SPS SPS FM Pure 2025 January — Question 2 8 marks

Exam BoardSPS
ModuleSPS FM Pure (SPS FM Pure)
Year2025
SessionJanuary
Marks8
TopicGeneralised Binomial Theorem and Partial Fractions
TypePartial fractions showing coefficient is zero
DifficultyStandard +0.3 This is a standard Further Maths partial fractions question followed by routine binomial expansion. Part (i) uses the cover-up method or substitution to find constants (straightforward algebra). Part (ii) applies standard binomial expansion formulas and combines series—all textbook techniques with no novel insight required. Slightly easier than average A-level due to its mechanical nature.
Spec1.02y Partial fractions: decompose rational functions1.04c Extend binomial expansion: rational n, |x|<1

  1. Given that $$\frac{3 + 2x^2}{(1 + x)^2(1 - 4x)} = \frac{A}{1 + x} + \frac{B}{(1 + x)^2} + \frac{C}{1 - 4x},$$ where \(A\), \(B\) and \(C\) are constants, find \(B\) and \(C\), and show that \(A = 0\). [4]
  2. Given that \(x\) is sufficiently small, find the first three terms of the binomial expansions of \((1 + x)^{-2}\) and \((1 - 4x)^{-1}\). Hence find the first three terms of the expansion of \(\frac{3 + 2x^2}{(1 + x)^2(1 - 4x)}\). [4]

\begin{enumerate}[label=(\roman*)]
\item Given that
$$\frac{3 + 2x^2}{(1 + x)^2(1 - 4x)} = \frac{A}{1 + x} + \frac{B}{(1 + x)^2} + \frac{C}{1 - 4x},$$
where $A$, $B$ and $C$ are constants, find $B$ and $C$, and show that $A = 0$. [4]

\item Given that $x$ is sufficiently small, find the first three terms of the binomial expansions of $(1 + x)^{-2}$ and $(1 - 4x)^{-1}$.

Hence find the first three terms of the expansion of $\frac{3 + 2x^2}{(1 + x)^2(1 - 4x)}$. [4]
\end{enumerate}

\hfill \mbox{\textit{SPS SPS FM Pure 2025 Q2 [8]}}