OCR MEI FP1 2006 January — Question 6 7 marks

Exam BoardOCR MEI
ModuleFP1 (Further Pure Mathematics 1)
Year2006
SessionJanuary
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof by induction
TypeProve summation with fractions
DifficultyStandard +0.3 This is a straightforward proof by induction with a simple algebraic formula. The partial fraction decomposition 1/(r(r+1)) = 1/r - 1/(r+1) makes the telescoping sum obvious, and the inductive step requires only basic algebraic manipulation. While it's a Further Maths topic, it's a standard textbook exercise requiring no novel insight, making it slightly easier than average overall.
Spec4.01a Mathematical induction: construct proofs

6 Prove by induction that \(\sum _ { r = 1 } ^ { n } \frac { 1 } { r ( r + 1 ) } = \frac { n } { n + 1 }\).

Question 6:
AnswerMarks Guidance
Base case: \(n=1\): LHS \(= \frac{1}{1\times2} = \frac{1}{2}\), RHS \(= \frac{1}{2}\) ✓B1
Inductive step: Assume true for \(n=k\): \(\sum_{r=1}^{k}\frac{1}{r(r+1)} = \frac{k}{k+1}\)M1
\(\sum_{r=1}^{k+1}\frac{1}{r(r+1)} = \frac{k}{k+1} + \frac{1}{(k+1)(k+2)}\)M1
\(= \frac{k(k+2)+1}{(k+1)(k+2)}\)M1
\(= \frac{k^2+2k+1}{(k+1)(k+2)} = \frac{(k+1)^2}{(k+1)(k+2)} = \frac{k+1}{k+2}\)A1 correct simplification
This is the result with \(n = k+1\)A1 conclusion
Therefore true for all \(n \geq 1\) by inductionB1 final conclusion stated
# Question 6:

**Base case:** $n=1$: LHS $= \frac{1}{1\times2} = \frac{1}{2}$, RHS $= \frac{1}{2}$ ✓ | B1 |

**Inductive step:** Assume true for $n=k$: $\sum_{r=1}^{k}\frac{1}{r(r+1)} = \frac{k}{k+1}$ | M1 |

$\sum_{r=1}^{k+1}\frac{1}{r(r+1)} = \frac{k}{k+1} + \frac{1}{(k+1)(k+2)}$ | M1 |

$= \frac{k(k+2)+1}{(k+1)(k+2)}$ | M1 |

$= \frac{k^2+2k+1}{(k+1)(k+2)} = \frac{(k+1)^2}{(k+1)(k+2)} = \frac{k+1}{k+2}$ | A1 | correct simplification

This is the result with $n = k+1$ | A1 | conclusion

Therefore true for all $n \geq 1$ by induction | B1 | final conclusion stated

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6 Prove by induction that $\sum _ { r = 1 } ^ { n } \frac { 1 } { r ( r + 1 ) } = \frac { n } { n + 1 }$.

\hfill \mbox{\textit{OCR MEI FP1 2006 Q6 [7]}}