| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Find constants from coefficient conditions on terms |
| Difficulty | Moderate -0.3 This is a straightforward application of binomial expansion formulas where students equate coefficients of x and x² to form simultaneous equations. The algebra is routine (solving nb=12 and n(n-1)b²/2=70), requiring only substitution and factorization. Slightly easier than average as it's a standard textbook exercise with clear methodology. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((1+bx)^n = 1 + nbx + \frac{n(n-1)(bx)^2}{2!} = 1+12x+(70x^2)\); comparing \(x\) terms: \(nb=12\) | B1* | Must show working; cannot just state \(nb=12\); accept \({}^nC_1 1^{n-1}(bx)=12x \Rightarrow nb=12\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Comparing \(x^2\) terms: \(\frac{n(n-1)b^2}{2} = 70 \Rightarrow n(n-1)b^2=140\) | M1 A1 | M1 for setting \(x^2\) term equal to 70; A1 for correct equation in \(b\) and \(n\) |
| Substitutes \(b=\frac{12}{n}\) into \(n(n-1)b^2=140 \Rightarrow \frac{n(n-1)144}{n^2}=140\) | M1 | Substitutes \(b=\frac{12}{n}\) or \(n=\frac{12}{b}\) or \(nb=12\) to get single variable equation |
| \(\Rightarrow 144(n-1)=140n \Rightarrow n=36\) | dM1 A1 | dM1 dependent on both previous M's; A1 correct \(n=36\) or \(b=\frac{1}{3}\) (cso) |
| Substitutes \(n=36\) into \(nb=12 \Rightarrow b=\frac{1}{3}\) | A1 | Both \(n=36\) and \(b=\frac{1}{3}\) with no other solutions; cso |
# Question 8(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(1+bx)^n = 1 + nbx + \frac{n(n-1)(bx)^2}{2!} = 1+12x+(70x^2)$; comparing $x$ terms: $nb=12$ | B1* | Must show working; cannot just state $nb=12$; accept ${}^nC_1 1^{n-1}(bx)=12x \Rightarrow nb=12$ |
# Question 8(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Comparing $x^2$ terms: $\frac{n(n-1)b^2}{2} = 70 \Rightarrow n(n-1)b^2=140$ | M1 A1 | M1 for setting $x^2$ term equal to 70; A1 for correct equation in $b$ and $n$ |
| Substitutes $b=\frac{12}{n}$ into $n(n-1)b^2=140 \Rightarrow \frac{n(n-1)144}{n^2}=140$ | M1 | Substitutes $b=\frac{12}{n}$ or $n=\frac{12}{b}$ or $nb=12$ to get single variable equation |
| $\Rightarrow 144(n-1)=140n \Rightarrow n=36$ | dM1 A1 | dM1 dependent on both previous M's; A1 correct $n=36$ or $b=\frac{1}{3}$ (cso) |
| Substitutes $n=36$ into $nb=12 \Rightarrow b=\frac{1}{3}$ | A1 | Both $n=36$ **and** $b=\frac{1}{3}$ with no other solutions; cso |
---8. Given that
$$1 + 12 x + 70 x ^ { 2 } + \ldots$$
is the binomial expansion, in ascending powers of $x$ of $( 1 + b x ) ^ { n }$, where $n \in \mathbb { N }$ and $b$ is a constant,
\begin{enumerate}[label=(\alph*)]
\item show that $n b = 12$
\item find the values of the constants $b$ and $n$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2014 Q8 [7]}}