Edexcel FP1 Specimen — Question 5 9 marks

Exam BoardEdexcel
ModuleFP1 (Further Pure Mathematics 1)
SessionSpecimen
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeSum from n+1 to 2n or similar range
DifficultyStandard +0.3 Part (a) requires standard application of sum formulas for r² and r, then algebraic simplification—routine for FP1. Part (b) uses the 'hence' to subtract cumulative sums, which is a straightforward technique once part (a) is established. This is a typical textbook exercise testing formula manipulation rather than problem-solving insight.
Spec4.06a Summation formulae: sum of r, r^2, r^3
5. (a) Show that \(\sum _ { r = 1 } ^ { n } \left( r ^ { 2 } - r - 1 \right) = \frac { 1 } { 3 } ( n - 2 ) n ( n + 2 )\).
(b) Hence calculate the value of \(\sum _ { r = 10 } ^ { 40 } \left( r ^ { 2 } - r - 1 \right)\).
Question 5:
Part (a):
AnswerMarks Guidance
Working/AnswerMarks Guidance
\(\sum_{r=1}^{n}(r^2-r-1) = \sum_{r=1}^{n}r^2 - \sum_{r=1}^{n}r - \sum_{r=1}^{n}1\)M1 Splitting the sum
\(\sum_{r=1}^{n}1 = n\)B1
\(\sum_{r=1}^{n}(r^2-r-1) = \frac{n}{6}(n+1)(2n+1) - \frac{1}{2}n(n+1) - n\)M1 Substituting standard formulae
\(= \frac{n}{6}(2n^2-8)\)M1 A1 Simplification
\(= \frac{1}{3}(n-2)n(n+2)\)A1 (6 marks)
Part (b):
AnswerMarks Guidance
Working/AnswerMarks Guidance
\(\sum_{r=10}^{40}(r^2-r-1) = \sum_{r=1}^{40}(r^2-r-1) - \sum_{r=1}^{9}(r^2-r-1)\)M1
\(= \frac{1}{3}\times 38\times 40\times 42 - \frac{1}{3}\times 7\times 9\times 11 = 21649\)M1 A1 (3 marks) 
# Question 5:

## Part (a):

| Working/Answer | Marks | Guidance |
|---|---|---|
| $\sum_{r=1}^{n}(r^2-r-1) = \sum_{r=1}^{n}r^2 - \sum_{r=1}^{n}r - \sum_{r=1}^{n}1$ | M1 | Splitting the sum |
| $\sum_{r=1}^{n}1 = n$ | B1 | |
| $\sum_{r=1}^{n}(r^2-r-1) = \frac{n}{6}(n+1)(2n+1) - \frac{1}{2}n(n+1) - n$ | M1 | Substituting standard formulae |
| $= \frac{n}{6}(2n^2-8)$ | M1 A1 | Simplification |
| $= \frac{1}{3}(n-2)n(n+2)$ | A1 | **(6 marks)** |

## Part (b):

| Working/Answer | Marks | Guidance |
|---|---|---|
| $\sum_{r=10}^{40}(r^2-r-1) = \sum_{r=1}^{40}(r^2-r-1) - \sum_{r=1}^{9}(r^2-r-1)$ | M1 | |
| $= \frac{1}{3}\times 38\times 40\times 42 - \frac{1}{3}\times 7\times 9\times 11 = 21649$ | M1 A1 | **(3 marks)** |

---
5. (a) Show that $\sum _ { r = 1 } ^ { n } \left( r ^ { 2 } - r - 1 \right) = \frac { 1 } { 3 } ( n - 2 ) n ( n + 2 )$.\\
(b) Hence calculate the value of $\sum _ { r = 10 } ^ { 40 } \left( r ^ { 2 } - r - 1 \right)$.\\

\hfill \mbox{\textit{Edexcel FP1  Q5 [9]}}
This paper (9 questions)
View full paper
Q1 6 Q2 7 Q3 5 Q4 3 Q5 9 Q6 10 Q7 12 Q8 9 Q9 14