CAIE Further Paper 1 2024 November — Question 4 8 marks

Exam BoardCAIE
ModuleFurther Paper 1 (Further Paper 1)
Year2024
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypePartial fractions then method of differences
DifficultyChallenging +1.2 This is a structured Further Maths question on method of differences with clear scaffolding. Part (a) requires partial fractions and telescoping series (standard FM technique), part (b) is straightforward substitution into a limit, and part (c) applies the result from (a). While it involves multiple steps and FM content, the method is signposted and each part follows logically from the previous one without requiring novel insight.
Spec4.06b Method of differences: telescoping series
  1. Use the method of differences to find \(\sum_{r=1}^{n} \frac{5k}{(5r+k)(5r+5+k)}\) in terms of \(n\) and \(k\), where \(k\) is a positive constant. [4]
It is given that \(\sum_{r=1}^{\infty} \frac{5k}{(5r+k)(5r+5+k)} = \frac{1}{3}\).
  1. Find the value of \(k\). [2]
  2. Hence find \(\sum_{r=1}^{n-1} \frac{5k}{(5r+k)(5r+5+k)}\) in terms of \(n\). [2]
Question 4:
AnswerMarks
4(a)5k  1 1 
=k  − 
AnswerMarks Guidance
(5r+k)(5r+5+k) 5r+k 5r+5+kM1 A1 Finds partial fractions. (Don’t allow a
substitution of a value for k.)
n 5k  1 1 1 1 1 1 
 =k  − + − + + − 
(5r+k)(5r+5+k) 5+k 10+k 10+k 15+k 5n+k 5n+5+k
AnswerMarks Guidance
r=1M1 Writes at least three terms, including last.
(Allow any value of k.)
 1 1 
=k  − 
AnswerMarks
5+k 5n+5+kA1
4
AnswerMarks
4(b)k 1 5
= 3k =5+kk =
AnswerMarks
5+k 3 2M1 A1
2
AnswerMarks
4(c)n2 5k n2 5k n−1 5k
 = −
(5r+k)(5r+5+k) (5r+k)(5r+5+k) (5r+k)(5r+5+k)
AnswerMarks Guidance
r=n r=1 r=1M1 Or applies the method of differences again.
 1 1   1 1   1 1 
=k − −k − =k − 
5+k 5n2 +5+k 5+k 5n+k 5n+k 5n2 +5+k
1 1
= −
AnswerMarks Guidance
2n+1 2n2 +3A1FT FT on their value of k (must be substituted in).
2
AnswerMarks Guidance
QuestionAnswer Marks 
Question 4:
--- 4(a) ---
4(a) | 5k  1 1 
=k  − 
(5r+k)(5r+5+k) 5r+k 5r+5+k | M1 A1 | Finds partial fractions. (Don’t allow a
substitution of a value for k.)
n 5k  1 1 1 1 1 1 
 =k  − + − + + − 
(5r+k)(5r+5+k) 5+k 10+k 10+k 15+k 5n+k 5n+5+k
r=1 | M1 | Writes at least three terms, including last.
(Allow any value of k.)
 1 1 
=k  − 
5+k 5n+5+k | A1
4
--- 4(b) ---
4(b) | k 1 5
= 3k =5+kk =
5+k 3 2 | M1 A1
2
--- 4(c) ---
4(c) | n2 5k n2 5k n−1 5k
 = −
(5r+k)(5r+5+k) (5r+k)(5r+5+k) (5r+k)(5r+5+k)
r=n r=1 r=1 | M1 | Or applies the method of differences again.
 1 1   1 1   1 1 
=k − −k − =k − 
5+k 5n2 +5+k 5+k 5n+k 5n+k 5n2 +5+k
1 1
= −
2n+1 2n2 +3 | A1FT | FT on their value of k (must be substituted in).
2
Question | Answer | Marks | Guidance
\begin{enumerate}[label=(\alph*)]
\item Use the method of differences to find $\sum_{r=1}^{n} \frac{5k}{(5r+k)(5r+5+k)}$ in terms of $n$ and $k$, where $k$ is a positive constant. [4]
\end{enumerate}

It is given that $\sum_{r=1}^{\infty} \frac{5k}{(5r+k)(5r+5+k)} = \frac{1}{3}$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the value of $k$. [2]

\item Hence find $\sum_{r=1}^{n-1} \frac{5k}{(5r+k)(5r+5+k)}$ in terms of $n$. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 1 2024 Q4 [8]}}
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