| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2014 |
| Session | January |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Expansion up to x^2 term |
| Difficulty | Easy -1.2 This is a straightforward application of the binomial theorem requiring only substitution into the formula and basic arithmetic simplification. It's a standard C2 question with no problem-solving element—students simply expand using (a+b)^n and simplify three terms, making it easier than average. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\left(2-\frac{x}{2}\right)^6 = 2^6 + \binom{6}{1}2^5\cdot\left(-\frac{x}{2}\right) + \binom{6}{2}2^4\cdot\left(-\frac{x}{2}\right)^2 + \ldots\) | M1 | Attempt at Binomial to get second and/or third term — needs correct binomial coefficient combined with correct power of \(x\). Ignore bracket errors or sign errors. Accept \(^6C_1\), \(^6C_2\), or 6 and 15 from Pascal's triangle. Mark may be given if no working shown but either/both terms including \(x\) are correct. |
| \(= 64\) | B1 | Must be simplified to 64 (writing \(2^6\) alone is B0). Must be the only constant term. |
| \(-96x\) | A1 | cao. The \(x\) is required. Allow \(+(-96x)\) |
| \(+60x^2 + \ldots\) | A1 | cao. Can follow omission of negative sign in working. Any extra terms in higher powers of \(x\) should be ignored. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\left[2^6\right]\!\left(1-\frac{x}{4}\right)^6 = \left[2^6\right]\!\left(1+\binom{6}{1}\!\left(-\frac{x}{4}\right)+\binom{6}{2}\!\left(\frac{-x}{4}\right)^2+\ldots\right)\) | M1 | Does not require power of 2 to be accurate |
| \(= 64, -96x, +60x^2 + \ldots\) | B1, A1, A1 | If answer left as \(64\!\left(1+\binom{6}{1}\!\left(-\frac{x}{4}\right)+\binom{6}{2}\!\left(\frac{-x}{4}\right)^2+\ldots\right)\) allow M1 B1 A0 A0 |
## Question 1:
**Main Method:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left(2-\frac{x}{2}\right)^6 = 2^6 + \binom{6}{1}2^5\cdot\left(-\frac{x}{2}\right) + \binom{6}{2}2^4\cdot\left(-\frac{x}{2}\right)^2 + \ldots$ | M1 | Attempt at Binomial to get second **and/or** third term — needs correct binomial coefficient combined with correct power of $x$. Ignore bracket errors or sign errors. Accept $^6C_1$, $^6C_2$, or 6 and 15 from Pascal's triangle. Mark may be given if no working shown but either/both terms including $x$ are correct. |
| $= 64$ | B1 | Must be simplified to 64 (writing $2^6$ alone is **B0**). Must be the only constant term. |
| $-96x$ | A1 | cao. The $x$ is required. Allow $+(-96x)$ |
| $+60x^2 + \ldots$ | A1 | cao. Can follow omission of negative sign in working. Any extra terms in higher powers of $x$ should be ignored. |
**Special case:** $= 64, -192\!\left(\frac{x}{2}\right), +240\!\left(\frac{x}{2}\right)^2 + \ldots$ This is correct but unsimplified — award M1B1A0A0
**Alternative Method:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left[2^6\right]\!\left(1-\frac{x}{4}\right)^6 = \left[2^6\right]\!\left(1+\binom{6}{1}\!\left(-\frac{x}{4}\right)+\binom{6}{2}\!\left(\frac{-x}{4}\right)^2+\ldots\right)$ | M1 | Does not require power of 2 to be accurate |
| $= 64, -96x, +60x^2 + \ldots$ | B1, A1, A1 | If answer left as $64\!\left(1+\binom{6}{1}\!\left(-\frac{x}{4}\right)+\binom{6}{2}\!\left(\frac{-x}{4}\right)^2+\ldots\right)$ allow M1 B1 A0 A0 |
**Total: [4 marks]**\begin{enumerate}
\item Find the first 3 terms in ascending powers of $x$ of
\end{enumerate}
$$\left( 2 - \frac { x } { 2 } \right) ^ { 6 }$$
giving each term in its simplest form.\\
\hfill \mbox{\textit{Edexcel C12 2014 Q1 [4]}}