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 standard Further Maths method of differences question with routine partial fractions decomposition. Part (a) requires applying a well-practiced technique, part (b) is straightforward limit evaluation, and part (c) involves simple index manipulation. While it requires multiple steps and is harder than typical A-level questions due to being Further Maths content, it follows a predictable template 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 \(\sum_{r=7}^{n+5} \frac{5k}{(5r+k)(5r+5+k)}\) 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 $\sum_{r=7}^{n+5} \frac{5k}{(5r+k)(5r+5+k)}$ in terms of $n$. [2]
\end{enumerate}

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