| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Single coefficient given directly |
| Difficulty | Moderate -0.8 This is a straightforward application of the binomial theorem requiring students to write out the first four terms using the formula, then solve a simple equation for k. The only steps are: apply (a+b)^n formula, simplify coefficients with powers of 2, then solve 35×8×k³=1890. No conceptual difficulty or novel insight required—pure routine calculation. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Answer/Working | Marks |
| (a) | \((2+kx)^7\) | |
| \(2^7 + ^7C_1 2^6(kx) + ^7C_2 2^5(kx)^2 + ^7C_3 2^4(kx)^3...\) | B1 M1 | B1: constant term should be 128 in expansion (should not be followed by other constant terms). M1: Two of three binomial coefficients must be correct and must be with correct power of \(x\). Accept \(^7C_1\) or \(\binom{7}{1}\) or 7 as coefficient, and \(^7C_2\) or \(\binom{7}{2}\) or 21 as another and \(^7C_3\) or \(\binom{7}{3}\) or 35 as another. Pascal's triangle may be used to establish coefficients. |
| First term of 128 | ||
| \((\hspace{0.5cm}^7C_1 x...x\times) + (\hspace{0.5cm}^7C_2 x...x\times^2) + (\hspace{0.5cm}^7C_3 x...x\times^3)...\) | ||
| \(=(128..) + 448kx + 672k^2x^2 + 560k^3x^3..\) | A1, A1 | A1: Two of final three terms correct (i.e. two of \(448kx + 672k^2x^2 + 560k^3x^3\)). A1: All three final terms correct. (Accept answers without + signs, can be listed with commas or appear on separate lines) |
| (4) | ||
| (b) | \(560k^3 = 1890\) | M1 |
| \(k^3 = \frac{1890}{560}\) so \(k =\) | dM1 | dM1: Divides then attempts cube root of their answer to give \(k\) – intention must be clear. (You may need to check on calculator) Correct answer implies this mark. |
| \(k = 1.5\) o.e. | A1 | A1: Any equivalent to 1.5. If they give \(-1.5\) as second answer this is A0 |
| (3) | ||
| (7 marks) |
| **Part** | **Answer/Working** | **Marks** | **Notes** |
|---|---|---|---|
| (a) | $(2+kx)^7$ | | |
| | $2^7 + ^7C_1 2^6(kx) + ^7C_2 2^5(kx)^2 + ^7C_3 2^4(kx)^3...$ | B1 M1 | B1: constant term should be 128 in expansion (should not be followed by other constant terms). M1: Two of three binomial coefficients must be correct and must be with correct power of $x$. Accept $^7C_1$ or $\binom{7}{1}$ or 7 as coefficient, and $^7C_2$ or $\binom{7}{2}$ or 21 as another and $^7C_3$ or $\binom{7}{3}$ or 35 as another. Pascal's triangle may be used to establish coefficients. |
| | First term of 128 | | |
| | $(\hspace{0.5cm}^7C_1 x...x\times) + (\hspace{0.5cm}^7C_2 x...x\times^2) + (\hspace{0.5cm}^7C_3 x...x\times^3)...$ | | |
| | $=(128..) + 448kx + 672k^2x^2 + 560k^3x^3..$ | A1, A1 | A1: Two of final three terms correct (i.e. two of $448kx + 672k^2x^2 + 560k^3x^3$). A1: All three final terms correct. (Accept answers without + signs, can be listed with commas or appear on separate lines) |
| | | **(4)** | |
| (b) | $560k^3 = 1890$ | M1 | M1: Sets their coefficient of $x^3$ equal to 1890. They should have an equation which does not include power of $x$. This mark may be recovered if they continue on to get $k=1.5$ |
| | $k^3 = \frac{1890}{560}$ so $k =$ | dM1 | dM1: Divides then attempts cube root of their answer to give $k$ – intention must be clear. (You may need to check on calculator) Correct answer implies this mark. |
| | $k = 1.5$ o.e. | A1 | A1: Any equivalent to 1.5. If they give $-1.5$ as second answer this is A0 |
| | | **(3)** | |
| | | **(7 marks)** | |
**Alternative method for (a):** $(2+kx)^7 = 2^7(1 + \frac{kx}{2})^7$; $2^7(1 + ^7C_1(\frac{x}{2}) + ^7C_2(\frac{x}{2})^2 + ^7C_3(\frac{x}{2})^3 ...)$. Scheme is applied exactly as before.
**Special case using Alternative Method:** Uses $2(1+\frac{kx}{2})^7$ is likely to result in maximum mark of B0M1A0A0 then M1M1A0; then M1M1A0.
If correct expansion seen award marks and isw.
**Notes on (b):**
M1: Sets their coefficient of $x^3$ equal to 1890. They should have equation which does not include power of $x$. This mark may be recovered if continue to get $k=1.5$.
dM1: Dependent on previous M mark. Divides then attempts cube root of their answer to give $k$ – intention must be clear. (You may need to check calculator) Correct answer implies this mark.
A1: Any equivalent to 1.5. If they give $-1.5$ as second answer this is A0.
---\begin{enumerate}
\item (a) Find the first 4 terms, in ascending powers of $x$, of the binomial expansion of
\end{enumerate}
$$( 2 + k x ) ^ { 7 }$$
where $k$ is a non-zero constant. Give each term in its simplest form.
Given that the coefficient of $x ^ { 3 }$ in this expansion is 1890\\
(b) find the value of $k$.\\
\hfill \mbox{\textit{Edexcel C2 2018 Q2 [7]}}