| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Find constants from coefficient conditions on terms |
| Difficulty | Standard +0.8 This is a multi-part binomial expansion question requiring algebraic manipulation of binomial coefficients, forming and solving simultaneous equations in two variables, and careful arithmetic. While the techniques are C2-standard, the problem-solving aspect (setting up coefficient equations, manipulating binomial coefficient expressions, solving the system) and the extended nature (11 marks total) place it above average difficulty. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(x) = \ldots + \left(\begin{array}{c}n\\2\end{array}\right)\frac{x^2}{k^2} + \left(\begin{array}{c}n\\3\end{array}\right)\frac{x^3}{k^3} + \ldots\) | M1 | |
| \(2 \times \frac{n(n-1)}{2k^2} = \frac{n(n-1)(n-2)}{6k^3}\) | M1 | |
| \(\Rightarrow 6k = n - 2\) or \(n = 6k + 2\) | * | A1 cso |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{n(n-1)(n-2)(n-3)}{4! \cdot k^4} = \frac{n(n-1)(n-2)(n-3)(n-4)}{5! \cdot k^5}\), \(\Rightarrow 5k = n - 4\) | (oe) | M1, A1 |
| Solving: \(5k = 6k + 2 - 4\), \(\Rightarrow k = 2\) and \(n = 14\) | * | M1, A1 cso |
| Answer | Marks | Guidance |
|---|---|---|
| \((1 + \frac{x}{2})^{14} = 1 + 7x + \left(\begin{array}{c}14\\2\end{array}\right)\frac{x^2}{4} + \left(\begin{array}{c}14\\3\end{array}\right)\frac{x^3}{8} + \left(\begin{array}{c}14\\4\end{array}\right)\frac{x^4}{16} + \left(\begin{array}{c}14\\5\end{array}\right)\frac{x^5}{32} + \ldots\) | M1 (≥2 correct) | |
| \(= 1 + 7x + \frac{91}{4}x^2 + \frac{91}{2}x^3 + \frac{1001}{16}x^4 + \frac{1001}{16}x^5 + \ldots\) | B1, A1, A1 | (4 marks) |
**Part (a)**
$f(x) = \ldots + \left(\begin{array}{c}n\\2\end{array}\right)\frac{x^2}{k^2} + \left(\begin{array}{c}n\\3\end{array}\right)\frac{x^3}{k^3} + \ldots$ | M1 |
$2 \times \frac{n(n-1)}{2k^2} = \frac{n(n-1)(n-2)}{6k^3}$ | M1 |
$\Rightarrow 6k = n - 2$ or $n = 6k + 2$ | * | A1 cso | (3 marks)
**Part (b)**
$\frac{n(n-1)(n-2)(n-3)}{4! \cdot k^4} = \frac{n(n-1)(n-2)(n-3)(n-4)}{5! \cdot k^5}$, $\Rightarrow 5k = n - 4$ | (oe) | M1, A1 |
Solving: $5k = 6k + 2 - 4$, $\Rightarrow k = 2$ and $n = 14$ | * | M1, A1 cso | (4 marks)
**Part (c)**
$(1 + \frac{x}{2})^{14} = 1 + 7x + \left(\begin{array}{c}14\\2\end{array}\right)\frac{x^2}{4} + \left(\begin{array}{c}14\\3\end{array}\right)\frac{x^3}{8} + \left(\begin{array}{c}14\\4\end{array}\right)\frac{x^4}{16} + \left(\begin{array}{c}14\\5\end{array}\right)\frac{x^5}{32} + \ldots$ | M1 (≥2 correct) |
$= 1 + 7x + \frac{91}{4}x^2 + \frac{91}{2}x^3 + \frac{1001}{16}x^4 + \frac{1001}{16}x^5 + \ldots$ | B1, A1, A1 | (4 marks)
**Total for Question 8: 11 marks**
---$$f(x) = \left(1 + \frac{x}{k}\right)^n, \quad k, n \in \mathbb{N}, \quad n > 2.$$
Given that the coefficient of $x^3$ is twice the coefficient of $x^2$ in the binomial expansion of f(x),
\begin{enumerate}[label=(\alph*)]
\item prove that $n = 6k + 2$. [3]
\end{enumerate}
Given also that the coefficients of $x^4$ and $x^5$ are equal and non-zero,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item form another equation in $n$ and $k$ and hence show that $k = 2$ and $n = 14$. [4]
\end{enumerate}
Using these values of $k$ and $n$,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item expand f(x) in ascending powers of $x$, up to and including the term in $x^5$. Give each coefficient as an exact fraction in its lowest terms [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q8 [11]}}