| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 2 (Further Pure Core 2) |
| Year | 2018 |
| Session | September |
| Marks | 8 |
| Topic | Sequences and series, recurrence and convergence |
| Type | Proving standard summation formulae |
| Difficulty | Challenging +1.2 This is a standard Further Maths technique for deriving summation formulas using telescoping sums, followed by a routine application. Part (i) requires expanding (r+1)^5, collecting terms, and algebraic manipulation over 6 marks—methodical but not conceptually difficult for FM students. Part (ii) is straightforward substitution. While above average A-level difficulty due to being FM content with algebraic complexity, it's a textbook exercise without novel insight required. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum_{r=1}^{n} ((r+1)^5 - r^5) = (n+1)^5 - 1\) | B1 | |
| \((r+1)^5 - r^5 = 5r^4 + 10r^3 + 10r^2 + 5r + 1\) | M1 | Binomial expansion of \((r+1)^5\) with \(r^5\) cancelling |
| Use of \(\sum_{r=1}^{n} 1 = n\) and \(\sum_{r=1}^{n} r = \frac{n(n+1)}{2}\) | B1 | |
| \((n+1)^5 - 1 = 5\sum r^4 + \frac{5n^2(n+1)^2}{2} + \frac{5n(n+1)(2n+1)}{3} + \frac{5n(n+1)}{2} + n\) | M1 | Correct cancellation of sum down to two terms and substitution |
| \(30\sum_{r=1}^{n} r^4 = 6n(n^4 + 5n^3 + 10n^2 + 10n + 4) - n(n+1)(15n(n+1) + 15 + 10(2n+1))\) | M1 | Expansion and rearrangement leading to (eg) \(n\) as being a common factor |
| Further factorisation leading to \(\sum_{r=1}^{n} r^4 = \frac{1}{30}n(n+1)(2n+1)(3n^2 + 3n - 1)\) | A1 | No gaps in argument for A1 AG or both expansions correctly and completely demonstrated as equal www. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum_{r=1}^{30} r^4 - \sum_{r=1}^{20} r^4\) | M1 | soi |
| \((676010664 - 59416665) = 616593999\) | A1 |
## (i)
$\sum_{r=1}^{n} ((r+1)^5 - r^5) = (n+1)^5 - 1$ | B1 |
$(r+1)^5 - r^5 = 5r^4 + 10r^3 + 10r^2 + 5r + 1$ | M1 | Binomial expansion of $(r+1)^5$ with $r^5$ cancelling
Use of $\sum_{r=1}^{n} 1 = n$ and $\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$ | B1 |
$(n+1)^5 - 1 = 5\sum r^4 + \frac{5n^2(n+1)^2}{2} + \frac{5n(n+1)(2n+1)}{3} + \frac{5n(n+1)}{2} + n$ | M1 | Correct cancellation of sum down to two terms and substitution
$30\sum_{r=1}^{n} r^4 = 6n(n^4 + 5n^3 + 10n^2 + 10n + 4) - n(n+1)(15n(n+1) + 15 + 10(2n+1))$ | M1 | Expansion and rearrangement leading to (eg) $n$ as being a common factor
Further factorisation leading to $\sum_{r=1}^{n} r^4 = \frac{1}{30}n(n+1)(2n+1)(3n^2 + 3n - 1)$ | A1 | No gaps in argument for A1 AG or both expansions correctly and completely demonstrated as equal www.
[6]
## (ii)
$\sum_{r=1}^{30} r^4 - \sum_{r=1}^{20} r^4$ | M1 | soi
$(676010664 - 59416665) = 616593999$ | A1 |
[2]
---\begin{enumerate}[label=(\roman*)]
\item By considering $\sum_{r=1}^n ((r+1)^5 - r^5)$ show that $\sum_{r=1}^n r^4 = \frac{1}{30}n(n+1)(2n+1)(3n^2+3n-1)$. [6]
\item Use the formula given in part (i) to find $50^4 + 51^4 + \ldots + 80^4$. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 2 2018 Q6 [8]}}