Question 9 - A-Level Maths

CAIE FP1 2008 November — Question 9 10 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2008
Session	November
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Proof by induction
Type	Prove summation with fractions
Difficulty	Challenging +1.2 This is a two-part Further Maths induction question. Part 1 is a standard induction proof with algebraic manipulation of fractions—routine for FP1 students. Part 2 requires establishing an inequality by bounding the sum, which adds modest problem-solving beyond pure mechanical induction. The algebra is involved but follows standard techniques, making this moderately above average difficulty.
Spec	4.01a Mathematical induction: construct proofs 4.06b Method of differences: telescoping series

9 Use induction to prove that $$\sum _ { n = 1 } ^ { N } \frac { 4 n + 1 } { n ( n + 1 ) ( 2 n - 1 ) ( 2 n + 1 ) } = 1 - \frac { 1 } { ( N + 1 ) ( 2 N + 1 ) }$$ Show that $$\sum _ { n = N + 1 } ^ { 2 N } \frac { 4 n + 1 } { n ( n + 1 ) ( 2 n - 1 ) ( 2 n + 1 ) } < \frac { 3 } { 8 N ^ { 2 } }$$

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Answer	Marks
Set up
$H_k: \sum_{n=1}^k \frac{4n + 1}{n(n+1)(2n-1)(2n+1)} = 1 - \frac{1}{(k+1)(2k+1)}$	B1
for some positive integer $k$
$H_k \Rightarrow \sum_{n=1}^{k+1} \frac{4n + 1}{n(n+1)(2n-1)(2n+1)} = 1 - \frac{1}{(k+1)(2k+1)} + \frac{4k + 5}{(k+1)(k+2)(2k+1)(2k+3)}$	M1
$= 1 - \frac{2k^2 + 3k + 1}{(k+1)(k+2)(2k+1)(2k+3)}$	A1
$= \ldots = 1 - \frac{1}{(k+2)(2k+3)}$	A1
Verifies $H_1$ is true	B1
Correct completion of induction argument	A1
$\sum_{n=N+1}^{2N} \frac{4n+1}{n(n+1)(2n-1)(2n+1)} = \ldots = \frac{1}{(N+1)(2N+1)} - \frac{1}{(2N+1)(4N+1)}$	M1A1
$= \frac{3N}{(N+1)(2N+1)(4N+1)} < \frac{3N}{N \cdot 2N \cdot 4N} = \frac{3}{8N^2}$	M1A1
OR
$= \frac{3N}{8N^3 + 14N^2 + 7N + 1} = \frac{3}{8N^2 + 14N + 7 + 1/N}$
Since $N \geq 1 \quad 14N + 7 + 1/N > 0$
$\therefore \sum < \frac{3}{8N^2}$

Set up | |

$H_k: \sum_{n=1}^k \frac{4n + 1}{n(n+1)(2n-1)(2n+1)} = 1 - \frac{1}{(k+1)(2k+1)}$ | B1 |

for some positive integer $k$ | |

$H_k \Rightarrow \sum_{n=1}^{k+1} \frac{4n + 1}{n(n+1)(2n-1)(2n+1)} = 1 - \frac{1}{(k+1)(2k+1)} + \frac{4k + 5}{(k+1)(k+2)(2k+1)(2k+3)}$ | M1 |

$= 1 - \frac{2k^2 + 3k + 1}{(k+1)(k+2)(2k+1)(2k+3)}$ | A1 |

$= \ldots = 1 - \frac{1}{(k+2)(2k+3)}$ | A1 |

Verifies $H_1$ is true | B1 |

Correct completion of induction argument | A1 |

$\sum_{n=N+1}^{2N} \frac{4n+1}{n(n+1)(2n-1)(2n+1)} = \ldots = \frac{1}{(N+1)(2N+1)} - \frac{1}{(2N+1)(4N+1)}$ | M1A1 |

$= \frac{3N}{(N+1)(2N+1)(4N+1)} < \frac{3N}{N \cdot 2N \cdot 4N} = \frac{3}{8N^2}$ | M1A1 |

OR |

$= \frac{3N}{8N^3 + 14N^2 + 7N + 1} = \frac{3}{8N^2 + 14N + 7 + 1/N}$ | |

Since $N \geq 1 \quad 14N + 7 + 1/N > 0$ | |

$\therefore \sum < \frac{3}{8N^2}$ | |

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9 Use induction to prove that

$$\sum _ { n = 1 } ^ { N } \frac { 4 n + 1 } { n ( n + 1 ) ( 2 n - 1 ) ( 2 n + 1 ) } = 1 - \frac { 1 } { ( N + 1 ) ( 2 N + 1 ) }$$

Show that

$$\sum _ { n = N + 1 } ^ { 2 N } \frac { 4 n + 1 } { n ( n + 1 ) ( 2 n - 1 ) ( 2 n + 1 ) } < \frac { 3 } { 8 N ^ { 2 } }$$

\hfill \mbox{\textit{CAIE FP1 2008 Q9 [10]}}

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Q1 5 Q2 6 Q3 6 Q4 6 Q5 7 Q6 7 Q7 8 Q8 9 Q9 10 Q10 10 Q11 12 Q12 EITHER Q12 OR

Answer	Marks
Set up
\(H_k: \sum_{n=1}^k \frac{4n + 1}{n(n+1)(2n-1)(2n+1)} = 1 - \frac{1}{(k+1)(2k+1)}\)	B1
for some positive integer \(k\)
\(H_k \Rightarrow \sum_{n=1}^{k+1} \frac{4n + 1}{n(n+1)(2n-1)(2n+1)} = 1 - \frac{1}{(k+1)(2k+1)} + \frac{4k + 5}{(k+1)(k+2)(2k+1)(2k+3)}\)	M1
\(= 1 - \frac{2k^2 + 3k + 1}{(k+1)(k+2)(2k+1)(2k+3)}\)	A1
\(= \ldots = 1 - \frac{1}{(k+2)(2k+3)}\)	A1
Verifies \(H_1\) is true	B1
Correct completion of induction argument	A1
\(\sum_{n=N+1}^{2N} \frac{4n+1}{n(n+1)(2n-1)(2n+1)} = \ldots = \frac{1}{(N+1)(2N+1)} - \frac{1}{(2N+1)(4N+1)}\)	M1A1
\(= \frac{3N}{(N+1)(2N+1)(4N+1)} < \frac{3N}{N \cdot 2N \cdot 4N} = \frac{3}{8N^2}\)	M1A1
OR
\(= \frac{3N}{8N^3 + 14N^2 + 7N + 1} = \frac{3}{8N^2 + 14N + 7 + 1/N}\)
Since \(N \geq 1 \quad 14N + 7 + 1/N > 0\)
\(\therefore \sum < \frac{3}{8N^2}\)