CAIE FP1 2012 June — Question 3 6 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2012
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeInfinite series convergence and sum
DifficultyStandard +0.3 This is a standard telescoping series question requiring algebraic manipulation to verify the given identity, then applying summation to find a finite sum formula, and finally taking a limit. While it involves multiple steps, each technique is routine for Further Maths students: partial fractions manipulation, telescoping series recognition, and basic limit evaluation. The structure is highly guided with the key identity provided.
Spec4.06b Method of differences: telescoping series
3 Given that \(\mathrm { f } ( r ) = \frac { 1 } { ( r + 1 ) ( r + 2 ) }\), show that $$\mathrm { f } ( r - 1 ) - \mathrm { f } ( r ) = \frac { 2 } { r ( r + 1 ) ( r + 2 ) }$$ Hence find \(\sum _ { r = 1 } ^ { n } \frac { 1 } { r ( r + 1 ) ( r + 2 ) }\). Deduce the value of \(\sum _ { r = 1 } ^ { \infty } \frac { 1 } { r ( r + 1 ) ( r + 2 ) }\).
Question 3:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(f(r-1)-f(r) = \frac{1}{r(r+1)} - \frac{1}{(r+1)(r+2)}\)M1 Proves initial result
\(= \frac{r+2-r}{r(r+1)(r+2)} = \frac{2}{r(r+1)(r+2)}\) (AG)A1
\(\displaystyle\sum_1^n \frac{1}{r(r+1)(r+2)} = \frac{1}{2}\left\{\frac{1}{1\times2} - \frac{1}{2\times3}\right\}\cdots\)M1A1 Sets up method of differences
\(+\frac{1}{2}\left\{\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)}\right\}\)
\(= \frac{1}{4} - \frac{1}{2}\left\{\frac{1}{(n+1)(n+2)}\right\}\) (OE)A1 Shows cancellation to get result
\(\therefore \displaystyle\sum_1^{\infty} \frac{1}{r(r+1)(r+2)} = \frac{1}{4}\)A1\(\sqrt{}\) States sum to infinity
Non-hence method (penalty of 1 mark):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{1}{r(r+1)(r-2)} = \frac{1}{2r} - \frac{1}{(r+1)} + \frac{1}{2(r+2)}\)(M1)
\(\frac{1}{2} - \frac{1}{2} + \frac{1}{4}\cdots + \frac{1}{2(n+1)} - \frac{1}{(n+1)} + \frac{1}{2(n+2)}\)(A1)
\(= \frac{1}{4} - \frac{1}{2}\left\{\frac{1}{(n+1)(n+2)}\right\}\) (OE)(A1) (3)
\(\therefore \displaystyle\sum_1^{\infty} \frac{1}{r(r+1)(r+2)} = \frac{1}{4}\)(A1\(\sqrt{}\)) (1)
Total: [6]
## Question 3:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(r-1)-f(r) = \frac{1}{r(r+1)} - \frac{1}{(r+1)(r+2)}$ | M1 | Proves initial result |
| $= \frac{r+2-r}{r(r+1)(r+2)} = \frac{2}{r(r+1)(r+2)}$ (AG) | A1 | |
| $\displaystyle\sum_1^n \frac{1}{r(r+1)(r+2)} = \frac{1}{2}\left\{\frac{1}{1\times2} - \frac{1}{2\times3}\right\}\cdots$ | M1A1 | Sets up method of differences |
| $+\frac{1}{2}\left\{\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)}\right\}$ | | |
| $= \frac{1}{4} - \frac{1}{2}\left\{\frac{1}{(n+1)(n+2)}\right\}$ (OE) | A1 | Shows cancellation to get result |
| $\therefore \displaystyle\sum_1^{\infty} \frac{1}{r(r+1)(r+2)} = \frac{1}{4}$ | A1$\sqrt{}$ | States sum to infinity |

**Non-hence method (penalty of 1 mark):**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{r(r+1)(r-2)} = \frac{1}{2r} - \frac{1}{(r+1)} + \frac{1}{2(r+2)}$ | (M1) | |
| $\frac{1}{2} - \frac{1}{2} + \frac{1}{4}\cdots + \frac{1}{2(n+1)} - \frac{1}{(n+1)} + \frac{1}{2(n+2)}$ | (A1) | |
| $= \frac{1}{4} - \frac{1}{2}\left\{\frac{1}{(n+1)(n+2)}\right\}$ (OE) | (A1) | (3) |
| $\therefore \displaystyle\sum_1^{\infty} \frac{1}{r(r+1)(r+2)} = \frac{1}{4}$ | (A1$\sqrt{}$) | (1) |

**Total: [6]**

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3 Given that $\mathrm { f } ( r ) = \frac { 1 } { ( r + 1 ) ( r + 2 ) }$, show that

$$\mathrm { f } ( r - 1 ) - \mathrm { f } ( r ) = \frac { 2 } { r ( r + 1 ) ( r + 2 ) }$$

Hence find $\sum _ { r = 1 } ^ { n } \frac { 1 } { r ( r + 1 ) ( r + 2 ) }$.

Deduce the value of $\sum _ { r = 1 } ^ { \infty } \frac { 1 } { r ( r + 1 ) ( r + 2 ) }$.

\hfill \mbox{\textit{CAIE FP1 2012 Q3 [6]}}
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Q1 5 Q2 5 Q3 6 Q4 9 Q5 9 Q6 9 Q7 10 Q8 11 Q9 11 Q10 11 Q11 EITHER Q11 OR