| Exam Board | OCR MEI |
|---|
| Module | FP1 (Further Pure Mathematics 1) |
|---|
| Year | 2013 |
|---|
| Session | June |
|---|
| Topic | Sequences and series, recurrence and convergence |
|---|
5 You are given that \(\frac { 4 } { ( 4 n - 3 ) ( 4 n + 1 ) } \equiv \frac { 1 } { 4 n - 3 } - \frac { 1 } { 4 n + 1 }\). Use the method of differences to show that
$$\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 4 r - 3 ) ( 4 r + 1 ) } = \frac { n } { 4 n + 1 }$$