5 Given that \(\frac { 3 } { ( 3 r - 1 ) ( 3 r + 2 ) } \equiv \frac { 1 } { 3 r - 1 } - \frac { 1 } { 3 r + 2 }\), find \(\sum _ { r = 1 } ^ { 20 } \frac { 1 } { ( 3 r - 1 ) ( 3 r + 2 ) }\), giving your answer as an exact fraction.
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Question 5:
Answer Marks
Guidance
Answer Mark
Guidance
\(\sum_{r=1}^{20} \frac{1}{(3r-1)(3r+2)} = \frac{1}{3}\sum_{r=1}^{20}\left[\frac{1}{3r-1} - \frac{1}{3r+2}\right]\) M1
Attempt to use identity – may be implied
A1 Correct use of \(\frac{1}{3}\) seen
\(= \frac{1}{3}\left[\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{8}\right)+\cdots+\left(\frac{1}{59}-\frac{1}{62}\right)\right]\) A1
Terms in full (at least first and last)
M1 Attempt at cancelling
\(= \frac{1}{3}\left(\frac{1}{2}-\frac{1}{62}\right) = \frac{5}{31}\) A1
c.a.o.
[5]
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# Question 5:
| Answer | Mark | Guidance |
|--------|------|----------|
| $\sum_{r=1}^{20} \frac{1}{(3r-1)(3r+2)} = \frac{1}{3}\sum_{r=1}^{20}\left[\frac{1}{3r-1} - \frac{1}{3r+2}\right]$ | M1 | Attempt to use identity – may be implied |
| | A1 | Correct use of $\frac{1}{3}$ seen |
| $= \frac{1}{3}\left[\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{8}\right)+\cdots+\left(\frac{1}{59}-\frac{1}{62}\right)\right]$ | A1 | Terms in full (at least first and last) |
| | M1 | Attempt at cancelling |
| $= \frac{1}{3}\left(\frac{1}{2}-\frac{1}{62}\right) = \frac{5}{31}$ | A1 | c.a.o. |
| | **[5]** | |
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5 Given that $\frac { 3 } { ( 3 r - 1 ) ( 3 r + 2 ) } \equiv \frac { 1 } { 3 r - 1 } - \frac { 1 } { 3 r + 2 }$, find $\sum _ { r = 1 } ^ { 20 } \frac { 1 } { ( 3 r - 1 ) ( 3 r + 2 ) }$, giving your answer as an exact fraction.
\hfill \mbox{\textit{OCR MEI FP1 2011 Q5 [5]}}