5 Use the result \(\frac { 1 } { 5 r - 1 } - \frac { 1 } { 5 r + 4 } \equiv \frac { 5 } { ( 5 r - 1 ) ( 5 r + 4 ) }\) and the method of differences to find
$$\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 5 r - 1 ) ( 5 r + 4 ) }$$
simplifying your answer.
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Question 5:
Answer Marks
Guidance
Answer Mark
Guidance
\(\sum_{r=1}^{n} \frac{1}{(5r-1)(5r+4)} = \frac{1}{5}\sum_{r=1}^{n}\left(\frac{1}{5r-1} - \frac{1}{5r+4}\right)\) M1
Attempt to use identity – may be implied
\(= \frac{1}{5}\left(\left(\frac{1}{4}-\frac{1}{9}\right)+\left(\frac{1}{9}-\frac{1}{14}\right)+\cdots+\left(\frac{1}{5n-1}-\frac{1}{5n+4}\right)\right)\) A1
Terms in full (at least first and last)
\(= \frac{1}{5}\left(\frac{1}{4} - \frac{1}{5n+4}\right) = \frac{1}{5}\left(\frac{5n+4-4}{4(5n+4)}\right) = \frac{n}{4(5n+4)}\) M1
Attempt at cancelling
A1 \(\left(\frac{1}{4} - \frac{1}{5n+4}\right)\)
A1 factor of \(\frac{1}{5}\)
A1 [6] Correct answer as a single algebraic fraction
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# Question 5:
| Answer | Mark | Guidance |
|--------|------|----------|
| $\sum_{r=1}^{n} \frac{1}{(5r-1)(5r+4)} = \frac{1}{5}\sum_{r=1}^{n}\left(\frac{1}{5r-1} - \frac{1}{5r+4}\right)$ | M1 | Attempt to use identity – may be implied |
| $= \frac{1}{5}\left(\left(\frac{1}{4}-\frac{1}{9}\right)+\left(\frac{1}{9}-\frac{1}{14}\right)+\cdots+\left(\frac{1}{5n-1}-\frac{1}{5n+4}\right)\right)$ | A1 | Terms in full (at least first and last) |
| $= \frac{1}{5}\left(\frac{1}{4} - \frac{1}{5n+4}\right) = \frac{1}{5}\left(\frac{5n+4-4}{4(5n+4)}\right) = \frac{n}{4(5n+4)}$ | M1 | Attempt at cancelling |
| | A1 | $\left(\frac{1}{4} - \frac{1}{5n+4}\right)$ |
| | A1 | factor of $\frac{1}{5}$ |
| | A1 **[6]** | Correct answer as a single algebraic fraction |
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5 Use the result $\frac { 1 } { 5 r - 1 } - \frac { 1 } { 5 r + 4 } \equiv \frac { 5 } { ( 5 r - 1 ) ( 5 r + 4 ) }$ and the method of differences to find
$$\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 5 r - 1 ) ( 5 r + 4 ) }$$
simplifying your answer.
\hfill \mbox{\textit{OCR MEI FP1 2010 Q5 [6]}}