtracts \(\displaystyle\sum_{r=0}^{\infty}\dfrac{1}{4^r} = \dfrac{1}{1 - \frac{1}{4}} = \dfrac{4}{3}\) from their \(\dfrac{16}{9}\); or writes \(\displaystyle\sum_{r=1}^{\infty}\dfrac{r}{4^r} = \dfrac{1}{4}\sum_{s=0}^{\infty}\dfrac{s + 1}{4^s}\)
The geometric series subtracted must start from \(r = 0\) (first term 1) to match; using \(\dfrac{1}{3}\) (the sum from \(r = 1\)) without adjusting is M0.
1 mark
A1 \(\dfrac{4}{9}\)
Exact. 0.444 on its own is A0.

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