Pre-U Pre-U 9795/1 2018 June — Question 1 5 marks

Exam BoardPre-U
ModulePre-U 9795/1 (Pre-U Further Mathematics Paper 1)
Year2018
SessionJune
Marks5
TopicSequences and series, recurrence and convergence
TypePartial fractions then method of differences
DifficultyModerate -0.3 This is a standard three-part question on partial fractions and telescoping series. Part (i) is routine A-level partial fractions, part (ii) applies the standard method of differences technique with clear cancellation, and part (iii) is a straightforward limit as nā†’āˆž. While it requires multiple techniques, each step follows a well-practiced procedure with no novel insight needed, making it slightly easier than average.
Spec4.06b Method of differences: telescoping series

  1. Express \(\frac{3}{(3r-1)(3r+2)}\) in partial fractions. [2]
  2. Using the method of differences, prove that \(\sum_{r=1}^{n} \frac{3}{(3r-1)(3r+2)} = \frac{1}{2} - \frac{1}{3n+2}\). [2]
  3. Deduce the value of \(\sum_{r=1}^{\infty} \frac{1}{(3r-1)(3r+2)}\). [1]

\begin{enumerate}[label=(\roman*)]
\item Express $\frac{3}{(3r-1)(3r+2)}$ in partial fractions. [2]

\item Using the method of differences, prove that $\sum_{r=1}^{n} \frac{3}{(3r-1)(3r+2)} = \frac{1}{2} - \frac{1}{3n+2}$. [2]

\item Deduce the value of $\sum_{r=1}^{\infty} \frac{1}{(3r-1)(3r+2)}$. [1]
\end{enumerate}

\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2018 Q1 [5]}}