CAIE FP1 2019 November — Question 5 9 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2019
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeStandard summation formulae application
DifficultyStandard +0.8 This is a multi-part Further Maths question requiring manipulation of summation formulae, method of differences with partial fractions, and limit evaluation. While each technique is standard for FP1, the combination across three parts and the final limit calculation involving the product of two series expressions requires careful algebraic manipulation and understanding of dominant terms, placing it moderately above average difficulty.
Spec4.06b Method of differences: telescoping series
Let \(S_N = \sum_{r=1}^{N} (5r + 1)(5r + 6)\) and \(T_N = \sum_{r=1}^{N} \frac{1}{(5r + 1)(5r + 6)}\).
  1. Use standard results from the List of Formulae (MF10) to show that $$S_N = \frac{1}{3}N(25N^2 + 90N + 83).$$ [3]
  2. Use the method of differences to express \(T_N\) in terms of \(N\). [4]
  3. Find \(\lim_{N \to \infty} (N^{-3} S_N T_N)\). [2]
Question 5:
AnswerMarks
5(i)N N N
∑( 5r+1 )( 5r+6 )=25∑r2 +35∑r+6N
AnswerMarks Guidance
r=1 r=1 r=1M1 Expands.
25  1 N ( N +1 )( 2N +1 ) +35  1 N ( N +1 ) +6N
   
AnswerMarks Guidance
6  2 M1 Substitutes formulae for ∑r and ∑r2.
25( ) 35 35  1 ( )
=N 2N2 +3N +1 + N + +6 = N 25N2 +90N +83
 
AnswerMarks Guidance
 6 2 2  3A1 Simplifies to the given answer (AG).
3
AnswerMarks Guidance
5(ii)1 1 1 1 
( 5r+1 )( 5r+6 ) = 5  5r+1 − 5r+6 M1 A1 Finds partial fractions.
11 1 1 1 1 1 
T = − + − +(cid:34)+ −
AnswerMarks Guidance
N 5  6 11 11 16 5N +1 5N +6  M1 Expresses terms as differences.
11 1  1 1
− = −
AnswerMarks Guidance
5  6 5N +6   30 5 ( 5N +6 )A1 At least 3 terms including last.
4
AnswerMarks
5(iii)S 25 1 5
N T → × =
AnswerMarks Guidance
N3 N 3 30 18M1 A1 Divides S by N3 and takes limits as N →∞
N
2
AnswerMarks Guidance
QuestionAnswer Marks 
Question 5:
--- 5(i) ---
5(i) | N N N
∑( 5r+1 )( 5r+6 )=25∑r2 +35∑r+6N
r=1 r=1 r=1 | M1 | Expands.
25  1 N ( N +1 )( 2N +1 ) +35  1 N ( N +1 ) +6N
   
6  2  | M1 | Substitutes formulae for ∑r and ∑r2.
25( ) 35 35  1 ( )
=N 2N2 +3N +1 + N + +6 = N 25N2 +90N +83
 
 6 2 2  3 | A1 | Simplifies to the given answer (AG).
3
--- 5(ii) ---
5(ii) | 1 1 1 1 
( 5r+1 )( 5r+6 ) = 5  5r+1 − 5r+6  | M1 A1 | Finds partial fractions.
11 1 1 1 1 1 
T = − + − +(cid:34)+ −
N 5  6 11 11 16 5N +1 5N +6   | M1 | Expresses terms as differences.
11 1  1 1
− = −
5  6 5N +6   30 5 ( 5N +6 ) | A1 | At least 3 terms including last.
4
--- 5(iii) ---
5(iii) | S 25 1 5
N T → × =
N3 N 3 30 18 | M1 A1 | Divides S by N3 and takes limits as N →∞
N
2
Question | Answer | Marks | Guidance
Let $S_N = \sum_{r=1}^{N} (5r + 1)(5r + 6)$ and $T_N = \sum_{r=1}^{N} \frac{1}{(5r + 1)(5r + 6)}$.

\begin{enumerate}[label=(\roman*)]
\item Use standard results from the List of Formulae (MF10) to show that
$$S_N = \frac{1}{3}N(25N^2 + 90N + 83).$$ [3]

\item Use the method of differences to express $T_N$ in terms of $N$. [4]

\item Find $\lim_{N \to \infty} (N^{-3} S_N T_N)$. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE FP1 2019 Q5 [9]}}
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