CAIE FP1 2013 June — Question 1 5 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2013
SessionJune
Marks5
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TopicSequences and series, recurrence and convergence
TypeFactorial or product method of differences
DifficultyChallenging +1.2 This is a Further Maths question requiring method of differences with factorials. Students must recognize the telescoping pattern after simplifying f(r+1) - f(r) = (r+1)!·r - r!(r-1) = r!·r², then manipulate the given sum to match this form. While the factorial context and algebraic manipulation are non-trivial, the technique is standard for FP1 and the question provides clear scaffolding through the 'hence' structure.
Spec4.06b Method of differences: telescoping series
1 Let \(\mathrm { f } ( r ) = r ! ( r - 1 )\). Simplify \(\mathrm { f } ( r + 1 ) - \mathrm { f } ( r )\) and hence find \(\sum _ { r = n + 1 } ^ { 2 n } r ! \left( r ^ { 2 } + 1 \right)\).
Question 1:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(f(r+1) - f(r) = r(r+1)! - (r-1)r!\)M1 Simplifies
\(= r!(r^2 + r - r + 1) = r!(r^2+1)\)A1
\(\sum_{1}^{n} = f(2)-f(1)+f(3)-f(2)+\ldots+f(n+1)-f(n)\)M1 Uses difference method
\(= n(n+1)! - 0 = n(n+1)!\)A1
\(\therefore \sum_{n+1}^{2n} = 2n(2n+1)! - n(n+1)!\)A1 5 marks total 
## Question 1:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(r+1) - f(r) = r(r+1)! - (r-1)r!$ | M1 | Simplifies |
| $= r!(r^2 + r - r + 1) = r!(r^2+1)$ | A1 | |
| $\sum_{1}^{n} = f(2)-f(1)+f(3)-f(2)+\ldots+f(n+1)-f(n)$ | M1 | Uses difference method |
| $= n(n+1)! - 0 = n(n+1)!$ | A1 | |
| $\therefore \sum_{n+1}^{2n} = 2n(2n+1)! - n(n+1)!$ | A1 | 5 marks total |

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1 Let $\mathrm { f } ( r ) = r ! ( r - 1 )$. Simplify $\mathrm { f } ( r + 1 ) - \mathrm { f } ( r )$ and hence find $\sum _ { r = n + 1 } ^ { 2 n } r ! \left( r ^ { 2 } + 1 \right)$.

\hfill \mbox{\textit{CAIE FP1 2013 Q1 [5]}}
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Q1 5 Q2 6 Q3 7 Q4 8 Q5 8 Q6 8 Q7 10 Q8 11 Q9 11 Q10 12 Q11 EITHER Q11 OR