Question 1 - A-Level Maths

Edexcel FP2 2018 June — Question 1 8 marks

Exam Board	Edexcel
Module	FP2 (Further Pure Mathematics 2)
Year	2018
Session	June
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Partial Fractions
Type	Two linear factors in denominator
Difficulty	Standard +0.3 This is a standard Further Pure 2 question testing routine partial fractions with two linear factors, followed by telescoping series (method of differences). Part (a) is straightforward decomposition, part (b) is a standard telescoping sum technique, and part (c) applies the formula. While it's Further Maths content, it follows a well-practiced template with no novel insight required, making it slightly easier than average overall.
Spec	4.05c Partial fractions: extended to quadratic denominators 4.06b Method of differences: telescoping series

(a) Express $\frac { 1 } { ( r + 3 ) ( r + 4 ) }$ in partial fractions.
(b) Hence, using the method of differences, show that

$$\sum _ { r = 1 } ^ { n } \frac { 1 } { ( r + 3 ) ( r + 4 ) } = \frac { n } { a ( n + a ) }$$ where $a$ is a constant to be found.
(c) Find the exact value of $\sum _ { r = 15 } ^ { 30 } \frac { 1 } { ( r + 3 ) ( r + 4 ) }$ uestion 1 continued \includegraphics[max width=\textwidth, alt={}, center]{5aa7f449-215b-4a21-9fdc-df55d26abc9d-05_29_40_182_1914} \includegraphics[max width=\textwidth, alt={}, center]{5aa7f449-215b-4a21-9fdc-df55d26abc9d-05_33_37_201_1914}
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Show mark scheme Show mark scheme source

Question 1:

Part (a)

Answer	Marks	Guidance
$\frac{1}{(r+3)(r+4)} \equiv \frac{1}{(r+3)} - \frac{1}{(r+4)}$	B1	Cao, no working needed – ignore any shown

Part (b)

Answer	Marks	Guidance
$r=1$: $\frac{1}{4} - \frac{1}{5}$
$r=2$: $\frac{1}{5} - \frac{1}{6}$
$...r=n-1$: $\frac{1}{(n+2)} - \frac{1}{(n+3)}$
$r=n$: $\frac{1}{(n+3)} - \frac{1}{(n+4)}$	M1	First 2 and last term or first and last 2 terms required. Must start at $r=1$. First term complete, 2nd and last may be partial or last term complete 1st and penultimate partial.
$\sum_{r=1}^{n} \frac{1}{(r+2)(r+3)} = \frac{1}{4} - \frac{1}{(n+4)}$	M1A1	Cancel terms
$\sum_{r=1}^{n} \frac{1}{(r+2)(r+3)} = \frac{n}{4(n+4)}$	dM1, A1cso	Find common denominator, dependent on second M mark. Need not be shown explicitly. $(a=4)$

NB1: All marks can be awarded if work done with values $1,2,...r$ and then $r$ replaced with $n$; if no replacement made, deduct final A mark.

NB2: $\frac{1}{4} - \frac{1}{(n+4)}$ with NO other working gets M0M1A1M1A0 max

Part (c)

Answer	Marks	Guidance
$\sum_{r=15}^{30} \frac{1}{(r+3)(r+4)} = \frac{30}{4(30+4)} - \frac{14}{4(14+4)}$	M1	Accept $n=30$ and $n=14$ only in their answer to (b). Must be subtracted.
$= \frac{4}{153}$ oe (exact)	A1	Exact answer $\frac{4}{153}$ implies method provided no incorrect work seen in (c).
ALT: Use method of differences again, starting at $r=15$ and ending at $r=30$	M1	Complete method
$= \frac{4}{153}$ oe (exact)	A1	Correct answer

## Question 1:

### Part (a)
$\frac{1}{(r+3)(r+4)} \equiv \frac{1}{(r+3)} - \frac{1}{(r+4)}$ | B1 | Cao, no working needed – ignore any shown

### Part (b)
$r=1$: $\frac{1}{4} - \frac{1}{5}$ | | |

$r=2$: $\frac{1}{5} - \frac{1}{6}$ | | |

$...r=n-1$: $\frac{1}{(n+2)} - \frac{1}{(n+3)}$ | | |

$r=n$: $\frac{1}{(n+3)} - \frac{1}{(n+4)}$ | M1 | First 2 and last term or first and last 2 terms required. Must start at $r=1$. First term complete, 2nd and last may be partial or last term complete 1st and penultimate partial.

$\sum_{r=1}^{n} \frac{1}{(r+2)(r+3)} = \frac{1}{4} - \frac{1}{(n+4)}$ | M1A1 | Cancel terms

$\sum_{r=1}^{n} \frac{1}{(r+2)(r+3)} = \frac{n}{4(n+4)}$ | dM1, A1cso | Find common denominator, dependent on second M mark. Need not be shown explicitly. $(a=4)$

**NB1:** All marks can be awarded if work done with values $1,2,...r$ and then $r$ replaced with $n$; if no replacement made, deduct final A mark.

**NB2:** $\frac{1}{4} - \frac{1}{(n+4)}$ with NO other working gets M0M1A1M1A0 max

### Part (c)
$\sum_{r=15}^{30} \frac{1}{(r+3)(r+4)} = \frac{30}{4(30+4)} - \frac{14}{4(14+4)}$ | M1 | Accept $n=30$ and $n=14$ only in their answer to (b). Must be subtracted.

$= \frac{4}{153}$ oe (exact) | A1 | Exact answer $\frac{4}{153}$ implies method provided no incorrect work seen in (c).

**ALT:** Use method of differences again, starting at $r=15$ and ending at $r=30$ | M1 | Complete method

$= \frac{4}{153}$ oe (exact) | A1 | Correct answer

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Show LaTeX source

\begin{enumerate}
  \item (a) Express $\frac { 1 } { ( r + 3 ) ( r + 4 ) }$ in partial fractions.\\
(b) Hence, using the method of differences, show that
\end{enumerate}

$$\sum _ { r = 1 } ^ { n } \frac { 1 } { ( r + 3 ) ( r + 4 ) } = \frac { n } { a ( n + a ) }$$

where $a$ is a constant to be found.\\
(c) Find the exact value of $\sum _ { r = 15 } ^ { 30 } \frac { 1 } { ( r + 3 ) ( r + 4 ) }$

uestion 1 continued\\
\includegraphics[max width=\textwidth, alt={}, center]{5aa7f449-215b-4a21-9fdc-df55d26abc9d-05_29_40_182_1914}

\includegraphics[max width=\textwidth, alt={}, center]{5aa7f449-215b-4a21-9fdc-df55d26abc9d-05_33_37_201_1914}\\
□ D D D " "\\

\hfill \mbox{\textit{Edexcel FP2 2018 Q1 [8]}}

This paper (8 questions)

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Q1 8 Q2 4 Q3 9 Q4 7 Q5 9 Q6 13 Q7 12 Q8 13

Answer	Marks	Guidance
\(r=1\): \(\frac{1}{4} - \frac{1}{5}\)
\(r=2\): \(\frac{1}{5} - \frac{1}{6}\)
\(...r=n-1\): \(\frac{1}{(n+2)} - \frac{1}{(n+3)}\)
\(r=n\): \(\frac{1}{(n+3)} - \frac{1}{(n+4)}\)	M1	First 2 and last term or first and last 2 terms required. Must start at \(r=1\). First term complete, 2nd and last may be partial or last term complete 1st and penultimate partial.
\(\sum_{r=1}^{n} \frac{1}{(r+2)(r+3)} = \frac{1}{4} - \frac{1}{(n+4)}\)	M1A1	Cancel terms
\(\sum_{r=1}^{n} \frac{1}{(r+2)(r+3)} = \frac{n}{4(n+4)}\)	dM1, A1cso	Find common denominator, dependent on second M mark. Need not be shown explicitly. \((a=4)\)