Binomial expansion
Sequences and series
- A-level
- Non-calculator
- 8 questions
- 37 marks
These are original practice questions for the Edexcel A-level Pure content, not a past paper. This site is independent and not affiliated with Pearson.
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Expand $(1 + 2x)^4$ fully.
3 marks
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Find the coefficient of $x^3$ in the expansion of $(2 - x)^5$.
3 marks
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Find the first three terms in the expansion of $(1 - 2x)^{-1}$ in ascending powers of $x$. State the values of $x$ for which the expansion is valid.
4 marks
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Find the first three terms in the expansion of $(1 + 4x)^{1/2}$ in ascending powers of $x$, and state the values of $x$ for which it is valid.
4 marks
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Find the coefficient of $x^2$ in the expansion of $(2 + 3x)^6$.
5 marks
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Expand $(1 + x)^{1/2}$ up to and including the term in $x^2$.
5 marks
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Use the binomial expansion of $(1 + x)^{1/2}$, up to the term in $x^2$, to estimate $\sqrt{1.02}$. Give your estimate to $5$ decimal places, and state why the expansion may be used.
6 marks
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Find the coefficient of $x^3$ in the expansion of $(2 - x)^4(1 + x)$.
7 marks
Mark scheme
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Answer. $1 + 8x + 24x^2 + 32x^3 + 16x^4$
- Use $\binom{4}{k}(2x)^k$.
- The terms are $1$, $8x$, $24x^2$, $32x^3$ and $16x^4$.
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Answer. $-40$
- The term is $\binom{5}{3}(2)^2(-x)^3$.
- $10 \times 4 \times (-1) = -40$.
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Answer. $1 + 2x + 4x^2$, valid for $|x| < \dfrac{1}{2}$.
- $(1 + y)^{-1} = 1 - y + y^2 + \cdots$ with $y = -2x$.
- The terms are $1 + 2x + 4x^2$.
- Validity requires $|-2x| < 1$, so $|x| < \dfrac{1}{2}$.
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Answer. $1 + 2x - 2x^2$, valid for $|x| < \dfrac{1}{4}$.
- $(1 + 4x)^{1/2} = 1 + \dfrac{1}{2}(4x) + \dfrac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2}(4x)^2 + \cdots$.
- This simplifies to $1 + 2x - 2x^2$.
- $|4x| < 1$, so $|x| < \dfrac{1}{4}$.
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Answer. $2160$
- The term is $\binom{6}{2}(2)^4(3x)^2$.
- $15 \times 16 \times 9 = 2160$.
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Answer. $1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2$
- The $x$ term is $\dfrac{1}{2}x$.
- The $x^2$ coefficient is $\dfrac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2} = -\dfrac{1}{8}$.
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Answer. $1.00995$. The expansion may be used because $|0.02| < 1$.
- $\sqrt{1.02} = (1 + 0.02)^{1/2}$.
- $1 + \dfrac{1}{2}(0.02) - \dfrac{1}{8}(0.02)^2 = 1 + 0.01 - 0.00005$.
- The estimate is $1.00995$.
- $|x| = 0.02 < 1$, so $x$ is inside the interval of validity.
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Answer. $16$
- $(2 - x)^4 = 16 - 32x + 24x^2 - 8x^3 + x^4$.
- Multiplying by $1 + x$, the $x^3$ terms come from $-8x^3$ and from $x \times 24x^2$.
- $-8 + 24 = 16$.