Standard +0.3 This is a standard binomial expansion question requiring factoring out 8^(2/3) first, then applying the formula with fractional index. Part (b) tests understanding of alternating series convergence, which is conceptual but straightforward. Slightly easier than average as it's a routine application with clear steps.
7. a. Use the binomial theorem to expand
$$( 8 - 3 x ) ^ { \frac { 2 } { 3 } }$$
in ascending powers of \(x\), up to and including the term \(x ^ { 3 }\), as a fully simplifying each term.
Edward, a student decides to use the expansion with \(x = \frac { 1 } { 3 }\) to find an approximation for \(( 7 ) ^ { \frac { 2 } { 3 } }\). Using the answer to part (a) and without doing any calculations, b. explain clearly whether Edward's approximation will be an overestimate, or, an underestimate.
Takes out a factor of 8 and writes \((8 - 3x)^{\frac{2}{3}} = 4(1 \pm \cdots)^{\frac{2}{3}}\) or \(8^{\frac{2}{3}}(1 \pm \cdots)^{\frac{2}{3}}\) or \(2^2(1 \pm \cdots)^{\frac{2}{3}}\)
M1
For an attempt at the binomial expansion of \((1 + ax)^{\frac{2}{3}}\), \(a \neq 1\) to form term 3 or term 4 with the correct structure. Look for the correct binomial coefficient multiplied by the corresponding power of x. Example: \(\frac{(\frac{2}{3})(\frac{2}{3}-1)}{2!}(\cdots x)^2\) or \(\frac{(\frac{2}{3})(\frac{2}{3}-1)(\frac{2}{3}-2)}{3!}(\cdots x)^3\) where \(\cdots \neq 1\)
A1
Correct expression for the expansion of \(\left(1 - \frac{3}{8}x\right)^{\frac{2}{3}}\)
which may left unsimplified as shown but the bracketing must be correct unless any missing brackets are implied by subsequent work. If the 4 outside this expansion is only partially applied to this expansion then scores A0 but if it is applied to all terms this A1 can be implied.
**OR at least 2 correct simplified terms for the final expansion from, \(-x\), \(-\frac{1}{16}x^2\), \(-\frac{1}{96}x^3\)
Answer
Marks
A1
\((8 - 3x)^{\frac{2}{3}} = 4 - x - \frac{1}{16}x^2 - \frac{1}{96}x^3\) or equivalent
OR Direct expansion in (a) can be marked in a similar way.
States that the approximation will be an overestimate due to the fact that all terms (after the first one) in the expansion are negative or equivalent statements e.g. Overestimate because the terms are negative; Overestimate as the terms are being taken away from 4
(5 marks)
**Part a.** | (4 marks)
**B1** | Takes out a factor of 8 and writes $(8 - 3x)^{\frac{2}{3}} = 4(1 \pm \cdots)^{\frac{2}{3}}$ or $8^{\frac{2}{3}}(1 \pm \cdots)^{\frac{2}{3}}$ or $2^2(1 \pm \cdots)^{\frac{2}{3}}$
**M1** | For an attempt at the binomial expansion of $(1 + ax)^{\frac{2}{3}}$, $a \neq 1$ to form term 3 or term 4 with the correct structure. Look for the correct binomial coefficient multiplied by the corresponding power of x. Example: $\frac{(\frac{2}{3})(\frac{2}{3}-1)}{2!}(\cdots x)^2$ or $\frac{(\frac{2}{3})(\frac{2}{3}-1)(\frac{2}{3}-2)}{3!}(\cdots x)^3$ where $\cdots \neq 1$
**A1** | Correct expression for the expansion of $\left(1 - \frac{3}{8}x\right)^{\frac{2}{3}}$
Example: $1 + \frac{2}{3}\left(-\frac{3}{8}x\right) + \frac{(\frac{2}{3})(\frac{2}{3}-1)}{2!} \times \left(+\frac{3}{8}x\right)^2 + \frac{(\frac{2}{3})(\frac{2}{3}-1)(\frac{2}{3}-2)}{3!}\left(-\frac{3}{8}x\right)^3$
which may left unsimplified as shown but the bracketing must be correct unless any missing brackets are implied by subsequent work. If the 4 outside this expansion is only partially applied to this expansion then scores A0 but if it is applied to all terms this A1 can be implied.
**OR at least 2 correct simplified terms for the final expansion from, $-x$, $-\frac{1}{16}x^2$, $-\frac{1}{96}x^3$
**A1** | $(8 - 3x)^{\frac{2}{3}} = 4 - x - \frac{1}{16}x^2 - \frac{1}{96}x^3$ or equivalent
**OR Direct expansion in (a) can be marked in a similar way.**
$(8 - 3x)^{\frac{2}{3}} = (8)^{\frac{2}{3}} + \frac{2}{3} \times (8)^{-\frac{1}{3}} \times (-3x) + \left(\frac{2}{3}\right)\left(\frac{2}{3}-1\right) \times (8)^{-\frac{4}{3}} \times \frac{(-3x)^2}{2!} + \left(\frac{2}{3}\right)\left(\frac{2}{3}-1\right)\left(\frac{2}{3}-2\right) \times (8)^{-\frac{7}{3}} \times \frac{(-3x)^3}{3!}$
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**Part b.** | (1 mark)
**B1** | States that the approximation will be an overestimate due to the fact that all terms (after the first one) in the expansion are negative or equivalent statements e.g. Overestimate because the terms are negative; Overestimate as the terms are being taken away from 4
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7. a. Use the binomial theorem to expand
$$( 8 - 3 x ) ^ { \frac { 2 } { 3 } }$$
in ascending powers of $x$, up to and including the term $x ^ { 3 }$, as a fully simplifying each term.
Edward, a student decides to use the expansion with $x = \frac { 1 } { 3 }$ to find an approximation for $( 7 ) ^ { \frac { 2 } { 3 } }$. Using the answer to part (a) and without doing any calculations, b. explain clearly whether Edward's approximation will be an overestimate, or, an underestimate.\\
\hfill \mbox{\textit{Edexcel PMT Mocks Q7 [5]}}