| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2017 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Infinite series convergence and sum |
| Difficulty | Standard +0.8 This is a telescoping series question requiring algebraic verification, summation using differences, and limit evaluation. While the method is signposted ('verify', 'hence', 'deduce'), it requires careful algebraic manipulation of complex fractions and understanding of telescoping series—more demanding than standard A-level but routine for Further Maths students who have practiced this technique. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{RHS} = \frac{1}{2}\left\{\frac{4r^3+8r^2+3r-(4r^2-1)(r+2)}{r(r+1)(r+2)}\right\}\) | M1 | |
| \(= \frac{1}{2}\left\{\frac{4r+2}{r(r+1)(r+2)}\right\} = \frac{(2r+1)}{r(r+1)(r+2)}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Sum to \(n\) terms shown via telescoping series | M1 | |
| \(= \frac{1}{2}\left\{\frac{(2n+1)(2n+3)}{(n+1)(n+2)} - \frac{3}{2}\right\}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(S_\infty = \frac{1}{2} \times 4 - \frac{3}{4} = 1\frac{1}{4}\) | M1A1 |
## Question 2:
**Part (i):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{RHS} = \frac{1}{2}\left\{\frac{4r^3+8r^2+3r-(4r^2-1)(r+2)}{r(r+1)(r+2)}\right\}$ | M1 | |
| $= \frac{1}{2}\left\{\frac{4r+2}{r(r+1)(r+2)}\right\} = \frac{(2r+1)}{r(r+1)(r+2)}$ | A1 | AG |
**Part (ii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Sum to $n$ terms shown via telescoping series | M1 | |
| $= \frac{1}{2}\left\{\frac{(2n+1)(2n+3)}{(n+1)(n+2)} - \frac{3}{2}\right\}$ | A1 | AG |
**Part (iii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $S_\infty = \frac{1}{2} \times 4 - \frac{3}{4} = 1\frac{1}{4}$ | M1A1 | |
---2 (i) Verify that $\frac { 2 r + 1 } { r ( r + 1 ) ( r + 2 ) } = \frac { 1 } { 2 } \left\{ \frac { ( 2 r + 1 ) ( 2 r + 3 ) } { ( r + 1 ) ( r + 2 ) } - \frac { ( 2 r - 1 ) ( 2 r + 1 ) } { r ( r + 1 ) } \right\}$.\\
(ii) Hence show that $\sum _ { r = 1 } ^ { n } \frac { 2 r + 1 } { r ( r + 1 ) ( r + 2 ) } = \frac { 1 } { 2 } \left\{ \frac { ( 2 n + 1 ) ( 2 n + 3 ) } { ( n + 1 ) ( n + 2 ) } - \frac { 3 } { 2 } \right\}$.\\
(iii) Deduce the value of $\sum _ { r = 1 } ^ { \infty } \frac { 2 r + 1 } { r ( r + 1 ) ( r + 2 ) }$.\\
\hfill \mbox{\textit{CAIE FP1 2017 Q2 [6]}}