| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2017 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Sum from n+1 to 2n or similar range |
| Difficulty | Challenging +1.2 Part (a) is routine algebraic verification. Part (b) is a standard telescoping series proof using the result from (a). Part (c) requires recognizing that the sum from n to 3n can be expressed using the formula from (b) with appropriate substitutions and factoring out constants, but follows a predictable pattern once the telescoping structure is understood. This is moderately above average for FP2 due to the multi-step nature and the non-standard summation range in part (c), but remains a fairly standard Further Maths series question. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{r^2} - \frac{1}{(r+1)^2} = \frac{(r+1)^2 - r^2}{r^2(r+1)^2} = \frac{2r+1}{r^2(r+1)^2}\) | B1 | Correct proof (minimum as shown); \((r+1)^2\) or \(r^2+2r+1\). Can be worked in either direction. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sum_{r=1}^{n}\left(\frac{1}{r^2} - \frac{1}{(r+1)^2}\right) = 1 - \frac{1}{4} + \frac{1}{4} - \frac{1}{9} + \ldots + \left(\frac{1}{n^2}\right) - \frac{1}{(n+1)^2}\) | M1 | Terms with \(r=1\), \(r=n\) and one of \(r=2\), \(r=n-1\) should be shown |
| \(1 - \frac{1}{(n+1)^2}\) | A1 | Extracts correct terms that do not cancel |
| \(\frac{(n+1)^2 - 1}{(n+1)^2} = \frac{n(n+2)}{(n+1)^2}\) | A1*cso | Correct completion with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sum_{r=n}^{3n}\frac{6r+3}{r^2(r+1)^2} = 3\left(\frac{3n(3n+2)}{(3n+1)^2} - \frac{(n-1)(n+1)}{n^2}\right)\) | M1 | Attempts to use \(f(3n) - (f(n-1)\) or \(f(n))\); 3 may be missing |
| \(= 3\left(\frac{3n^3(3n+2) - (3n+1)^2(n^2-1)}{n^2(3n+1)^2}\right)\) | dM1 | Attempt at common denominator; denom to be \(n^2(3n+1)^2\) or \((n+1)^2(3n+1)^2\); numerator to be difference of 2 quartics; 3 may be missing |
| \(= \frac{24n^2 + 18n + 3}{n^2(3n+1)^2}\) | A1cao | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sum_{r=n}^{3n}\frac{6r+3}{r^2(r+1)^2} = 3\left(\frac{1}{n^2} - \frac{1}{(3n+1)^2}\right)\) or \(3\left(\frac{1}{(n+1)^2} - \frac{1}{(3n+1)^2}\right)\) | M1 | Attempts the difference of 2 terms (either difference accepted); 3 may be missing |
| \(= 3\left(\frac{(3n+1)^2 - n^2}{n^2(3n+1)^2}\right)\) | dM1 | Valid attempt at common denominator for their fractions; 3 may be missing |
| \(= \frac{24n^2 + 18n + 3}{n^2(3n+1)^2}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Assume true for \(n = k\) | ||
| \(\sum_{r=1}^{k+1} \frac{2r+1}{r^2(r+1)^2} = \frac{k(k+2)}{(k+1)^2} + \frac{1}{(k+1)^2} - \frac{1}{(k+2)^2}\) | M1 | Uses \(\sum_{r=1}^{k}\) together with the \((k+1)\)th term as 2 fractions (see part (a)) |
| \(= \frac{k^2+2k+1}{(k+1)^2} - \frac{1}{(k+2)^2}\) | ||
| \(1 - \frac{1}{(k+2)^2} = \frac{k^2+4k+3}{(k+2)^2} = \frac{(k+1)(k+3)}{(k+2)^2}\) | A1 | Combines the 3 fractions to obtain a single fraction. Must be correct but numerator need not be factorised. |
| Show true for \(n=1\) | This must be seen somewhere | |
| Hence proved by induction | A1 | Complete proof with no errors and a concluding statement |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{r^2} - \frac{1}{(r+1)^2} = \frac{(r+1)^2 - r^2}{r^2(r+1)^2} = \frac{2r+1}{r^2(r+1)^2}$ | B1 | Correct proof (minimum as shown); $(r+1)^2$ or $r^2+2r+1$. Can be worked in either direction. |
**Total: (1)**
---
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sum_{r=1}^{n}\left(\frac{1}{r^2} - \frac{1}{(r+1)^2}\right) = 1 - \frac{1}{4} + \frac{1}{4} - \frac{1}{9} + \ldots + \left(\frac{1}{n^2}\right) - \frac{1}{(n+1)^2}$ | M1 | Terms with $r=1$, $r=n$ and one of $r=2$, $r=n-1$ should be shown |
| $1 - \frac{1}{(n+1)^2}$ | A1 | Extracts correct terms that do not cancel |
| $\frac{(n+1)^2 - 1}{(n+1)^2} = \frac{n(n+2)}{(n+1)^2}$ | A1*cso | Correct completion with no errors |
**Total: (3)**
---
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sum_{r=n}^{3n}\frac{6r+3}{r^2(r+1)^2} = 3\left(\frac{3n(3n+2)}{(3n+1)^2} - \frac{(n-1)(n+1)}{n^2}\right)$ | M1 | Attempts to use $f(3n) - (f(n-1)$ or $f(n))$; 3 may be missing |
| $= 3\left(\frac{3n^3(3n+2) - (3n+1)^2(n^2-1)}{n^2(3n+1)^2}\right)$ | dM1 | Attempt at common denominator; denom to be $n^2(3n+1)^2$ or $(n+1)^2(3n+1)^2$; numerator to be difference of 2 quartics; 3 may be missing |
| $= \frac{24n^2 + 18n + 3}{n^2(3n+1)^2}$ | A1cao | cao |
**Total: (3)**
---
## Alternative for Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sum_{r=n}^{3n}\frac{6r+3}{r^2(r+1)^2} = 3\left(\frac{1}{n^2} - \frac{1}{(3n+1)^2}\right)$ or $3\left(\frac{1}{(n+1)^2} - \frac{1}{(3n+1)^2}\right)$ | M1 | Attempts the difference of 2 terms (either difference accepted); 3 may be missing |
| $= 3\left(\frac{(3n+1)^2 - n^2}{n^2(3n+1)^2}\right)$ | dM1 | Valid attempt at common denominator for their fractions; 3 may be missing |
| $= \frac{24n^2 + 18n + 3}{n^2(3n+1)^2}$ | A1 | cao |
> **Note:** If (b) and/or (c) are worked with $r$ instead of $n$, do **NOT** award the final A mark for the parts affected. This applies even if $r$ is changed to $n$ at the end.
**Question 1 Total: 7**
# Question 1 (Induction Alternative for part b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Assume true for $n = k$ | | |
| $\sum_{r=1}^{k+1} \frac{2r+1}{r^2(r+1)^2} = \frac{k(k+2)}{(k+1)^2} + \frac{1}{(k+1)^2} - \frac{1}{(k+2)^2}$ | M1 | Uses $\sum_{r=1}^{k}$ together with the $(k+1)$th term as 2 fractions (see part (a)) |
| $= \frac{k^2+2k+1}{(k+1)^2} - \frac{1}{(k+2)^2}$ | | |
| $1 - \frac{1}{(k+2)^2} = \frac{k^2+4k+3}{(k+2)^2} = \frac{(k+1)(k+3)}{(k+2)^2}$ | A1 | Combines the 3 fractions to obtain a single fraction. Must be correct but numerator need not be factorised. |
| Show true for $n=1$ | | This must be seen somewhere |
| Hence proved by induction | A1 | Complete proof with no errors and a concluding statement |
---\begin{enumerate}
\item (a) Show that, for $r > 0$
\end{enumerate}
$$\frac { 1 } { r ^ { 2 } } - \frac { 1 } { ( r + 1 ) ^ { 2 } } \equiv \frac { 2 r + 1 } { r ^ { 2 } ( r + 1 ) ^ { 2 } }$$
(b) Hence prove that, for $n \in \mathbb { N }$
$$\sum _ { r = 1 } ^ { n } \frac { 2 r + 1 } { r ^ { 2 } ( r + 1 ) ^ { 2 } } = \frac { n ( n + 2 ) } { ( n + 1 ) ^ { 2 } }$$
(c) Show that, for $n \in \mathbb { N } , n > 1$
$$\sum _ { r = n } ^ { 3 n } \frac { 6 r + 3 } { r ^ { 2 } ( r + 1 ) ^ { 2 } } = \frac { a n ^ { 2 } + b n + c } { n ^ { 2 } ( 3 n + 1 ) ^ { 2 } }$$
where $a , b$ and $c$ are constants to be found.
\hfill \mbox{\textit{Edexcel FP2 2017 Q1 [7]}}