Edexcel F2 2021 January — Question 2 6 marks

Exam BoardEdexcel
ModuleF2 (Further Pure Mathematics 2)
Year2021
SessionJanuary
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeMethod of differences with given identity
DifficultyStandard +0.3 Part (a) is straightforward algebraic verification requiring common denominators and simplification. Part (b) is a standard telescoping series application where the identity is given—students just need to recognize the pattern and sum the collapsing terms. This is a typical Further Maths method of differences question, slightly easier than average due to the identity being provided rather than derived.
Spec4.06b Method of differences: telescoping series

2. (a) Show that, for \(r > 0\) $$\frac { r + 2 } { r ( r + 1 ) } - \frac { r + 3 } { ( r + 1 ) ( r + 2 ) } = \frac { r + 4 } { r ( r + 1 ) ( r + 2 ) }$$ (b) Hence show that $$\sum _ { r = 1 } ^ { n } \frac { r + 4 } { r ( r + 1 ) ( r + 2 ) } = \frac { n ( a n + b ) } { c ( n + 1 ) ( n + 2 ) }$$ where \(a\), \(b\) and \(c\) are integers to be determined.

Question 2:
Part (a):
AnswerMarks Guidance
Working/AnswerMark Guidance
\(\frac{r+2}{r(r+1)} - \frac{r+3}{(r+1)(r+2)} = \frac{(r+2)^2 - r(r+3)}{r(r+1)(r+2)}\)M1 Attempt single fraction with correct denominator
\(= \frac{r^2+4r+4-r^2-3r}{r(r+1)(r+2)} = \frac{r+4}{r(r+1)(r+2)}\) *A1* Correct result, no errors. Must include LHS
Part (b):
AnswerMarks Guidance
Working/AnswerMark Guidance
Show telescoping terms from \(r=1\) to \(r=n\)M1 Show sufficient terms to demonstrate cancelling, min 3 at start and 1 at end or 2 at start and 2 at end
\(\sum_{r=1}^{n} \frac{r+4}{r(r+1)(r+2)} = \frac{3}{2} - \frac{n+3}{(n+1)(n+2)}\)A1 Extract the correct 2 remaining terms
\(= \frac{3(n+1)(n+2)-2n-6}{2(n+1)(n+2)} = \frac{n(3n+7)}{2(n+1)(n+2)}\)dM1, A1cao Attempt common denominator of form \(k(n+1)(n+2)\); correct result, no need to show \(a\), \(b\), \(c\) explicitly
## Question 2:

### Part (a):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{r+2}{r(r+1)} - \frac{r+3}{(r+1)(r+2)} = \frac{(r+2)^2 - r(r+3)}{r(r+1)(r+2)}$ | M1 | Attempt single fraction with correct denominator |
| $= \frac{r^2+4r+4-r^2-3r}{r(r+1)(r+2)} = \frac{r+4}{r(r+1)(r+2)}$ * | A1* | Correct result, no errors. Must include LHS |

### Part (b):

| Working/Answer | Mark | Guidance |
|---|---|---|
| Show telescoping terms from $r=1$ to $r=n$ | M1 | Show sufficient terms to demonstrate cancelling, min 3 at start and 1 at end or 2 at start and 2 at end |
| $\sum_{r=1}^{n} \frac{r+4}{r(r+1)(r+2)} = \frac{3}{2} - \frac{n+3}{(n+1)(n+2)}$ | A1 | Extract the correct 2 remaining terms |
| $= \frac{3(n+1)(n+2)-2n-6}{2(n+1)(n+2)} = \frac{n(3n+7)}{2(n+1)(n+2)}$ | dM1, A1cao | Attempt common denominator of form $k(n+1)(n+2)$; correct result, no need to show $a$, $b$, $c$ explicitly |

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2. (a) Show that, for $r > 0$

$$\frac { r + 2 } { r ( r + 1 ) } - \frac { r + 3 } { ( r + 1 ) ( r + 2 ) } = \frac { r + 4 } { r ( r + 1 ) ( r + 2 ) }$$

(b) Hence show that

$$\sum _ { r = 1 } ^ { n } \frac { r + 4 } { r ( r + 1 ) ( r + 2 ) } = \frac { n ( a n + b ) } { c ( n + 1 ) ( n + 2 ) }$$

where $a$, $b$ and $c$ are integers to be determined.

\hfill \mbox{\textit{Edexcel F2 2021 Q2 [6]}}