| Exam Board | OCR MEI |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2010 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Infinite series convergence and sum |
| Difficulty | Standard +0.3 This is a standard method of differences question with clear scaffolding. Part (i) is routine verification by combining fractions, part (ii) applies the telescoping technique directly, part (iii) is immediate limit evaluation, and part (iv) uses subtraction of partial sums. All steps follow established procedures with no novel insight required, making it slightly easier than average. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{2}{r}-\frac{3}{r+1}+\frac{1}{r+2} = \frac{2(r+1)(r+2)-3r(r+2)+r(r+1)}{r(r+1)(r+2)}\) | M1 | Attempt a common denominator |
| \(= \frac{2r^2+6r+4-3r^2-6r+r^2+r}{r(r+1)(r+2)} = \frac{4+r}{r(r+1)(r+2)}\) | A1 [2] | Convincingly shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\sum_{r=1}^{n}\frac{4+r}{r(r+1)(r+2)} = \sum_{r=1}^{n}\left[\frac{2}{r}-\frac{3}{r+1}+\frac{1}{r+2}\right]\) | M1 | Use of given result (may be implied) |
| Writing terms in full (at least first and one other) | M1 | |
| At least 3 consecutive terms correct | A2 | −1 each error |
| Attempt to cancel, including algebraic terms | M1 | |
| \(= \frac{3}{2}-\frac{2}{n+1}+\frac{1}{n+2}\) | A1 [6] | Convincingly shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{3}{2}\) | B1 [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\sum_{r=50}^{100}\frac{4+r}{r(r+1)(r+2)} = \sum_{r=1}^{100}\frac{4+r}{r(r+1)(r+2)} - \sum_{r=1}^{49}\frac{4+r}{r(r+1)(r+2)}\) | M1 | Splitting into two parts |
| \(= \left(\frac{3}{2}-\frac{2}{101}+\frac{1}{102}\right)-\left(\frac{3}{2}-\frac{2}{50}+\frac{1}{51}\right)\) | M1 | Use of result from (ii) |
| \(= 0.0104\) (3 s.f.) | A1 [3] | c.a.o. |
# Question 9:
**Part (i):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{2}{r}-\frac{3}{r+1}+\frac{1}{r+2} = \frac{2(r+1)(r+2)-3r(r+2)+r(r+1)}{r(r+1)(r+2)}$ | M1 | Attempt a common denominator |
| $= \frac{2r^2+6r+4-3r^2-6r+r^2+r}{r(r+1)(r+2)} = \frac{4+r}{r(r+1)(r+2)}$ | A1 **[2]** | Convincingly shown |
**Part (ii):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\sum_{r=1}^{n}\frac{4+r}{r(r+1)(r+2)} = \sum_{r=1}^{n}\left[\frac{2}{r}-\frac{3}{r+1}+\frac{1}{r+2}\right]$ | M1 | Use of given result (may be implied) |
| Writing terms in full (at least first and one other) | M1 | |
| At least 3 consecutive terms correct | A2 | −1 each error |
| Attempt to cancel, including algebraic terms | M1 | |
| $= \frac{3}{2}-\frac{2}{n+1}+\frac{1}{n+2}$ | A1 **[6]** | Convincingly shown |
**Part (iii):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{3}{2}$ | B1 **[1]** | |
**Part (iv):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\sum_{r=50}^{100}\frac{4+r}{r(r+1)(r+2)} = \sum_{r=1}^{100}\frac{4+r}{r(r+1)(r+2)} - \sum_{r=1}^{49}\frac{4+r}{r(r+1)(r+2)}$ | M1 | Splitting into two parts |
| $= \left(\frac{3}{2}-\frac{2}{101}+\frac{1}{102}\right)-\left(\frac{3}{2}-\frac{2}{50}+\frac{1}{51}\right)$ | M1 | Use of result from (ii) |
| $= 0.0104$ (3 s.f.) | A1 **[3]** | c.a.o. |9 (i) Verify that $\frac { 4 + r } { r ( r + 1 ) ( r + 2 ) } = \frac { 2 } { r } - \frac { 3 } { r + 1 } + \frac { 1 } { r + 2 }$.\\
(ii) Use the method of differences to show that
$$\sum _ { r = 1 } ^ { n } \frac { 4 + r } { r ( r + 1 ) ( r + 2 ) } = \frac { 3 } { 2 } - \frac { 2 } { n + 1 } + \frac { 1 } { n + 2 } .$$
(iii) Write down the limit to which $\sum _ { r = 1 } ^ { n } \frac { 4 + r } { r ( r + 1 ) ( r + 2 ) }$ converges as $n$ tends to infinity.\\
(iv) Find $\sum _ { r = 50 } ^ { 100 } \frac { 4 + r } { r ( r + 1 ) ( r + 2 ) }$, giving your answer to 3 significant figures.
\hfill \mbox{\textit{OCR MEI FP1 2010 Q9 [12]}}