| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2021 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Proving standard summation formulae |
| Difficulty | Standard +0.8 This is a standard Further Maths method of differences question requiring binomial expansion, telescoping sum manipulation, and algebraic simplification to derive a summation formula. While mechanical, it demands careful algebraic handling across multiple steps and is more sophisticated than typical A-level questions, placing it moderately above average difficulty. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(n^5 - (n-1)^5 = n^5 - (n^5 - 5n^4 + 10n^3 - 10n^2 + 5n - 1) = \ldots\) | M1 | Starts proof by expanding the bracket |
| \(5n^4 - 10n^3 + 10n^2 - 5n + 1\) | A1* | Correct proof with no errors. Full expansion of \((n-1)^5\) must be shown |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| Writing out telescoping sum from \(r=1\) to \(n\), minimum 3 lines shown, leading to: \(n^5 = 5\sum_{r=1}^{n}r^4 - 10\sum_{r=1}^{n}r^3 + 10\sum_{r=1}^{n}r^2 - 5\sum_{r=1}^{n}r + n\) | M1A1 | M1: Applies result from (a) between 1 and \(n\) and sums both sides, min 3 lines shown. A1: Correct equation — if only last line seen, award M1A1. Marks can be implied by correct following stage |
| \(n^5 = 5\sum_{r=1}^{n}r^4 - 10 \times \frac{1}{4}n^2(n+1)^2 + 10 \times \frac{1}{6}n(n+1)(2n+1) - 5 \times \frac{1}{2}n(n+1) + n\) | M1A1 | M1: Introduces at least 2 correct summation formulae. A1: Correct equation |
| \(5\sum_{r=1}^{n}r^4 = n(n+1)\left[\frac{5}{2}n(n+1) - \frac{5}{3}(2n+1) + \frac{5}{2} + n^3 - n^2 + n - 1\right]\) | M1 | Makes \(5\sum r^4\) or \(\sum r^4\) the subject and takes out a factor of \(n(n+1)\) |
| \(\sum_{r=1}^{n}r^4 = \frac{1}{30}n(n+1)\left[15n(n+1) - 10(2n+1) + 15 + 6(n^3 - n^2 + n - 1)\right]\) \(= \frac{1}{30}n(n+1)\left[6n^3 + 9n^2 + n - 1\right] = \frac{1}{30}n(n+1)(2n+1)(\ldots)\) | dM1 | Takes out factor of \(n(n+1)(2n+1)\). Depends on all previous method marks |
| \(= \dfrac{1}{30}n(n+1)(2n+1)(3n^2+3n-1)\) | A1 | cao |
## Question 9:
### Part (a):
| Working | Mark | Notes |
|---------|------|-------|
| $n^5 - (n-1)^5 = n^5 - (n^5 - 5n^4 + 10n^3 - 10n^2 + 5n - 1) = \ldots$ | M1 | Starts proof by expanding the bracket |
| $5n^4 - 10n^3 + 10n^2 - 5n + 1$ | A1* | Correct proof with no errors. Full expansion of $(n-1)^5$ must be shown |
**Total: (2)**
---
### Part (b):
| Working | Mark | Notes |
|---------|------|-------|
| Writing out telescoping sum from $r=1$ to $n$, minimum 3 lines shown, leading to: $n^5 = 5\sum_{r=1}^{n}r^4 - 10\sum_{r=1}^{n}r^3 + 10\sum_{r=1}^{n}r^2 - 5\sum_{r=1}^{n}r + n$ | M1A1 | M1: Applies result from (a) between 1 and $n$ and sums **both** sides, min 3 lines shown. A1: Correct equation — if only last line seen, award M1A1. Marks can be implied by correct following stage |
| $n^5 = 5\sum_{r=1}^{n}r^4 - 10 \times \frac{1}{4}n^2(n+1)^2 + 10 \times \frac{1}{6}n(n+1)(2n+1) - 5 \times \frac{1}{2}n(n+1) + n$ | M1A1 | M1: Introduces at least 2 correct summation formulae. A1: Correct equation |
| $5\sum_{r=1}^{n}r^4 = n(n+1)\left[\frac{5}{2}n(n+1) - \frac{5}{3}(2n+1) + \frac{5}{2} + n^3 - n^2 + n - 1\right]$ | M1 | Makes $5\sum r^4$ or $\sum r^4$ the subject and takes out a factor of $n(n+1)$ |
| $\sum_{r=1}^{n}r^4 = \frac{1}{30}n(n+1)\left[15n(n+1) - 10(2n+1) + 15 + 6(n^3 - n^2 + n - 1)\right]$ $= \frac{1}{30}n(n+1)\left[6n^3 + 9n^2 + n - 1\right] = \frac{1}{30}n(n+1)(2n+1)(\ldots)$ | dM1 | Takes out factor of $n(n+1)(2n+1)$. Depends on all previous method marks |
| $= \dfrac{1}{30}n(n+1)(2n+1)(3n^2+3n-1)$ | A1 | cao |
**Total: (7)**
**Question Total: 9**\begin{enumerate}
\item (a) Show that
\end{enumerate}
$$n ^ { 5 } - ( n - 1 ) ^ { 5 } \equiv 5 n ^ { 4 } - 10 n ^ { 3 } + 10 n ^ { 2 } - 5 n + 1$$
(b) Hence, using the method of differences, show that for all integer values of $n$,
$$\sum _ { r = 1 } ^ { n } r ^ { 4 } = \frac { 1 } { 30 } n ( n + 1 ) ( 2 n + 1 ) \left( a n ^ { 2 } + b n + c \right)$$
where $a$, $b$ and $c$ are integers to be determined.
\hfill \mbox{\textit{Edexcel F2 2021 Q9 [9]}}