AS October 2020 Paper 1 Q5

EdexcelCurrent spec7 marksSeries

Figure 2: a cuboid block with length (r + 2), width (r + 1) and height r — Figure 2

A block has length \((r + 2)\) cm, width \((r + 1)\) cm and height \(r\) cm, as shown in Figure 2.

In a set of \(n\) such blocks, the first block has a height of 1 cm, the second block has a height of 2 cm, the third block has a height of 3 cm and so on.

(a) Use the standard results for \(\displaystyle\sum_{r=1}^{n} r^3, \sum_{r=1}^{n} r^2\) and \(\displaystyle\sum_{r=1}^{n} r\) to show that the total volume, \(V\), of all \(n\) blocks in the set is given by\[V = \frac{n}{4}(n + 1)(n + 2)(n + 3) \qquad n \geqslant 1\] (5)

Given that the total volume of all \(n\) blocks is

\[\left(n^4 + 6n^3 - 11\,710\right)\text{ cm}^3\]

(b) determine how many blocks make up the set. (2)

Mark scheme (a)

Scheme	Marks	AO
Volume = \(r \times (r + 1) \times (r + 2)\)	B1	1.1b
A complete method for finding the total volume of \(n\) blocks and expressing it in sigma notation. This can be implied by later work. \(\displaystyle\sum_{r=1}^{n}\left(r^3 + 3r^2 + 2r\right)\)	M1	3.1b
\(V = \dfrac{1}{4}n^2(n + 1)^2 + 3 \times \dfrac{1}{6}n(n + 1)(2n + 1) + 2 \times \dfrac{n}{2}(n + 1)\)	M1	2.1
\(V = \dfrac{1}{4}n(n + 1)\left[n(n + 1) + 2(2n + 1) + 4\right]\)	dM1	1.1b
\(V = \dfrac{1}{4}n(n + 1)\left[n^2 + 5n + 6\right]\) \(\Rightarrow V = \dfrac{1}{4}n(n + 1)(n + 2)(n + 3)\)*	A1*	1.1b
	(5)

Notes

B1: Correct volume of a block

M1: Expressing the total volume of all \(n\) blocks as a series in terms of \(r\), \(r^2\) and \(r^3\)

M1: Substitutes at least one of the standard formulae into their volume.

dM1: Attempts to factorise \(\dfrac{1}{4}n(n + 1)\) having used at least one standard formula correctly. Each term must contain a factor of \(n(n + 1)\)

A1*: Obtains the printed result with no errors seen, no bracketing errors and following from \(V = \dfrac{1}{4}n(n + 1)\left[n^2 + 5n + 6\right]\) o.e.

Note: Going from \(\dfrac{1}{4}n\left(n^3 + 6n^2 + 11n + 6\right)\) to \(\dfrac{1}{4}n(n + 1)(n + 2)(n + 3)\) with no reasoning shown scores dM0 A0

Mark scheme (b)

Scheme	Marks	AO
Sets \(\dfrac{1}{4}n(n + 1)(n + 2)(n + 3) = n^4 + 6n^3 - 11710\) \(\dfrac{1}{4}n^4 + \dfrac{3}{2}n^3 + \dfrac{11}{4}n^2 + \dfrac{3}{2}n = n^4 + 6n^3 - 11710\) simplifies \(\left(3n^4 + 18n^3 - 11n^2 - 6n - 46840 = 0\right)\) and solves \(n = \ldots\)	M1	1.1b
There are 10 blocks or \(n = 10\)	A1	3.2a
	(2)
	(7 marks)

Notes

M1: Sets the printed answer \(= n^4 + 6n^3 - 11710\), simplifies, collects terms and uses their calculator to solve a quartic equation to find a value for \(n\).

A1: Selects \(n = 10\) or states that there are 10 blocks from a correct equation