| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Ratio of coefficients condition |
| Difficulty | Moderate -0.3 This is a straightforward binomial expansion question requiring standard application of the formula. Part (a) involves routine substitution into C(n,r) terms with simple arithmetic (powers of 1/4). Part (b) requires equating two coefficients and solving a linear equation for n. While it has two parts and requires careful algebraic manipulation, it's a standard textbook exercise with no conceptual challenges beyond basic binomial theorem application, making it slightly easier than average. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(1 = 1 + n\left(\frac{1}{3}x\right) + \frac{n(n-1)}{2}\left(\frac{1}{3}x\right)^2 + \ldots\) | B1 M1 | |
| \(= 1 + \frac{1}{3}nx + \frac{1}{32}n(n-1)x^2 + \ldots\) | A1 | |
| (b) \(\frac{4}{3}n = \frac{1}{32}n(n-1)\) | M1 | |
| \(8n = n(n-1)\) | M1 | |
| \(n[8 - (n-1)] = 0\) | M1 | |
| \(n \neq 0 \quad \therefore n = 9\) | A1 | (6 marks) |
**(a)** $1 = 1 + n\left(\frac{1}{3}x\right) + \frac{n(n-1)}{2}\left(\frac{1}{3}x\right)^2 + \ldots$ | B1 M1 |
$= 1 + \frac{1}{3}nx + \frac{1}{32}n(n-1)x^2 + \ldots$ | A1 |
**(b)** $\frac{4}{3}n = \frac{1}{32}n(n-1)$ | M1 |
$8n = n(n-1)$ | M1 |
$n[8 - (n-1)] = 0$ | M1 |
$n \neq 0 \quad \therefore n = 9$ | A1 | **(6 marks)**3. For the binomial expansion in ascending powers of $x$ of $\left( 1 + \frac { 1 } { 4 } x \right) ^ { n }$, where $n$ is an integer and $n \geq 2$,
\begin{enumerate}[label=(\alph*)]
\item find and simplify the first three terms,
\item find the value of $n$ for which the coefficient of $x$ is equal to the coefficient of $x ^ { 2 }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q3 [6]}}