CAIE FP1 2010 November — Question 4 5 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2010
SessionNovember
Marks5
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Mark schemeDownload PDF ↗
TopicProof by induction
TypeProve divisibility
DifficultyStandard +0.8 This is a standard proof by induction for divisibility, requiring students to manipulate algebraic expressions involving powers and factor out 44. While it involves Further Maths content and requires careful algebraic manipulation in the inductive step (particularly handling 7^{2n+3} and 5^{n+4}), it follows a well-established template that students practice extensively. The algebraic manipulation is moderately challenging but routine for FP1 students.
Spec4.01a Mathematical induction: construct proofs
4 Prove by mathematical induction that, for all non-negative integers \(n , 7 ^ { 2 n + 1 } + 5 ^ { n + 3 }\) is divisible by 44 .
AnswerMarks Guidance
\(n = 0: 7^1 + 5^3 = 132\) which is divisible by \(44\)B1
\(\text{Assume } 7^{2k+1} + 5^{k+3}\) is divisible by \(44\)B1
\(\text{Consider } 7^{3(k+1)+1} + 5^{(k+1)+3} = 7^{7k+4} + 5.5^{k+3} = 49(7^{2k+1} + 5^{k+3}) - 44.5^{k+3}\)M1 \((k+1)\)th term
M1in appropriate form
A1convincing argument [5]
Alternative solution for final three marks:
AnswerMarks
\(\text{Consider } (7^{2k+3} + 5^{k+4}) - (7^{2k+1} + 5^{k+3}) = 48(7^{2k+1} + 5^{k+3}) - 44.5^{k+3}\)M1
M1in appropriate form
A1convincing argument 
$n = 0: 7^1 + 5^3 = 132$ which is divisible by $44$ | B1

$\text{Assume } 7^{2k+1} + 5^{k+3}$ is divisible by $44$ | B1

$\text{Consider } 7^{3(k+1)+1} + 5^{(k+1)+3} = 7^{7k+4} + 5.5^{k+3} = 49(7^{2k+1} + 5^{k+3}) - 44.5^{k+3}$ | M1 | $(k+1)$th term
| | M1 | in appropriate form
| | A1 | convincing argument | [5]

**Alternative solution for final three marks:**

$\text{Consider } (7^{2k+3} + 5^{k+4}) - (7^{2k+1} + 5^{k+3}) = 48(7^{2k+1} + 5^{k+3}) - 44.5^{k+3}$ | M1
| | M1 | in appropriate form
| | A1 | convincing argument

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4 Prove by mathematical induction that, for all non-negative integers $n , 7 ^ { 2 n + 1 } + 5 ^ { n + 3 }$ is divisible by 44 .

\hfill \mbox{\textit{CAIE FP1 2010 Q4 [5]}}
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Q1 4 Q2 5 Q3 5 Q4 5 Q5 8 Q6 8 Q7 9 Q8 10 Q9 10 Q10 10 Q11 12 Q12 EITHER Q12 OR