| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Direct single expansion substitution |
| Difficulty | Standard +0.3 This is a standard C4 binomial expansion question with straightforward algebraic manipulation. Part (a) requires simplifying a numerical expression, part (b) is routine application of the binomial series formula with negative/fractional index, and part (c) connects the expansion to approximation. While it requires careful arithmetic and understanding of the binomial series validity, it follows a very common exam pattern with no novel problem-solving required. |
| Spec | 1.02b Surds: manipulation and rationalising denominators1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\left(\frac{25}{24}\right)^{-1} = \sqrt{\frac{24}{25}} = \frac{\sqrt{24}}{5} \times 6 = \frac{2}{5}\sqrt{6}\) | M1 A1 | [\(k = \frac{2}{5}\)] |
| (b) \(= 1 + \left(-\frac{1}{2}\right)\left(\frac{1}{2}x\right) + \frac{(-1)(-\frac{1}{2})(-\frac{3}{2})}{2!}\left(\frac{1}{2}x\right)^2 + \frac{(-1)(-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})}{3!}\left(\frac{1}{2}x\right)^3 + \ldots\) | M1 | |
| \(= 1 - \frac{1}{4}x + \frac{3}{32}x^2 - \frac{5}{128}x^3 + \ldots\) | A3 | |
| (c) \(x = \frac{1}{12} \Rightarrow (1 + \frac{1}{2})^{-\frac{1}{2}} = \left(\frac{13}{24}\right)^{-\frac{1}{2}} = \frac{2}{3}\sqrt{6}\) | M1 | |
| \(x = \frac{1}{12} \Rightarrow (1 + \frac{1}{2})^{-1} = 1 - \frac{1}{2}\left(\frac{1}{12}\right) + \frac{3}{32}\left(\frac{1}{12}\right)^2 - \frac{5}{128}\left(\frac{1}{12}\right)^3\) | M1 | |
| \(= 0.97979510\) | ||
| \(\therefore \sqrt{6} \approx \frac{2}{3} \times 0.97979510 = 2.44949\) (5dp) | M1 A1 | (9 marks total) |
**(a)** $\left(\frac{25}{24}\right)^{-1} = \sqrt{\frac{24}{25}} = \frac{\sqrt{24}}{5} \times 6 = \frac{2}{5}\sqrt{6}$ | M1 A1 | [$k = \frac{2}{5}$] |
**(b)** $= 1 + \left(-\frac{1}{2}\right)\left(\frac{1}{2}x\right) + \frac{(-1)(-\frac{1}{2})(-\frac{3}{2})}{2!}\left(\frac{1}{2}x\right)^2 + \frac{(-1)(-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})}{3!}\left(\frac{1}{2}x\right)^3 + \ldots$ | M1 |
$= 1 - \frac{1}{4}x + \frac{3}{32}x^2 - \frac{5}{128}x^3 + \ldots$ | A3 |
**(c)** $x = \frac{1}{12} \Rightarrow (1 + \frac{1}{2})^{-\frac{1}{2}} = \left(\frac{13}{24}\right)^{-\frac{1}{2}} = \frac{2}{3}\sqrt{6}$ | M1 |
$x = \frac{1}{12} \Rightarrow (1 + \frac{1}{2})^{-1} = 1 - \frac{1}{2}\left(\frac{1}{12}\right) + \frac{3}{32}\left(\frac{1}{12}\right)^2 - \frac{5}{128}\left(\frac{1}{12}\right)^3$ | M1 |
$= 0.97979510$ | |
$\therefore \sqrt{6} \approx \frac{2}{3} \times 0.97979510 = 2.44949$ (5dp) | M1 A1 | (9 marks total) |\begin{enumerate}[label=(\alph*)]
\item Show that $(1 + \frac{1}{24})^{-\frac{1}{2}} = k\sqrt{6}$, where $k$ is rational. [2]
\item Expand $(1 + \frac{1}{4}x)^{-\frac{1}{2}}$, $|x| < 2$, in ascending powers of $x$ up to and including the term in $x^3$, simplifying each coefficient. [4]
\item Use your answer to part $(b)$ with $x = \frac{1}{6}$ to find an approximate value for $\sqrt{6}$, giving your answer to 5 decimal places. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q3 [9]}}