| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Expansion with algebraic manipulation |
| Difficulty | Moderate -0.3 This is a straightforward application of the binomial expansion formula for negative/fractional powers. Part (a) requires direct substitution into the formula with careful arithmetic. Part (b) is trivial index law manipulation (n - 3/2 = 1/2). Part (c) either repeats the expansion process or uses multiplication of series from part (a). While it requires accuracy with fractions and negative exponents, it involves no problem-solving insight—just methodical application of a standard P4 technique, making it slightly easier than average. |
| Spec | 1.02a Indices: laws of indices for rational exponents1.04c Extend binomial expansion: rational n, |x|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left(\frac{1}{4}-\frac{1}{2}x\right)^{-\frac{3}{2}} = 8(1-2x)^{-\frac{3}{2}}\) | B1 | Takes out correct and simplified common factor to obtain \(8(1\pm...)^{-\frac{3}{2}}\); implied by expansion \(8+...\) |
| \((1-2x)^{-\frac{3}{2}} = 1+\left(-\frac{3}{2}\right)(-2x)+\frac{-\frac{3}{2}\left(-\frac{3}{2}-1\right)}{2!}(-2x)^2+\frac{-\frac{3}{2}\left(-\frac{3}{2}-1\right)\left(-\frac{3}{2}-2\right)}{3!}(-2x)^3+...\) | M1 A1 | Attempts binomial expansion of \((1\pm kx)^{-\frac{3}{2}}\) to get third and/or fourth term with acceptable structure; correct binomial coefficient combined with correct power of \(x\); look for \(\frac{-\frac{3}{2}\left(-\frac{3}{2}-1\right)}{2!}(\pm kx)^2\) or \(\frac{-\frac{3}{2}\left(-\frac{3}{2}-1\right)\left(-\frac{3}{2}-2\right)}{3!}(\pm kx)^3\); even allow \(k=1\) |
| \(\left(\frac{1}{4}-\frac{1}{2}x\right)^{-\frac{3}{2}} = 8+24x+60x^2+140x^3+...\) | A1, A1 | A1: correct simplified expansion for \((1-2x)^{-\frac{3}{2}}\) (simplified is \(1+3x+\frac{15}{2}x^2+\frac{35}{2}x^3+...\)); A1: two correct simplified terms of \(8+24x+60x^2+140x^3\); A1: all correct and simplified; ignore extra terms of \(x^4\) and above |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(n=2\) | B1 | Correct value; do NOT allow incomplete answers such as \(\frac{4}{2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left(\frac{1}{4}-\frac{1}{2}x\right)^2 = \frac{1}{16}-\frac{1}{4}x+\frac{1}{4}x^2\) | B1 | Correct expansion of \(\left(\frac{1}{4}-\frac{1}{2}x\right)^2\), simplified or unsimplified |
| \(\left(\frac{1}{4}-\frac{1}{2}x\right)^{\frac{1}{2}} = \left(\frac{1}{16}-\frac{1}{4}x+\frac{1}{4}x^2\right)\left(8+24x+60x^2+140x^3+...\right)\) | M1 | Attempts correct strategy; terms do not need to be collected; look for attempt to find 4 or more terms from \((A+Bx+Cx^2)(E+Fx+Gx^2+Hx^3+...)= AE+AFx+AGx^2+BEx+BFx^2+CEx^2+...\) |
| \(= \frac{1}{2}-\frac{1}{2}x-\frac{1}{4}x^2+...\) | A1 | Correct simplified expansion; ignore extra terms of \(x^3\) and above |
# Question 1:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(\frac{1}{4}-\frac{1}{2}x\right)^{-\frac{3}{2}} = 8(1-2x)^{-\frac{3}{2}}$ | B1 | Takes out correct and simplified common factor to obtain $8(1\pm...)^{-\frac{3}{2}}$; implied by expansion $8+...$ |
| $(1-2x)^{-\frac{3}{2}} = 1+\left(-\frac{3}{2}\right)(-2x)+\frac{-\frac{3}{2}\left(-\frac{3}{2}-1\right)}{2!}(-2x)^2+\frac{-\frac{3}{2}\left(-\frac{3}{2}-1\right)\left(-\frac{3}{2}-2\right)}{3!}(-2x)^3+...$ | M1 A1 | Attempts binomial expansion of $(1\pm kx)^{-\frac{3}{2}}$ to get third and/or fourth term with acceptable structure; correct binomial coefficient combined with correct power of $x$; look for $\frac{-\frac{3}{2}\left(-\frac{3}{2}-1\right)}{2!}(\pm kx)^2$ or $\frac{-\frac{3}{2}\left(-\frac{3}{2}-1\right)\left(-\frac{3}{2}-2\right)}{3!}(\pm kx)^3$; even allow $k=1$ |
| $\left(\frac{1}{4}-\frac{1}{2}x\right)^{-\frac{3}{2}} = 8+24x+60x^2+140x^3+...$ | A1, A1 | A1: correct simplified expansion for $(1-2x)^{-\frac{3}{2}}$ (simplified is $1+3x+\frac{15}{2}x^2+\frac{35}{2}x^3+...$); A1: two correct simplified terms of $8+24x+60x^2+140x^3$; A1: all correct and simplified; ignore extra terms of $x^4$ and above |
**Total: 5 marks**
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $n=2$ | B1 | Correct value; do NOT allow incomplete answers such as $\frac{4}{2}$ |
**Total: 1 mark**
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(\frac{1}{4}-\frac{1}{2}x\right)^2 = \frac{1}{16}-\frac{1}{4}x+\frac{1}{4}x^2$ | B1 | Correct expansion of $\left(\frac{1}{4}-\frac{1}{2}x\right)^2$, simplified or unsimplified |
| $\left(\frac{1}{4}-\frac{1}{2}x\right)^{\frac{1}{2}} = \left(\frac{1}{16}-\frac{1}{4}x+\frac{1}{4}x^2\right)\left(8+24x+60x^2+140x^3+...\right)$ | M1 | Attempts correct strategy; terms do not need to be collected; look for attempt to find 4 or more terms from $(A+Bx+Cx^2)(E+Fx+Gx^2+Hx^3+...)= AE+AFx+AGx^2+BEx+BFx^2+CEx^2+...$ |
| $= \frac{1}{2}-\frac{1}{2}x-\frac{1}{4}x^2+...$ | A1 | Correct simplified expansion; ignore extra terms of $x^3$ and above |
**Total: 3 marks**
---\begin{enumerate}
\item (a) Find the first 4 terms of the binomial expansion, in ascending powers of $x$, of
\end{enumerate}
$$\left( \frac { 1 } { 4 } - \frac { 1 } { 2 } x \right) ^ { - \frac { 3 } { 2 } } \quad | x | < \frac { 1 } { 2 }$$
giving each term in simplest form.
Given that
$$\left( \frac { 1 } { 4 } - \frac { 1 } { 2 } x \right) ^ { n } \left( \frac { 1 } { 4 } - \frac { 1 } { 2 } x \right) ^ { - \frac { 3 } { 2 } } = \left( \frac { 1 } { 4 } - \frac { 1 } { 2 } x \right) ^ { \frac { 1 } { 2 } }$$
(b) write down the value of $n$.\\
(c) Hence, or otherwise, find the first 3 terms of the binomial expansion, in ascending powers of $x$, of
$$\left( \frac { 1 } { 4 } - \frac { 1 } { 2 } x \right) ^ { \frac { 1 } { 2 } } \quad | x | < \frac { 1 } { 2 }$$
giving each term in simplest form.
\hfill \mbox{\textit{Edexcel P4 2023 Q1 [9]}}