Question 5 - A-Level Maths

Edexcel FP1 2011 January — Question 5 7 marks

Exam Board	Edexcel
Module	FP1 (Further Pure Mathematics 1)
Year	2011
Session	January
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	Standard +0.8 This is a Further Maths FP1 question requiring expansion of a cubic product, application of standard summation formulas, algebraic manipulation to reach a specific form, then evaluation of a sum from r=20 to 50 using the difference of two sums. While methodical, it demands careful algebra across multiple steps and is more demanding than typical A-level questions due to the Further Maths context and the non-trivial manipulation required.
Spec	4.06a Summation formulae: sum of r, r^2, r^3

5. (a) Use the results for $\sum _ { r = 1 } ^ { n } r , \sum _ { r = 1 } ^ { n } r ^ { 2 }$ and $\sum _ { r = 1 } ^ { n } r ^ { 3 }$, to prove that $$\sum _ { r = 1 } ^ { n } r ( r + 1 ) ( r + 5 ) = \frac { 1 } { 4 } n ( n + 1 ) ( n + 2 ) ( n + 7 )$$ for all positive integers $n$.
(b) Hence, or otherwise, find the value of $$\sum _ { r = 20 } ^ { 50 } r ( r + 1 ) ( r + 5 )$$

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Question 5:

Part (a):

Answer	Marks	Guidance
$\sum_{r=1}^{n} r(r+1)(r+5) = \sum_{r=1}^{n} r^3 + 6r^2 + 5r$	M1	Multiplying out brackets and an attempt to use at least one standard formula correctly
$= \frac{1}{4}n^2(n+1)^2 + 6 \cdot \frac{1}{6}n(n+1)(2n+1) + 5 \cdot \frac{1}{2}n(n+1)$	A1	Correct expression
$= \frac{1}{4}n(n+1)\bigl(n(n+1) + 4(2n+1) + 10\bigr)$	dM1	Factorising out at least $n(n+1)$
$= \frac{1}{4}n(n+1)\bigl(n^2 + n + 8n + 4 + 10\bigr)$
$= \frac{1}{4}n(n+1)\bigl(n^2 + 9n + 14\bigr)$	A1	Correct 3 term quadratic factor
$= \frac{1}{4}n(n+1)(n+2)(n+7)$	A1	Correct proof. No errors seen.

(5 marks)

Part (b):

Answer	Marks	Guidance
$S_n = \sum_{r=20}^{50} r(r+1)(r+5) = S_{50} - S_{19}$	M1	Use of $S_{50} - S_{19}$
$= \frac{1}{4}(50)(51)(52)(57) - \frac{1}{4}(19)(20)(21)(26)$
$= 1889550 - 51870$
$= 1837680$	A1	Correct answer only 2/2

(2 marks)

## Question 5:

### Part (a):
| $\sum_{r=1}^{n} r(r+1)(r+5) = \sum_{r=1}^{n} r^3 + 6r^2 + 5r$ | M1 | Multiplying out brackets and an attempt to use at least one standard formula correctly |
|---|---|---|
| $= \frac{1}{4}n^2(n+1)^2 + 6 \cdot \frac{1}{6}n(n+1)(2n+1) + 5 \cdot \frac{1}{2}n(n+1)$ | A1 | Correct expression |
| $= \frac{1}{4}n(n+1)\bigl(n(n+1) + 4(2n+1) + 10\bigr)$ | dM1 | Factorising out at least $n(n+1)$ |
| $= \frac{1}{4}n(n+1)\bigl(n^2 + n + 8n + 4 + 10\bigr)$ | | |
| $= \frac{1}{4}n(n+1)\bigl(n^2 + 9n + 14\bigr)$ | A1 | Correct 3 term quadratic factor |
| $= \frac{1}{4}n(n+1)(n+2)(n+7)$ | A1 | Correct proof. No errors seen. |
(5 marks)

### Part (b):
| $S_n = \sum_{r=20}^{50} r(r+1)(r+5) = S_{50} - S_{19}$ | M1 | Use of $S_{50} - S_{19}$ |
|---|---|---|
| $= \frac{1}{4}(50)(51)(52)(57) - \frac{1}{4}(19)(20)(21)(26)$ | | |
| $= 1889550 - 51870$ | | |
| $= 1837680$ | A1 | Correct answer only 2/2 |
(2 marks)

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5. (a) Use the results for $\sum _ { r = 1 } ^ { n } r , \sum _ { r = 1 } ^ { n } r ^ { 2 }$ and $\sum _ { r = 1 } ^ { n } r ^ { 3 }$, to prove that

$$\sum _ { r = 1 } ^ { n } r ( r + 1 ) ( r + 5 ) = \frac { 1 } { 4 } n ( n + 1 ) ( n + 2 ) ( n + 7 )$$

for all positive integers $n$.\\
(b) Hence, or otherwise, find the value of

$$\sum _ { r = 20 } ^ { 50 } r ( r + 1 ) ( r + 5 )$$

\hfill \mbox{\textit{Edexcel FP1 2011 Q5 [7]}}

This paper (10 questions)

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Q1 5 Q2 6 Q3 10 Q4 4 Q5 7 Q6 8 Q7 9 Q8 9 Q9 5 Q10 12

Answer	Marks	Guidance
\(\sum_{r=1}^{n} r(r+1)(r+5) = \sum_{r=1}^{n} r^3 + 6r^2 + 5r\)	M1	Multiplying out brackets and an attempt to use at least one standard formula correctly
\(= \frac{1}{4}n^2(n+1)^2 + 6 \cdot \frac{1}{6}n(n+1)(2n+1) + 5 \cdot \frac{1}{2}n(n+1)\)	A1	Correct expression
\(= \frac{1}{4}n(n+1)\bigl(n(n+1) + 4(2n+1) + 10\bigr)\)	dM1	Factorising out at least \(n(n+1)\)
\(= \frac{1}{4}n(n+1)\bigl(n^2 + n + 8n + 4 + 10\bigr)\)
\(= \frac{1}{4}n(n+1)\bigl(n^2 + 9n + 14\bigr)\)	A1	Correct 3 term quadratic factor
\(= \frac{1}{4}n(n+1)(n+2)(n+7)\)	A1	Correct proof. No errors seen.

Answer	Marks	Guidance
\(S_n = \sum_{r=20}^{50} r(r+1)(r+5) = S_{50} - S_{19}\)	M1	Use of \(S_{50} - S_{19}\)
\(= \frac{1}{4}(50)(51)(52)(57) - \frac{1}{4}(19)(20)(21)(26)\)
\(= 1889550 - 51870\)
\(= 1837680\)	A1	Correct answer only 2/2