Edexcel C34 2014 June — Question 5 5 marks

Exam BoardEdexcel
ModuleC34 (Core Mathematics 3 & 4)
Year2014
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeneralised Binomial Theorem
TypeNon-zero terms only
DifficultyStandard +0.3 This is a straightforward application of the binomial expansion with a fractional power. Students must factor out 8^(1/3)=2, recognize the form (1+u)^(1/3) where u=27x³/8, and apply the standard binomial series formula. While it requires careful algebraic manipulation and simplification of fractions, it's a routine C3/C4 question with no conceptual surprises—slightly easier than average due to its mechanical nature.
Spec1.04c Extend binomial expansion: rational n, |x|<1
5. $$f ( x ) = \left( 8 + 27 x ^ { 3 } \right) ^ { \frac { 1 } { 3 } } , \quad | x | < \frac { 2 } { 3 }$$ Find the first three non-zero terms of the binomial expansion of \(\mathrm { f } ( x )\) in ascending powers of \(x\). Give each coefficient as a simplified fraction.
Question 5:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\((8+27x^3)^{\frac{1}{3}} = (8)^{\frac{1}{3}}\left(1+\frac{27x^3}{8}\right)^{\frac{1}{3}} = 2\left(1+\frac{27x^3}{8}\right)^{\frac{1}{3}}\)B1 \((8)^{\frac{1}{3}}\) or \(2\) outside brackets; or \((8)^{\frac{1}{3}}\) or \(2\) as candidate's constant term in expansion
\(= \{2\}\left[1 + \left(\frac{1}{3}\right)(kx^3) + \frac{(\frac{1}{3})(-\frac{2}{3})}{2!}(kx^3)^2 + \ldots\right]\)M1 Expands \((\ldots + kx^3)^{\frac{1}{3}}\) to give any 2 terms out of 3 correct for their \(k\), simplified or unsimplified; \(k \neq 1\)
\(= \{2\}\left[1 + \left(\frac{1}{3}\right)\left(\frac{27x^3}{8}\right) + \frac{(\frac{1}{3})(-\frac{2}{3})}{2!}\left(\frac{27x^3}{8}\right)^2 + \ldots\right]\)A1 Correct simplified or unsimplified expansion with consistent \((kx^3)\); all three terms correct for their \(k\)
\(= 2\left[1 + \frac{9}{8}x^3;\; -\frac{81}{64}x^6 + \ldots\right]\)
\(= 2 + \frac{9}{4}x^3;\; -\frac{81}{32}x^6 + \ldots\)A1; A1 \(2 + \frac{9}{4}x^3\) (allow \(2 + 2.25x^3\) or \(2 + 2\frac{1}{4}x^3\)); \(-\frac{81}{32}x^6\) (allow \(-2.53125x^6\) or \(-2\frac{17}{32}x^6\)); ignore extra higher power terms 
# Question 5:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $(8+27x^3)^{\frac{1}{3}} = (8)^{\frac{1}{3}}\left(1+\frac{27x^3}{8}\right)^{\frac{1}{3}} = 2\left(1+\frac{27x^3}{8}\right)^{\frac{1}{3}}$ | B1 | $(8)^{\frac{1}{3}}$ or $2$ outside brackets; or $(8)^{\frac{1}{3}}$ or $2$ as candidate's constant term in expansion |
| $= \{2\}\left[1 + \left(\frac{1}{3}\right)(kx^3) + \frac{(\frac{1}{3})(-\frac{2}{3})}{2!}(kx^3)^2 + \ldots\right]$ | M1 | Expands $(\ldots + kx^3)^{\frac{1}{3}}$ to give any 2 terms out of 3 correct for their $k$, simplified or unsimplified; $k \neq 1$ |
| $= \{2\}\left[1 + \left(\frac{1}{3}\right)\left(\frac{27x^3}{8}\right) + \frac{(\frac{1}{3})(-\frac{2}{3})}{2!}\left(\frac{27x^3}{8}\right)^2 + \ldots\right]$ | A1 | Correct simplified or unsimplified expansion with consistent $(kx^3)$; all three terms correct for their $k$ |
| $= 2\left[1 + \frac{9}{8}x^3;\; -\frac{81}{64}x^6 + \ldots\right]$ | | |
| $= 2 + \frac{9}{4}x^3;\; -\frac{81}{32}x^6 + \ldots$ | A1; A1 | $2 + \frac{9}{4}x^3$ (allow $2 + 2.25x^3$ or $2 + 2\frac{1}{4}x^3$); $-\frac{81}{32}x^6$ (allow $-2.53125x^6$ or $-2\frac{17}{32}x^6$); ignore extra higher power terms |
5.

$$f ( x ) = \left( 8 + 27 x ^ { 3 } \right) ^ { \frac { 1 } { 3 } } , \quad | x | < \frac { 2 } { 3 }$$

Find the first three non-zero terms of the binomial expansion of $\mathrm { f } ( x )$ in ascending powers of $x$. Give each coefficient as a simplified fraction.\\

\hfill \mbox{\textit{Edexcel C34 2014 Q5 [5]}}
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