| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2017 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Standard binomial expansion |
| Difficulty | Easy -1.2 This is a straightforward application of the binomial theorem with n=5, requiring only substitution into the formula and simplification of coefficients. It's a routine C2 question testing basic recall and arithmetic with no problem-solving element, making it easier than average but not trivial due to the fractional coefficient requiring careful calculation. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(3-\frac{1}{3}x\right)^{5}\) - Binomial expansion | B1 | First term of 243 |
| \(3^{5}+^{5}C_{1}3^{4}\left(-\frac{1}{3}x\right)+^{5}C_{2}3^{3}\left(-\frac{1}{3}x\right)^{2}+^{5}C_{3}3^{2}\left(-\frac{1}{3}x\right)^{3}\ldots\) | M1 | Two of the three binomial coefficients must be correct and must be with the correct power of \(x\) |
| \(\left(^{5}C_{1}\times...x\right)+\left(^{5}C_{2}\times...x^{2}\right)\ldots\) | M1 | |
| \(=(243\ldots)-\frac{405}{3}x+\frac{270}{9}x^{2}-\frac{90}{27}x^{3}\ldots\) | A1 | Two of the final three terms correct – may be unsimplified |
| \(=(243\ldots)-135x+30x^{2}-\frac{10}{3}x^{3}\ldots\) | A1 | All three final terms correct and simplified |
| Alternative method: \(\left(3-\frac{1}{3}x\right)^{5}=3^{5}(1-\frac{1}{9}x)^{5}\) | Scheme applied exactly as before |
| $\left(3-\frac{1}{3}x\right)^{5}$ - Binomial expansion | B1 | First term of 243 |
|---|---|---|
| $3^{5}+^{5}C_{1}3^{4}\left(-\frac{1}{3}x\right)+^{5}C_{2}3^{3}\left(-\frac{1}{3}x\right)^{2}+^{5}C_{3}3^{2}\left(-\frac{1}{3}x\right)^{3}\ldots$ | M1 | Two of the three binomial coefficients must be correct and must be with the correct power of $x$ |
| $\left(^{5}C_{1}\times...x\right)+\left(^{5}C_{2}\times...x^{2}\right)\ldots$ | M1 | |
| $=(243\ldots)-\frac{405}{3}x+\frac{270}{9}x^{2}-\frac{90}{27}x^{3}\ldots$ | A1 | Two of the final three terms correct – may be unsimplified |
| $=(243\ldots)-135x+30x^{2}-\frac{10}{3}x^{3}\ldots$ | A1 | All three final terms correct and simplified |
| **Alternative method:** $\left(3-\frac{1}{3}x\right)^{5}=3^{5}(1-\frac{1}{9}x)^{5}$ | | Scheme applied exactly as before |\begin{enumerate}
\item Find the first 4 terms, in ascending powers of $x$, of the binomial expansion of
\end{enumerate}
$$\left( 3 - \frac { 1 } { 3 } x \right) ^ { 5 }$$
giving each term in its simplest form.\\
\includegraphics[max width=\textwidth, alt={}, center]{752efc6c-8d0e-46a6-b75d-5125956969d8-03_104_107_2631_1774}\\
\hfill \mbox{\textit{Edexcel C2 2017 Q1 [4]}}