CAIE FP1 2018 November — Question 7 10 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2018
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypePartial fractions then method of differences
DifficultyStandard +0.8 This is a multi-part Further Maths question requiring standard summation formulas, method of differences (partial fractions), and limit evaluation. While each individual part uses known techniques, the combination of skills and the final parts requiring algebraic manipulation and limit reasoning place it moderately above average difficulty for A-level, though still within reach of well-prepared FM students.
Spec4.06b Method of differences: telescoping series
Let $$S_N = \sum_{r=1}^{N}(3r + 1)(3r + 4) \quad \text{and} \quad T_N = \sum_{r=1}^{N}\frac{1}{(3r + 1)(3r + 4)}.$$
  1. Use standard results from the List of Formulae (MF10) to show that $$S_N = N(3N^2 + 12N + 13).$$ [3]
  2. Use the method of differences to show that $$T_N = \frac{1}{12} - \frac{1}{3(3N + 4)}.$$ [3]
  3. Deduce that \(\frac{S_N}{T_N}\) is an integer. [2]
  4. Find \(\lim_{N \to \infty} \frac{S_N}{N^3 T_N}\). [2]
Question 7:
AnswerMarks
7(i)N N N
∑( 3r+1 )( 3r+4 )=9∑r2 +15∑r+4N
AnswerMarks Guidance
r=1 r=1 r=1M1 Expands
9 1 N ( N +1 )( 2N +1 )   +15   1 N ( N +1 )   +4N
AnswerMarks Guidance
6  2 M1 Substitutes formulae for ∑r and ∑r2.
9( ) 15 15 
= N 2N2 +3N +1 + N + +4
6 2 2 
( )
AnswerMarks Guidance
= N 3N2 +12N +13A1 Shows simplification to the given answer
(AG).
3
AnswerMarks
7(ii)1 1 1 1 
= −
AnswerMarks Guidance
( 3r+1 )( 3r+4 ) 3  3r+1 3r+4 B1 Finds partial fractions.
11 1 1 1 1 1 
T N = 3   4 − 7 + 7 − 10 +(cid:34)+ 3 ( N −1 )+1 − 3N +4  
AnswerMarks Guidance
 M1 Expresses terms as differences.
11 1  1 1
− = −
AnswerMarks Guidance
3  4 3N +4  12 3 ( 3N +4 )A1 Cancels to given answer (AG).
3
AnswerMarks
7(iii)T = N ⇒ S N =4 ( 3N +4 ) ( 3N2 +12N +13 )
N 4 ( 3N +4 ) T
AnswerMarks Guidance
NM1 S
Writes N as a polynomial
T
N
S
So N is an integer because all terms are integers
T
AnswerMarks Guidance
NA1 Justifies expression being integer
2
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
7(iv)( )
S 4 ( 3N +4 ) 3N2 +16N +9
N =
N3T N3
AnswerMarks Guidance
NM1 Divides expression in (iii) by N3 and
takes limit
( )( )=36
AnswerMarks
→4 3 3A1
2
AnswerMarks Guidance
QuestionAnswer Marks 
Question 7:
--- 7(i) ---
7(i) | N N N
∑( 3r+1 )( 3r+4 )=9∑r2 +15∑r+4N
r=1 r=1 r=1 | M1 | Expands
9 1 N ( N +1 )( 2N +1 )   +15   1 N ( N +1 )   +4N
6  2  | M1 | Substitutes formulae for ∑r and ∑r2.
9( ) 15 15 
= N 2N2 +3N +1 + N + +4
6 2 2 
( )
= N 3N2 +12N +13 | A1 | Shows simplification to the given answer
(AG).
3
--- 7(ii) ---
7(ii) | 1 1 1 1 
= −
( 3r+1 )( 3r+4 ) 3  3r+1 3r+4  | B1 | Finds partial fractions.
11 1 1 1 1 1 
T N = 3   4 − 7 + 7 − 10 +(cid:34)+ 3 ( N −1 )+1 − 3N +4  
  | M1 | Expresses terms as differences.
11 1  1 1
− = −
3  4 3N +4  12 3 ( 3N +4 ) | A1 | Cancels to given answer (AG).
3
--- 7(iii) ---
7(iii) | T = N ⇒ S N =4 ( 3N +4 ) ( 3N2 +12N +13 )
N 4 ( 3N +4 ) T
N | M1 | S
Writes N as a polynomial
T
N
S
So N is an integer because all terms are integers
T
N | A1 | Justifies expression being integer
2
Question | Answer | Marks | Guidance
--- 7(iv) ---
7(iv) | ( )
S 4 ( 3N +4 ) 3N2 +16N +9
N =
N3T N3
N | M1 | Divides expression in (iii) by N3 and
takes limit
( )( )=36
→4 3 3 | A1
2
Question | Answer | Marks | Guidance
Let
$$S_N = \sum_{r=1}^{N}(3r + 1)(3r + 4) \quad \text{and} \quad T_N = \sum_{r=1}^{N}\frac{1}{(3r + 1)(3r + 4)}.$$

\begin{enumerate}[label=(\roman*)]
\item Use standard results from the List of Formulae (MF10) to show that
$$S_N = N(3N^2 + 12N + 13).$$ [3]

\item Use the method of differences to show that
$$T_N = \frac{1}{12} - \frac{1}{3(3N + 4)}.$$ [3]

\item Deduce that $\frac{S_N}{T_N}$ is an integer. [2]

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

\hfill \mbox{\textit{CAIE FP1 2018 Q7 [10]}}
This paper (11 questions)
View full paper
Q1 5 Q2 6 Q3 8 Q4 8 Q5 9 Q6 8 Q7 10 Q8 10 Q9 10 Q10 12 Q11 28