| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2021 |
| Session | January |
| Marks | 7 |
| 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 application of the binomial expansion requiring factoring out 1/4, then substituting a value. The technique of factoring constants is standard P4 material, and the substitution in part (b) is mechanical arithmetic. Slightly above average difficulty due to the fractional power and need for careful algebraic manipulation, but remains a routine textbook exercise. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left(\frac{1}{4}-5x\right)^{\frac{1}{2}} = \frac{1}{2}(\ldots)\) | B1 | For taking out factor of \(\left(\frac{1}{4}\right)^{\frac{1}{2}}\) or \(\frac{1}{2}\) or 0.5 |
| \(=(1-20x)^{\frac{1}{2}} = 1+\left(\frac{1}{2}\right)\times(-20x)+\frac{\left(\frac{1}{2}\right)\times\left(-\frac{1}{2}\right)}{2!}\times(-20x)^2+\frac{\left(\frac{1}{2}\right)\times\left(-\frac{1}{2}\right)\times\left(-\frac{3}{2}\right)}{3!}\times(-20x)^3\ldots\) | M1A1 | M1: Expands \((1+kx)^{\frac{1}{2}}, k\neq\pm 1\) with correct structure for third or fourth term. A1: Either term three or four correct in any form |
| \(=\frac{1}{2}-5x-25x^2-250x^3+\ldots\) | A1 A1 | First A1: Two terms correct and simplified. Second A1: All four terms correct and simplified. If any \(-\) signs written as \(+-\) score A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left(\frac{1}{4}-\frac{5}{100}\right)^{\frac{1}{2}}=\left(\frac{1}{5}\right)^{\frac{1}{2}}=\frac{1}{2}-5\times\frac{1}{100}-25\left(\frac{1}{100}\right)^2-250\left(\frac{1}{100}\right)^3+\ldots\) | M1 | Attempts to substitute \(x=\frac{1}{100}\) into part (a) and either multiplies by 5 or finds reciprocal |
| \(\frac{\sqrt{5}}{5}\approx\frac{1789}{4000}\) or \(\frac{1}{\sqrt{5}}\approx\frac{1789}{4000}\) leading to \(\sqrt{5}\approx\frac{1789}{800}\) or \(\frac{4000}{1789}\) | A1 | \(\left(\sqrt{5}=\right)\frac{1789}{800}\) or \(\frac{4000}{1789}\) |
## Question 1:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(\frac{1}{4}-5x\right)^{\frac{1}{2}} = \frac{1}{2}(\ldots)$ | B1 | For taking out factor of $\left(\frac{1}{4}\right)^{\frac{1}{2}}$ or $\frac{1}{2}$ or 0.5 |
| $=(1-20x)^{\frac{1}{2}} = 1+\left(\frac{1}{2}\right)\times(-20x)+\frac{\left(\frac{1}{2}\right)\times\left(-\frac{1}{2}\right)}{2!}\times(-20x)^2+\frac{\left(\frac{1}{2}\right)\times\left(-\frac{1}{2}\right)\times\left(-\frac{3}{2}\right)}{3!}\times(-20x)^3\ldots$ | M1A1 | M1: Expands $(1+kx)^{\frac{1}{2}}, k\neq\pm 1$ with correct structure for third or fourth term. A1: Either term three or four correct in any form |
| $=\frac{1}{2}-5x-25x^2-250x^3+\ldots$ | A1 A1 | First A1: Two terms correct and simplified. Second A1: All four terms correct and simplified. If any $-$ signs written as $+-$ score A0 |
**Special case:** If final answer left as $\frac{1}{2}(1-10x-50x^2-500x^3+\ldots)$ award SC B1M1A1A1A0
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### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(\frac{1}{4}-\frac{5}{100}\right)^{\frac{1}{2}}=\left(\frac{1}{5}\right)^{\frac{1}{2}}=\frac{1}{2}-5\times\frac{1}{100}-25\left(\frac{1}{100}\right)^2-250\left(\frac{1}{100}\right)^3+\ldots$ | M1 | Attempts to substitute $x=\frac{1}{100}$ into part (a) and either multiplies by 5 or finds reciprocal |
| $\frac{\sqrt{5}}{5}\approx\frac{1789}{4000}$ or $\frac{1}{\sqrt{5}}\approx\frac{1789}{4000}$ leading to $\sqrt{5}\approx\frac{1789}{800}$ or $\frac{4000}{1789}$ | A1 | $\left(\sqrt{5}=\right)\frac{1789}{800}$ or $\frac{4000}{1789}$ |
---\begin{enumerate}
\item (a) Find the first 4 terms, in ascending powers of $x$, of the binomial expansion of
\end{enumerate}
$$\left( \frac { 1 } { 4 } - 5 x \right) ^ { \frac { 1 } { 2 } } \quad | x | < \frac { 1 } { 20 }$$
giving each coefficient in its simplest form.
By substituting $x = \frac { 1 } { 100 }$ into the answer for (a),\\
(b) find an approximation for $\sqrt { 5 }$
Give your answer in the form $\frac { a } { b }$ where $a$ and $b$ are integers to be found.
\hfill \mbox{\textit{Edexcel P4 2021 Q1 [7]}}