| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2022 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Factoring out constants before expansion |
| Difficulty | Moderate -0.3 This is a straightforward application of the binomial expansion requiring factoring out constants to get the form (1+x)^n, then a simple substitution. The algebraic manipulation is routine for P4 students, and part (b) is a standard follow-up requiring only substitution and calculator work. Slightly easier than average due to its mechanical nature with no conceptual surprises. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Obtains \(\sqrt{9-2x} = 3\sqrt{1-...}\) | B1 | |
| \(\left(1-\frac{2}{9}x\right)^{-\frac{1}{2}} = 1+\left(-\frac{1}{2}\right)\left(-\frac{2}{9}x\right)+\frac{-\frac{1}{2}\left(-\frac{1}{2}-1\right)}{2!}\left(-\frac{2}{9}x\right)^2+\frac{-\frac{1}{2}\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right)}{3!}\left(-\frac{2}{9}x\right)^3+...\) | M1 A1 | M1: Attempts binomial expansion of \((1+kx)^n\) to get third and/or fourth term with acceptable structure. Correct binomial coefficient combined with correct power of \(x\) and correct power of 2. A1: Correct simplified or unsimplified expansion |
| \(\frac{2}{\sqrt{9-2x}} = \frac{2}{3}+\frac{2}{27}x+\frac{1}{81}x^2+\frac{5}{2187}x^3+...\) | A1 | 2 correct simplified terms |
| All correct | A1 | NB simplified is \(= 1+\frac{1}{9}x+\frac{1}{54}x^2+\frac{5}{1458}x^3+...\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x=1 \Rightarrow \frac{2}{\sqrt{7}} \approx \frac{2}{3}+\frac{2}{27}+\frac{1}{81}+\frac{5}{2187}+...\) \(\Rightarrow \sqrt{7} \approx 2 \div \text{"}\frac{1652}{2187}\text{"}\) or \(2\times\text{"}\frac{2187}{1652}\text{"}\) | M1 | Substitutes \(x=1\) and divides into 2 or equivalent |
| \(= 2.6477\) | A1 | Correct approximation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{2}{\sqrt{7}} = \frac{2\sqrt{7}}{7} \Rightarrow \sqrt{7} \approx \frac{7}{2}\times\text{"}\frac{1652}{2187}\text{"}\) | M1 | Substitutes \(x=1\) and multiplies by \(\frac{7}{2}\) |
| \(= 2.6438\) | A1 | Correct approximation |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Obtains $\sqrt{9-2x} = 3\sqrt{1-...}$ | B1 | |
| $\left(1-\frac{2}{9}x\right)^{-\frac{1}{2}} = 1+\left(-\frac{1}{2}\right)\left(-\frac{2}{9}x\right)+\frac{-\frac{1}{2}\left(-\frac{1}{2}-1\right)}{2!}\left(-\frac{2}{9}x\right)^2+\frac{-\frac{1}{2}\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right)}{3!}\left(-\frac{2}{9}x\right)^3+...$ | M1 A1 | M1: Attempts binomial expansion of $(1+kx)^n$ to get third and/or fourth term with acceptable structure. Correct binomial coefficient combined with correct power of $x$ and correct power of 2. A1: Correct simplified or unsimplified expansion |
| $\frac{2}{\sqrt{9-2x}} = \frac{2}{3}+\frac{2}{27}x+\frac{1}{81}x^2+\frac{5}{2187}x^3+...$ | A1 | 2 correct simplified terms |
| All correct | A1 | NB simplified is $= 1+\frac{1}{9}x+\frac{1}{54}x^2+\frac{5}{1458}x^3+...$ |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=1 \Rightarrow \frac{2}{\sqrt{7}} \approx \frac{2}{3}+\frac{2}{27}+\frac{1}{81}+\frac{5}{2187}+...$ $\Rightarrow \sqrt{7} \approx 2 \div \text{"}\frac{1652}{2187}\text{"}$ or $2\times\text{"}\frac{2187}{1652}\text{"}$ | M1 | Substitutes $x=1$ and divides into 2 or equivalent |
| $= 2.6477$ | A1 | Correct approximation |
**Alternative for (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{2}{\sqrt{7}} = \frac{2\sqrt{7}}{7} \Rightarrow \sqrt{7} \approx \frac{7}{2}\times\text{"}\frac{1652}{2187}\text{"}$ | M1 | Substitutes $x=1$ and multiplies by $\frac{7}{2}$ |
| $= 2.6438$ | A1 | Correct approximation |
---\begin{enumerate}
\item (a) Find the first 4 terms of the binomial expansion, in ascending powers of $x$, of
\end{enumerate}
$$\frac { 2 } { \sqrt { 9 - 2 x } } \quad | x | < \frac { 9 } { 2 }$$
giving each coefficient as a simplified fraction.
By substituting $x = 1$ into the answer to part (a),\\
(b) find an approximation for $\sqrt { 7 }$, giving your answer to 4 decimal places.\\
\hfill \mbox{\textit{Edexcel P4 2022 Q1 [7]}}