Question 7 - A-Level Maths

Edexcel FP1 2013 June — Question 7 8 marks

Exam Board	Edexcel
Module	FP1 (Further Pure Mathematics 1)
Year	2013
Session	June
Marks	8
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	Challenging +1.2 This is a standard FP1 summation question requiring algebraic manipulation of known formulas and telescoping sum technique. Part (a) is routine expansion and simplification using given results. Part (b) requires recognizing the pattern r²(r-1) and applying the formula over a shifted range (r=10 to 50), which is a common exam technique but requires careful index manipulation beyond basic recall.
Spec	4.06a Summation formulae: sum of r, r^2, r^3

7. (a) Use the standard results for $\sum _ { r = 1 } ^ { n } r ^ { 2 }$ and $\sum _ { r = 1 } ^ { n } r ^ { 3 }$ to show that $$\sum _ { r = 1 } ^ { n } r ^ { 2 } ( r - 1 ) = \frac { n ( n + 1 ) ( 3 n + 2 ) ( n - 1 ) } { 12 }$$ for all positive integers $n$.
(b) Hence find the sum of the series $$10 ^ { 2 } \times 9 + 11 ^ { 2 } \times 10 + 12 ^ { 2 } \times 11 + \ldots + 50 ^ { 2 } \times 49$$

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Question 7:

Part 7(a):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$\sum_{r=1}^{n} r^2(r-1) = \sum_{r=1}^{n} r^3 - \sum_{r=1}^{n} r^2$	M1	Expanding brackets
$= \frac{n^2(n+1)^2}{4} - \frac{n(n+1)(2n+1)}{6}$	A1	Correct expressions for $\sum r^3$ and $\sum r^2$
$= \frac{n(n+1)}{12}\big(3n(n+1)-2(2n+1)\big)$	M1	Factorising by $n(n+1)$
$= \frac{n(n+1)(3n^2-n-2)}{12}$	A1	$(3n^2-n-2)$ or equivalent factor
$= \frac{n(n+1)(3n+2)(n-1)}{12}$	A1cso

Part 7(b):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$\sum_{r=10}^{50} r^2(r-1) = \sum_{r=1}^{50} r^2(r-1) - \sum_{r=1}^{9} r^2(r-1)$	M1	$f(49$ or $50) - f(9$ or $10)$ and attempt to use part (a)
$= \frac{1}{12}(50\times51\times152\times49) - \frac{1}{12}(9\times10\times29\times8)$	A1
$= 1582700 - 1740 = 1580960$	A1

# Question 7:

## Part 7(a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum_{r=1}^{n} r^2(r-1) = \sum_{r=1}^{n} r^3 - \sum_{r=1}^{n} r^2$ | M1 | Expanding brackets |
| $= \frac{n^2(n+1)^2}{4} - \frac{n(n+1)(2n+1)}{6}$ | A1 | Correct expressions for $\sum r^3$ and $\sum r^2$ |
| $= \frac{n(n+1)}{12}\big(3n(n+1)-2(2n+1)\big)$ | M1 | Factorising by $n(n+1)$ |
| $= \frac{n(n+1)(3n^2-n-2)}{12}$ | A1 | $(3n^2-n-2)$ or equivalent factor |
| $= \frac{n(n+1)(3n+2)(n-1)}{12}$ | A1cso | |

## Part 7(b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum_{r=10}^{50} r^2(r-1) = \sum_{r=1}^{50} r^2(r-1) - \sum_{r=1}^{9} r^2(r-1)$ | M1 | $f(49$ or $50) - f(9$ or $10)$ and attempt to use part (a) |
| $= \frac{1}{12}(50\times51\times152\times49) - \frac{1}{12}(9\times10\times29\times8)$ | A1 | |
| $= 1582700 - 1740 = 1580960$ | A1 | |

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

$$\sum _ { r = 1 } ^ { n } r ^ { 2 } ( r - 1 ) = \frac { n ( n + 1 ) ( 3 n + 2 ) ( n - 1 ) } { 12 }$$

for all positive integers $n$.\\
(b) Hence find the sum of the series

$$10 ^ { 2 } \times 9 + 11 ^ { 2 } \times 10 + 12 ^ { 2 } \times 11 + \ldots + 50 ^ { 2 } \times 49$$

\hfill \mbox{\textit{Edexcel FP1 2013 Q7 [8]}}

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