| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2024 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Binomial times linear coefficient |
| Difficulty | Moderate -0.3 Part (a) is straightforward application of the binomial theorem with positive integer n=9, requiring routine calculation of four terms. Part (b) requires multiplying the expansion by a linear factor and collecting x³ terms—a standard 'hence' extension that adds one extra step but remains a textbook exercise with no novel insight required. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\left(1-\frac{1}{6}x\right)^9 = 1+9\left(-\frac{1}{6}x\right)^1 + \frac{9\times8}{2}\left(-\frac{1}{6}x\right)^2 + \frac{9\times8\times7}{3!}\left(-\frac{1}{6}x\right)^3+...\) | M1 | Attempt at binomial expansion; score for correct attempt at any of term 2, 3 or 4. Accept sight of \(^9C_1\left(\pm\frac{1}{6}x\right)^1\) or \(^9C_2\left(\pm\frac{1}{6}x\right)^2\) or \(^9C_3\left(\pm\frac{1}{6}x\right)^3\) condoning omission of brackets |
| \(= 1-\frac{3}{2}x+x^2-\frac{7}{18}x^3\) | A1, A1 | A1 for any two simplified correct terms of \(-\frac{3}{2}x+x^2-\frac{7}{18}x^3\); A1 for fully correct expression. Fractions must be simplified. Accept \(1x^2\). ISW following correct expression |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((10x+3)\left(1-\frac{3}{2}x+x^2-\frac{7}{18}x^3\right)\) | ||
| Coefficient of \(x^3 = 10\times1 + 3\times-\frac{7}{18} = \frac{53}{6}\) | M1, A1 | M1 for attempting \(10b\pm3c\) combined to single term for \((10x+3)(1+ax+bx^2+cx^3+)\); A1 cao for \(\frac{53}{6}\), condone \(\frac{53}{6}x^3\). Accept embedded in expansion as long as simplified to correct single \(x^3\) term |
# Question 1:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left(1-\frac{1}{6}x\right)^9 = 1+9\left(-\frac{1}{6}x\right)^1 + \frac{9\times8}{2}\left(-\frac{1}{6}x\right)^2 + \frac{9\times8\times7}{3!}\left(-\frac{1}{6}x\right)^3+...$ | M1 | Attempt at binomial expansion; score for correct attempt at any of term 2, 3 or 4. Accept sight of $^9C_1\left(\pm\frac{1}{6}x\right)^1$ or $^9C_2\left(\pm\frac{1}{6}x\right)^2$ or $^9C_3\left(\pm\frac{1}{6}x\right)^3$ condoning omission of brackets |
| $= 1-\frac{3}{2}x+x^2-\frac{7}{18}x^3$ | A1, A1 | A1 for any two simplified correct terms of $-\frac{3}{2}x+x^2-\frac{7}{18}x^3$; A1 for fully correct expression. Fractions must be simplified. Accept $1x^2$. ISW following correct expression |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(10x+3)\left(1-\frac{3}{2}x+x^2-\frac{7}{18}x^3\right)$ | | |
| Coefficient of $x^3 = 10\times1 + 3\times-\frac{7}{18} = \frac{53}{6}$ | M1, A1 | M1 for attempting $10b\pm3c$ combined to single term for $(10x+3)(1+ax+bx^2+cx^3+)$; A1 cao for $\frac{53}{6}$, condone $\frac{53}{6}x^3$. Accept embedded in expansion as long as simplified to correct single $x^3$ term |
---\begin{enumerate}
\item (a) Find the first four terms, in ascending powers of $x$, of the binomial expansion of
\end{enumerate}
$$\left( 1 - \frac { 1 } { 6 } x \right) ^ { 9 }$$
giving each term in simplest form.\\
(b) Hence find the coefficient of $x ^ { 3 }$ in the expansion of
$$( 10 x + 3 ) \left( 1 - \frac { 1 } { 6 } x \right) ^ { 9 }$$
giving the answer in simplest form.
\hfill \mbox{\textit{Edexcel P2 2024 Q1 [5]}}