| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2018 |
| Session | Specimen |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Series expansion of rational function |
| Difficulty | Moderate -0.3 This is a straightforward application of the binomial series requiring factoring out the constant, applying the formula with n=-3, and simplifying coefficients. It's slightly easier than average because it's a direct textbook-style exercise with clear steps and no problem-solving insight required, though the arithmetic with negative indices and fractions requires care. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions |
| VIIIV SIHI NI JIIIM ION OC | VIIV SIHI NI JAHAM ION OO | VJ4V SIHIL NI JIIIM IONOO |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left | \frac{1}{(2+5x)^3}\right | \) |
# Question 1
## Scheme
$\left|\frac{1}{(2+5x)^3}\right|$
M1
$= (2)^{-3}\left(1+\frac{5x}{2}\right)^{-3}$
$= \frac{1}{8}\left(1+\frac{5x}{2}\right)^{-3}$
B1
$= \frac{1}{8}\left[1+(-3)(kx)+\frac{(-3)(-4)}{2!}(kx)^2+\frac{(-3)(-4)(-5)}{3!}(kx)^3+\ldots\right]$
M1
$= \frac{1}{8}\left[1+(-3)\left(\frac{5x}{2}\right)+\frac{(-3)(-4)}{2!}\left(\frac{5x}{2}\right)^2+\frac{(-3)(-4)(-5)}{3!}\left(\frac{5x}{2}\right)^3+\ldots\right]$
$= \frac{1}{8}\left[1-\frac{15x}{2}+\frac{75x^2}{4}-\frac{625x^3}{8}+\ldots\right]$
M1 A1
$= \frac{1}{8}-\frac{15x}{16}+\frac{75x^2}{16}-\frac{625x^3}{32}+\ldots$
A1 A1
**(6 marks)**
## Notes
M1: Mark can be implied by a constant term of $(2)^{-3}$ or $\frac{1}{8}$.
B1: $2^{-3}$ or $\frac{1}{8}$ outside brackets or as candidate's constant term in their binomial expansion.
M1: Expands $\left(\ldots+kx\right)^{-3}$, where $k =$ a value $\neq 1$ to give any 2 terms out of 4 terms simplified or unsimplified.
Examples: $1+(-3)(kx)$ or $\frac{(-3)(-4)}{2!}(kx)^2+\frac{(-3)(-4)(-5)}{3!}(kx)^3$ or $1+(-3)\ldots+\frac{(-3)(-4)}{2!}(kx)^2$ or $\frac{(-3)(-4)}{2!}(kx)^2+\frac{(-3)(-4)(-5)}{3!}(kx)^3$ are acceptable for M1.
A1: A correct simplified or unsimplified $1+(-3)(kx)+\frac{(-3)(-4)}{2!}(kx)^2+\frac{(-3)(-4)(-5)}{3!}(kx)^3$ expansion with consistent $(kx)$. Note that $(kx)$ must be consistent and $k =$ a value $\neq 1$ (on the RHS, not necessarily the LHS) in a candidate's expansion.
A1: For $\frac{1}{8}-\frac{15x}{16}$ (simplified) or also allow $0.125-0.9375x$.
A1: Accept only $\frac{75x^2}{16}-\frac{625x^3}{32}$ or $4\frac{11}{16}x^2-19\frac{17}{32}x^3$ or $4.6875x^2-19.53125x^3$\begin{enumerate}
\item Use the binomial series to find the expansion of
\end{enumerate}
$$\frac { 1 } { ( 2 + 5 x ) ^ { 3 } } \quad | \boldsymbol { x } | < \frac { 2 } { 5 }$$
in ascending powers of $x$, up to and including the term in $x ^ { 3 }$\\
Give each coefficient as a fraction in its simplest form.\\
\begin{center}
\begin{tabular}{|l|l|l|}
\hline
VIIIV SIHI NI JIIIM ION OC & VIIV SIHI NI JAHAM ION OO & VJ4V SIHIL NI JIIIM IONOO \\
\hline
\end{tabular}
\end{center}
\begin{center}
\end{center}
\hfill \mbox{\textit{Edexcel P4 2018 Q1 [6]}}