| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2018 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Find constants from coefficient conditions on terms |
| Difficulty | Standard +0.3 This is a straightforward binomial expansion problem requiring students to equate coefficients and solve simultaneous equations. Part (a) is guided, and parts (b)-(c) follow mechanically from standard binomial coefficient formulas. The algebra is routine with no conceptual challenges beyond basic binomial theorem application. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{n(n-1)}{2}k^2 = 126k\) or \(\frac{n(n-1)}{2}k = 126k\) or \(^nC_2k^2 = 126k\) or \(^nC_2k = 126k\) | M1 | Compares \(x^2\) terms using one of these forms, with or without the \(x^2\) |
| \(kn(n-1) = 252\) | A1* | Obtains the printed equation from \(\frac{n(n-1)}{2}k^2 = 126k\) or \(\frac{n(n-1)}{2}k^2x^2 = 126kx^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(nk = 36\) | B1 | Correct equation. Can score anywhere. |
| \(36(n-1) = 252\) or \(36\left(\frac{36}{k}-1\right) = 252\) | M1 | Uses a valid method with \(nk=36\) and given equation to obtain equation in \(n\) or \(k\) only. Must be correct algebraic method allowing sign/arithmetic slips only. |
| \(36n - 36 = 252 \Rightarrow n = 8\) or \(\frac{36}{k}-1 = 7 \Rightarrow k = 4.5\) | dM1A1 | dM1: Solves using correct method to obtain value for \(n\) or \(k\). A1: Correct value for \(n\) or \(k\) |
| \(n=8 \Rightarrow k=4.5\) or \(k=4.5 \Rightarrow n=8\) | A1 | Correct values for both \(n\) and \(k\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{n(n-1)(n-2)}{3!}k^3\left(x^3\right)\) | B1ft | Correct coefficient. May be implied by \(56k^3\) or \(^8C_3\,''k''^3\) with or without \(x^3\). If no working shown, may need to check values. |
| \(= \frac{8(8-1)(8-2)}{3!}4.5^3 = \ldots\) | M1 | Substitutes values correctly including integer \(n\), \(n>3\), to obtain value for coefficient of \(x^3\). Must be correct calculation for \(x^3\) coefficient for their values. |
| \(= 5103\) | A1 | Allow \(5103x^3\) |
## Question 15:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{n(n-1)}{2}k^2 = 126k$ or $\frac{n(n-1)}{2}k = 126k$ or $^nC_2k^2 = 126k$ or $^nC_2k = 126k$ | M1 | Compares $x^2$ terms using one of these forms, with or without the $x^2$ |
| $kn(n-1) = 252$ | A1* | Obtains the printed equation from $\frac{n(n-1)}{2}k^2 = 126k$ or $\frac{n(n-1)}{2}k^2x^2 = 126kx^2$ |
**Total: (2)**
---
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $nk = 36$ | B1 | Correct equation. Can score anywhere. |
| $36(n-1) = 252$ or $36\left(\frac{36}{k}-1\right) = 252$ | M1 | Uses a valid method with $nk=36$ and given equation to obtain equation in $n$ or $k$ only. Must be correct algebraic method allowing sign/arithmetic slips only. |
| $36n - 36 = 252 \Rightarrow n = 8$ or $\frac{36}{k}-1 = 7 \Rightarrow k = 4.5$ | dM1A1 | dM1: Solves using correct method to obtain value for $n$ or $k$. A1: Correct value for $n$ or $k$ |
| $n=8 \Rightarrow k=4.5$ or $k=4.5 \Rightarrow n=8$ | A1 | Correct values for both $n$ and $k$ |
**Note:** Special case where second term is $nx$ giving $n=36$, then solving $kn(n-1)=252$ to give $k=0.2$ scores special case of B1. Generally candidates must be solving 2 equations in $n$ and $k$ for method marks.
**Total: (5)**
---
### Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{n(n-1)(n-2)}{3!}k^3\left(x^3\right)$ | B1ft | Correct coefficient. May be implied by $56k^3$ or $^8C_3\,''k''^3$ with or without $x^3$. If no working shown, may need to check values. |
| $= \frac{8(8-1)(8-2)}{3!}4.5^3 = \ldots$ | M1 | Substitutes values correctly including integer $n$, $n>3$, to obtain value for coefficient of $x^3$. Must be correct calculation for $x^3$ coefficient for their values. |
| $= 5103$ | A1 | Allow $5103x^3$ |
**Note:** Answer only of 5103 scores B1M1A1
**Total: (3)**
**Question Total: 10**15. The binomial expansion, in ascending powers of $x$, of $( 1 + k x ) ^ { n }$ is
$$1 + 36 x + 126 k x ^ { 2 } + \ldots$$
where $k$ is a non-zero constant and $n$ is a positive integer.
\begin{enumerate}[label=(\alph*)]
\item Show that $n k ( n - 1 ) = 252$
\item Find the value of $k$ and the value of $n$.
\item Using the values of $k$ and $n$, find the coefficient of $x ^ { 3 }$ in the binomial expansion of $( 1 + k x ) ^ { n }$\\
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2018 Q15 [10]}}