| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2007 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Sum of first n terms |
| Difficulty | Moderate -0.8 This is a straightforward binomial expansion question requiring routine application of the binomial theorem formula. Part (i) is direct calculation with no problem-solving, and part (ii) involves simple algebraic manipulation to find 'a' by equating coefficients. Both parts are standard textbook exercises with clear methods and minimal steps, making this easier than average for A-level. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \((1 + 4x)^7 = 1 + 28x + 336x^2 + 2240x^3\) | B1, M1, A1, A1 (4 marks) | Obtain \(1 + 28x\); Attempt binomial expansion of at least 1 more term, with each term the product of binomial coeff and power of \(4x\); Obtain \(336x^2\); Obtain \(2240x^3\) |
| (ii) \(28a + 1008 = 1001\) Hence \(a = -\frac{1}{4}\) | M1, A1∇, A1 (3 marks) | Multiply together two relevant pairs of terms; Obtain \(28a + 1008 = 1001\); Obtain \(a = -\frac{1}{4}\) |
**(i)** $(1 + 4x)^7 = 1 + 28x + 336x^2 + 2240x^3$ | B1, M1, A1, A1 (4 marks) | Obtain $1 + 28x$; Attempt binomial expansion of at least 1 more term, with each term the product of binomial coeff and power of $4x$; Obtain $336x^2$; Obtain $2240x^3$
**(ii)** $28a + 1008 = 1001$ Hence $a = -\frac{1}{4}$ | M1, A1∇, A1 (3 marks) | Multiply together two relevant pairs of terms; Obtain $28a + 1008 = 1001$; Obtain $a = -\frac{1}{4}$
**Total: 7 marks**\begin{enumerate}[label=(\roman*)]
\item Find and simplify the first four terms in the expansion of $(1 + 4x)^7$ in ascending powers of $x$. [4]
\item In the expansion of
$$(3 + ax)(1 + 4x)^7,$$
the coefficient of $x^2$ is 1001. Find the value of $a$. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR C2 2007 Q6 [7]}}