| Exam Board | Edexcel |
|---|---|
| Module | PMT Mocks (PMT Mocks) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Factoring out constants before expansion |
| Difficulty | Standard +0.3 This is a straightforward binomial expansion requiring factoring out 1/9 to get the standard form, then applying the formula with n=1/2. Part (b) tests understanding of how the expansion approximates values, but requires no calculation. Slightly above average due to the fractional power and negative term, but still a standard textbook exercise. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{3} - 3x - \frac{27}{2}x^2 - \frac{243}{2}x^3\) | B1 | For taking out a factor of \(\left(\frac{1}{9}\right)^{\frac{1}{2}}\) |
| M1 | For the form of the binomial expansion with \(n = \frac{1}{2}\) and a term of \((kx)\) | |
| A1 | Three of the four terms are correct | |
| A1 cso | All terms are correct: \(\frac{1}{3} - 3x - \frac{27}{2}x^2 - \frac{243}{2}x^3\) |
| Answer | Marks | Guidance |
|---|---|---|
| If \(x = \frac{1}{36}\), \(\left(\frac{1}{9} - 2x\right)^{\frac{1}{2}} = \frac{\sqrt{2}}{6}\) so \(\sqrt{2}\) can be approximated by substituting \(x = \frac{1}{36}\) into the expansion and multiplying by 6 | M1 | Score for substituting \(x = \frac{1}{36}\) into \(\left(\frac{1}{9} - 2x\right)^{\frac{1}{2}}\) to obtain \(\frac{\sqrt{2}}{6}\) or such as \(\sqrt{\frac{2}{36}}\) |
| A1 | Explains that \(x = \frac{1}{36}\) is substituted into both sides and you multiply the result by 6 |
**Part a:** Find the first four terms, in ascending powers of x, of the binomial expansion of $\left(\frac{1}{9} - 2x\right)^{\frac{1}{2}}$
| $\frac{1}{3} - 3x - \frac{27}{2}x^2 - \frac{243}{2}x^3$ | B1 | For taking out a factor of $\left(\frac{1}{9}\right)^{\frac{1}{2}}$ |
| --- | --- | --- |
| | M1 | For the form of the binomial expansion with $n = \frac{1}{2}$ and a term of $(kx)$ |
| | A1 | Three of the four terms are correct |
| | A1 cso | All terms are correct: $\frac{1}{3} - 3x - \frac{27}{2}x^2 - \frac{243}{2}x^3$ |
**Part b:** Explain how you could use $x = \frac{1}{36}$ in the expansion to find an approximation for $\sqrt{2}$
| If $x = \frac{1}{36}$, $\left(\frac{1}{9} - 2x\right)^{\frac{1}{2}} = \frac{\sqrt{2}}{6}$ so $\sqrt{2}$ can be approximated by substituting $x = \frac{1}{36}$ into the expansion and multiplying by 6 | M1 | Score for substituting $x = \frac{1}{36}$ into $\left(\frac{1}{9} - 2x\right)^{\frac{1}{2}}$ to obtain $\frac{\sqrt{2}}{6}$ or such as $\sqrt{\frac{2}{36}}$ |
| --- | --- | --- |
| | A1 | Explains that $x = \frac{1}{36}$ is substituted into both sides and you multiply the result by 6 |
**(4) marks total for Question 1**
---\begin{enumerate}
\item a. Find the first four terms, in ascending powers of $x$, of the binomial expansion of
\end{enumerate}
$$\left( \frac { 1 } { 9 } - 2 x \right) ^ { \frac { 1 } { 2 } }$$
giving each coefficient in its simplest form.\\
b. Explain how you could use $x = \frac { 1 } { 36 }$ in the expansion to find an approximation for $\sqrt { 2 }$.
There is no need to carry out the calculation.\\
\hfill \mbox{\textit{Edexcel PMT Mocks Q1 [6]}}