| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2020 |
| Session | October |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Single coefficient given directly |
| Difficulty | Moderate -0.5 This is a straightforward binomial theorem question requiring students to write the general term, identify the coefficient of x^4, set it equal to 15120, and solve for a. It's slightly easier than average as it's a direct application of the binomial formula with no additional complications, though it does require careful algebraic manipulation. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(^7C_4 a^3(2x)^4\) | M1 | Attempt at correct coefficient of \(x^4\); must have correct binomial coefficient, correct power of \(a\), and \(2\) or \(2^4\) |
| \(\frac{7!}{4!3!}a^3 \times 2^4 = 15120 \Rightarrow a = \ldots\) | dM1 | For "560"\(a^3 = 15120 \Rightarrow a=\ldots\); must attempt cube root of \(\frac{15120}{"560"}\); depends on first mark |
| \(a = 3\) | A1 | \(a=3\) only; \(\pm 3\) scores A0 |
# Question 4 (Binomial Expansion):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $^7C_4 a^3(2x)^4$ | M1 | Attempt at correct coefficient of $x^4$; must have correct binomial coefficient, correct power of $a$, and $2$ or $2^4$ |
| $\frac{7!}{4!3!}a^3 \times 2^4 = 15120 \Rightarrow a = \ldots$ | dM1 | For "560"$a^3 = 15120 \Rightarrow a=\ldots$; must attempt cube root of $\frac{15120}{"560"}$; depends on first mark |
| $a = 3$ | A1 | $a=3$ only; $\pm 3$ scores A0 |
---\begin{enumerate}
\item In the binomial expansion of\\
$( a + 2 x ) ^ { 7 } \quad$ where $a$ is a constant\\
the coefficient of $x ^ { 4 }$ is 15120\\
Find the value of $a$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 2 2020 Q4 [3]}}