OCR MEI FP1 2016 June — Question 6 6 marks

Exam BoardOCR MEI
ModuleFP1 (Further Pure Mathematics 1)
Year2016
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof by induction
TypeProve recurrence relation formula
DifficultyStandard +0.8 This is a standard proof by induction for a recurrence relation formula in Further Maths. While it requires proper induction structure (base case, assumption, inductive step) and algebraic manipulation of the recurrence relation, it follows a well-established template. The algebra is moderately involved but straightforward for FP1 students who have practiced this technique.
Spec1.04e Sequences: nth term and recurrence relations4.01a Mathematical induction: construct proofs
6 A sequence is defined by \(u _ { 1 } = 8\) and \(u _ { n + 1 } = 3 u _ { n } + 2 n + 5\). Prove by induction that \(u _ { n } = 4 \left( 3 ^ { n } \right) - n - 3\).
Question 6:
AnswerMarks
Base case: \(n=1\): \(u_1 = 4(3)-1-3 = 8\) ✓B1
Assume: \(u_k = 4(3^k)-k-3\)M1
Inductive step: \(u_{k+1} = 3u_k+2k+5\)M1
\(= 3(4\cdot3^k - k-3)+2k+5 = 12\cdot3^k - 3k-9+2k+5\)A1
\(= 4\cdot3^{k+1}-(k+1)-3\) ✓A1
Conclusion stated properlyB1 
## Question 6:

**Base case:** $n=1$: $u_1 = 4(3)-1-3 = 8$ ✓ | B1 |
**Assume:** $u_k = 4(3^k)-k-3$ | M1 |
**Inductive step:** $u_{k+1} = 3u_k+2k+5$ | M1 |
$= 3(4\cdot3^k - k-3)+2k+5 = 12\cdot3^k - 3k-9+2k+5$ | A1 |
$= 4\cdot3^{k+1}-(k+1)-3$ ✓ | A1 |
**Conclusion** stated properly | B1 |

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6 A sequence is defined by $u _ { 1 } = 8$ and $u _ { n + 1 } = 3 u _ { n } + 2 n + 5$. Prove by induction that $u _ { n } = 4 \left( 3 ^ { n } \right) - n - 3$.

\hfill \mbox{\textit{OCR MEI FP1 2016 Q6 [6]}}
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