5 Use the method of differences to show that \(\sum _ { r = 1 } ^ { N } \frac { 1 } { ( 2 r + 1 ) ( 2 r + 3 ) } = \frac { 1 } { 6 } - \frac { 1 } { 2 ( 2 N + 3 ) }\).
Deduce that \(\sum _ { r = N + 1 } ^ { 2 N } \frac { 1 } { ( 2 r + 1 ) ( 2 r + 3 ) } < \frac { 1 } { 8 N }\).
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Question 5:
Answer Marks
Guidance
Answer/Working Marks
Guidance
\(\frac{1}{(2r+1)(2r+3)} = \frac{1}{2}\left\{\frac{1}{2r+1}-\frac{1}{2r+3}\right\}\) M1A1
Finds partial fractions
\(= \frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}\right)+\ldots+\frac{1}{2}\left(\frac{1}{2N+1}-\frac{1}{2N+3}\right)\) M1A1
Expresses terms as differences
\(= \frac{1}{6} - \frac{1}{2(2N+3)}\) (AG) A1
Shows cancellation
\(\sum_{N+1}^{2N} = \left(\frac{1}{6}-\frac{1}{2(4N+3)}\right)-\left(\frac{1}{6}-\frac{1}{2(2N+3)}\right)\) M1
Uses \(\sum_{N+1}^{2N}=\sum_1^{2N}-\sum_1^N\)
\(= \frac{1}{2}\left(\frac{1}{2N+3}-\frac{1}{4N+3}\right)\) A1
Applies result
\(= \frac{N}{(2N+3)(4N+3)}\) M1
Simplifies
\(< \frac{N}{2N\cdot4N} = \frac{1}{8N}\) (AG) A1
Deduces inequality
Total: [9]
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## Question 5:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{(2r+1)(2r+3)} = \frac{1}{2}\left\{\frac{1}{2r+1}-\frac{1}{2r+3}\right\}$ | M1A1 | Finds partial fractions |
| $= \frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}\right)+\ldots+\frac{1}{2}\left(\frac{1}{2N+1}-\frac{1}{2N+3}\right)$ | M1A1 | Expresses terms as differences |
| $= \frac{1}{6} - \frac{1}{2(2N+3)}$ (AG) | A1 | Shows cancellation |
| $\sum_{N+1}^{2N} = \left(\frac{1}{6}-\frac{1}{2(4N+3)}\right)-\left(\frac{1}{6}-\frac{1}{2(2N+3)}\right)$ | M1 | Uses $\sum_{N+1}^{2N}=\sum_1^{2N}-\sum_1^N$ |
| $= \frac{1}{2}\left(\frac{1}{2N+3}-\frac{1}{4N+3}\right)$ | A1 | Applies result |
| $= \frac{N}{(2N+3)(4N+3)}$ | M1 | Simplifies |
| $< \frac{N}{2N\cdot4N} = \frac{1}{8N}$ (AG) | A1 | Deduces inequality |
**Total: [9]**
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5 Use the method of differences to show that $\sum _ { r = 1 } ^ { N } \frac { 1 } { ( 2 r + 1 ) ( 2 r + 3 ) } = \frac { 1 } { 6 } - \frac { 1 } { 2 ( 2 N + 3 ) }$.
Deduce that $\sum _ { r = N + 1 } ^ { 2 N } \frac { 1 } { ( 2 r + 1 ) ( 2 r + 3 ) } < \frac { 1 } { 8 N }$.
\hfill \mbox{\textit{CAIE FP1 2013 Q5 [9]}}