| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2019 |
| Session | October |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Product with reciprocal term binomial |
| Difficulty | Moderate -0.3 Part (a) is straightforward application of binomial theorem with positive integer power requiring routine calculation of coefficients. Part (b) requires identifying which terms from the expansion combine to give x^0, involving algebraic manipulation but following a standard method. This is slightly easier than average due to the mechanical nature of both parts, though part (b) requires some care in tracking powers of x. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\left(1+\frac{x}{4}\right)^{12} = 1+12\left(\frac{x}{4}\right)^1 + \frac{12\times11}{2}\left(\frac{x}{4}\right)^2 + \frac{12\times11\times10}{3!}\left(\frac{x}{4}\right)^3+...\) | M1 | Correct attempt at binomial expansion; accept sight of \(^{12}C_2\left(\frac{x}{4}\right)^2\) or \(^{12}C_3\left(\frac{x}{4}\right)^3\) condoning omission of brackets. FYI \(^{12}C_2=66\), \(^{12}C_3=220\) |
| \(= 1+3x+\frac{33}{8}x^2+\frac{55}{16}x^3\) | B1, A1 | B1 for \(1+3x\) must be simplified; A1 for \(+\frac{33}{8}x^2+\frac{55}{16}x^3\), accept \(+4.125x^2+3.4375x^3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Sight of term independent of \(x\): \("\frac{55}{16}"\) or \(8\times{}^{12}C_5\left(\frac{1}{4}\right)^5\left(=\frac{99}{16}\right)\) | B1ft | Must be independent of \(x\); most candidates do not find the additional term so typically score max 1/3 in (b) |
| \("\frac{55}{16}"+8\times{}^{12}C_5\left(\frac{1}{4}\right)^5 = \frac{55}{16}+\frac{99}{16}=\frac{77}{8}\) | M1, A1 | M1 for attempting to add \("\frac{55}{16}"\) to \(8\times{}^{12}C_5\left(\frac{1}{4}\right)^5\); A1 for \(\frac{77}{8}\) oe e.g. 9.625 |
## Question 3:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left(1+\frac{x}{4}\right)^{12} = 1+12\left(\frac{x}{4}\right)^1 + \frac{12\times11}{2}\left(\frac{x}{4}\right)^2 + \frac{12\times11\times10}{3!}\left(\frac{x}{4}\right)^3+...$ | M1 | Correct attempt at binomial expansion; accept sight of $^{12}C_2\left(\frac{x}{4}\right)^2$ or $^{12}C_3\left(\frac{x}{4}\right)^3$ condoning omission of brackets. FYI $^{12}C_2=66$, $^{12}C_3=220$ |
| $= 1+3x+\frac{33}{8}x^2+\frac{55}{16}x^3$ | B1, A1 | B1 for $1+3x$ must be simplified; A1 for $+\frac{33}{8}x^2+\frac{55}{16}x^3$, accept $+4.125x^2+3.4375x^3$ |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Sight of term independent of $x$: $"\frac{55}{16}"$ or $8\times{}^{12}C_5\left(\frac{1}{4}\right)^5\left(=\frac{99}{16}\right)$ | B1ft | Must be independent of $x$; most candidates do not find the additional term so typically score max 1/3 in (b) |
| $"\frac{55}{16}"+8\times{}^{12}C_5\left(\frac{1}{4}\right)^5 = \frac{55}{16}+\frac{99}{16}=\frac{77}{8}$ | M1, A1 | M1 for attempting to add $"\frac{55}{16}"$ to $8\times{}^{12}C_5\left(\frac{1}{4}\right)^5$; A1 for $\frac{77}{8}$ oe e.g. 9.625 |
---3.\\
\begin{enumerate}[label=(\alph*)]
\item Find the first 4 terms, in ascending powers of $x$, in the binomial expansion of
$$\left( 1 + \frac { x } { 4 } \right) ^ { 12 }$$
giving each coefficient in its simplest form.
\item Find the term independent of $x$ in the expansion of
$$\left( \frac { x ^ { 2 } + 8 } { x ^ { 5 } } \right) \left( 1 + \frac { x } { 4 } \right) ^ { 12 }$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel P2 2019 Q3 [6]}}