Edexcel FP1 2011 June — Question 7 10 marks

Exam BoardEdexcel
ModuleFP1 (Further Pure Mathematics 1)
Year2011
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeSum from n+1 to 2n or similar range
DifficultyStandard +0.8 This is a multi-step Further Maths question requiring algebraic manipulation of standard summation formulas, then applying the result to a shifted range (n+1 to 3n). Part (a) requires expanding and combining standard results, while part (b) requires the insight to use subtraction of sums. More demanding than typical A-level questions but follows a structured path with clear guidance.
Spec4.06a Summation formulae: sum of r, r^2, r^3
7. (a) Use the results for \(\sum _ { r = 1 } ^ { n } r\) and \(\sum _ { r = 1 } ^ { n } r ^ { 2 }\) to show that $$\sum _ { r = 1 } ^ { n } ( 2 r - 1 ) ^ { 2 } = \frac { 1 } { 3 } n ( 2 n + 1 ) ( 2 n - 1 )$$ for all positive integers \(n\).
(b) Hence show that $$\sum _ { r = n + 1 } ^ { 3 n } ( 2 r - 1 ) ^ { 2 } = \frac { 2 } { 3 } n \left( a n ^ { 2 } + b \right)$$ where \(a\) and \(b\) are integers to be found.
Question 7:
Part (a):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\sum_{r=1}^{n}(2r-1)^2 = \sum_{r=1}^{n}4r^2 - 4r + 1\)M1 Multiplying out brackets and attempt to use at least one of the two standard formulae correctly
\(= 4\cdot\frac{1}{6}n(n+1)(2n+1) - 4\cdot\frac{1}{2}n(n+1) + n\)A1 First two terms correct
B1\(+n\)
\(= \frac{1}{3}n\{2(2n+1)(2n+1) - 6(n+1) + 3\}\)M1 Attempt to factorise out \(\frac{1}{3}n\)
A1Correct expression with \(\frac{1}{3}n\) factored out, no errors seen
\(= \frac{1}{3}n(2n+1)(2n-1)\)A1* Correct proof, no errors seen
(6 marks)
Part (b):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\sum_{r=n+1}^{3n}(2r-1)^2 = S_{3n} - S_n\)—
\(= \frac{1}{3}\cdot3n(6n+1)(6n-1) - \frac{1}{3}n(2n+1)(2n-1)\)M1 Use of \(S_{3n} - S_n\) with result from (a) used at least once
A1Correct unsimplified expression. Allow \(2(3n)\) for \(6n\)
\(= n(36n^2-1) - \frac{1}{3}n(4n^2-1)\)—
\(= \frac{1}{3}n(108n^2 - 3 - 4n^2 + 1)\)dM1 Factorising out \(\frac{1}{3}n\) (or \(\frac{2}{3}n\))
\(= \frac{2}{3}n(52n^2 - 1)\)A1 \(\frac{2}{3}n(52n^2-1)\)
\(a = 52,\ b = -1\)—
(4 marks)
## Question 7:

### Part (a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum_{r=1}^{n}(2r-1)^2 = \sum_{r=1}^{n}4r^2 - 4r + 1$ | M1 | Multiplying out brackets and attempt to use at least one of the two standard formulae correctly |
| $= 4\cdot\frac{1}{6}n(n+1)(2n+1) - 4\cdot\frac{1}{2}n(n+1) + n$ | A1 | First two terms correct |
| | B1 | $+n$ |
| $= \frac{1}{3}n\{2(2n+1)(2n+1) - 6(n+1) + 3\}$ | M1 | Attempt to factorise out $\frac{1}{3}n$ |
| | A1 | Correct expression with $\frac{1}{3}n$ factored out, no errors seen |
| $= \frac{1}{3}n(2n+1)(2n-1)$ | A1* | Correct proof, no errors seen |

**(6 marks)**

### Part (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum_{r=n+1}^{3n}(2r-1)^2 = S_{3n} - S_n$ | — | |
| $= \frac{1}{3}\cdot3n(6n+1)(6n-1) - \frac{1}{3}n(2n+1)(2n-1)$ | M1 | Use of $S_{3n} - S_n$ with result from (a) used at least once |
| | A1 | Correct unsimplified expression. Allow $2(3n)$ for $6n$ |
| $= n(36n^2-1) - \frac{1}{3}n(4n^2-1)$ | — | |
| $= \frac{1}{3}n(108n^2 - 3 - 4n^2 + 1)$ | dM1 | Factorising out $\frac{1}{3}n$ (or $\frac{2}{3}n$) |
| $= \frac{2}{3}n(52n^2 - 1)$ | A1 | $\frac{2}{3}n(52n^2-1)$ |
| $a = 52,\ b = -1$ | — | |

**(4 marks)**

---
7. (a) Use the results for $\sum _ { r = 1 } ^ { n } r$ and $\sum _ { r = 1 } ^ { n } r ^ { 2 }$ to show that

$$\sum _ { r = 1 } ^ { n } ( 2 r - 1 ) ^ { 2 } = \frac { 1 } { 3 } n ( 2 n + 1 ) ( 2 n - 1 )$$

for all positive integers $n$.\\
(b) Hence show that

$$\sum _ { r = n + 1 } ^ { 3 n } ( 2 r - 1 ) ^ { 2 } = \frac { 2 } { 3 } n \left( a n ^ { 2 } + b \right)$$

where $a$ and $b$ are integers to be found.\\

\hfill \mbox{\textit{Edexcel FP1 2011 Q7 [10]}}
This paper (9 questions)
View full paper
Q1 5 Q2 8 Q3 9 Q4 6 Q5 8 Q6 7 Q7 10 Q8 10 Q9 12