Standard +0.3 This is a straightforward application of the method of differences with the identity already provided. Students need only write out telescoping terms and simplify the remaining first and last terms—a standard Further Maths technique requiring minimal problem-solving beyond pattern recognition and algebraic manipulation.
4 Use the identity \(\frac { 1 } { 2 r + 3 } - \frac { 1 } { 2 r + 5 } \equiv \frac { 2 } { ( 2 r + 3 ) ( 2 r + 5 ) }\) and the method of differences to find \(\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 2 r + 3 ) ( 2 r + 5 ) }\), expressing your answer as a single fraction.
Cancel to first minus last term, must be in terms of \(n\), oe single fraction
## Question 4:
$\displaystyle\sum_{r=1}^{n}\frac{1}{(2r+3)(2r+5)} = \frac{1}{2}\sum_{r=1}^{n}\left[\frac{1}{2r+3}-\frac{1}{2r+5}\right]$ | M1 | Split to partial fractions. Allow missing $\frac{1}{2}$
$= \frac{1}{2}\left[\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{7}\cdots\right)+\cdots+\left(\cdots-\frac{1}{2n+5}\right)\right]$ | M1 A1 | Expand to show pattern of cancelling, at least 4 fractions. All correct, allow missing $\frac{1}{2}$, condone $r$
$= \frac{1}{2}\left[\frac{1}{5}-\frac{1}{2n+5}\right] = \dfrac{n}{5(2n+5)}$ | M1 A1 **[5]** | Cancel to first minus last term, must be in terms of $n$, oe single fraction
---
4 Use the identity $\frac { 1 } { 2 r + 3 } - \frac { 1 } { 2 r + 5 } \equiv \frac { 2 } { ( 2 r + 3 ) ( 2 r + 5 ) }$ and the method of differences to find $\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 2 r + 3 ) ( 2 r + 5 ) }$, expressing your answer as a single fraction.
\hfill \mbox{\textit{OCR MEI FP1 2014 Q4 [5]}}