| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Finding unknown power and constant |
| Difficulty | Standard +0.3 This is a straightforward application of the binomial expansion requiring students to equate coefficients to find n and a, then substitute values. The algebra is routine (solving an + n(n-1)a²/2 = -24 and 270), and part (ii) is a standard numerical application. Slightly above average difficulty due to the two-part algebraic setup, but still a textbook exercise with no novel insight required. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<1 |
| Answer | Marks |
|---|---|
| \((1+ax)^n = 1 + n(ax) + \frac{n(n-1)}{2}(ax)^2 + \ldots\) | B2 |
| \(\therefore\ an = -24 \quad (1)\) and \(\frac{1}{2}a^2n(n-1) = 270 \quad (2)\) | M1 |
| \((1) \Rightarrow a = \frac{-24}{n}\) | |
| sub. \((2)\): \(\frac{288}{n}(n-1) = 270\) | M1 |
| \(288n - 288 = 270n\), \(18n = 288\) | M1 |
| \(n = \frac{288}{18} = 16\), \(a = -\frac{3}{2}\) | A2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(1 - \frac{3}{2}x = 0.9985\ \therefore\ x = 0.001\) | B1 | |
| \(\therefore\ (0.9985)^{16} \approx 1 - 0.024 + 0.000\,270 = 0.97627\) (5dp) | M1 A1 | (10) |
# Question 8:
## Part (i):
$(1+ax)^n = 1 + n(ax) + \frac{n(n-1)}{2}(ax)^2 + \ldots$ | B2 |
$\therefore\ an = -24 \quad (1)$ and $\frac{1}{2}a^2n(n-1) = 270 \quad (2)$ | M1 |
$(1) \Rightarrow a = \frac{-24}{n}$ | |
sub. $(2)$: $\frac{288}{n}(n-1) = 270$ | M1 |
$288n - 288 = 270n$, $18n = 288$ | M1 |
$n = \frac{288}{18} = 16$, $a = -\frac{3}{2}$ | A2 |
## Part (ii):
$1 - \frac{3}{2}x = 0.9985\ \therefore\ x = 0.001$ | B1 |
$\therefore\ (0.9985)^{16} \approx 1 - 0.024 + 0.000\,270 = 0.97627$ (5dp) | M1 A1 | **(10)**
---8. Given that for small values of $x$
$$( 1 + a x ) ^ { n } \approx 1 - 24 x + 270 x ^ { 2 }$$
where $n$ is an integer and $n > 1$,\\
(i) show that $n = 16$ and find the value of $a$,\\
(ii) use your value of $a$ and a suitable value of $x$ to estimate the value of (0.9985) ${ } ^ { 16 }$, giving your answer to 5 decimal places.\\
\hfill \mbox{\textit{OCR C2 Q8 [10]}}