AS June 2023 Paper 1 Q8

EdexcelCurrent spec8 marksSeries

(a) Use the standard results for \(\displaystyle\sum_{r=1}^{n} r^2\) and \(\displaystyle\sum_{r=1}^{n} r\) to show that, for all positive integers \(n\),\[\sum_{r=1}^{n} (2r - 1)^2 = \frac{n}{3}\left(an^2 - 1\right)\]where \(a\) is a constant to be determined. (5)

(b) Hence determine the sum of the squares of all positive odd three-digit integers. (3)

Mark scheme (a)

Scheme	Marks	AO
\((2r - 1)^2 = 4r^2 - 4r + 1\)	B1	1.1b
\[\begin{aligned}\sum_{r=1}^{n} (2r - 1)^2 &= 4\sum_{r=1}^{n} r^2 - 4\sum_{r=1}^{n} r + \sum_{r=1}^{n} 1\\ &= 4\frac{n}{6}(n + 1)(2n + 1) - 4\frac{n}{2}(n + 1) + n\end{aligned}\]	M1 A1	1.1b 1.1b
\(= \dfrac{n}{3}\left[2(n + 1)(2n + 1) - 6(n + 1) + 3\right]\) Or \(= n\left[\dfrac{2}{3}(n + 1)(2n + 1) - 2(n + 1) + 1\right]\)	dM1	1.1b
\(\left\{\dfrac{n}{3}\left(4n^2 + 6n + 2 - 6n - 6 + 3\right)\right\}\) \(= \dfrac{n}{3}\left(4n^2 - 1\right)\) cso	A1	2.1
	(5)

Notes

B1: Correct expanded expression.
M1: Substitutes at least one of the standard formulae into their expanded expression.
A1: Fully correct unsimplified expression.

dM1: Dependent on previous method. Attempts to factorises out \(n\). Must have a \(n\) in every term. Condone a slip with one term as long as the intention is clear.

A1: Achieves the correct answer, with a correct intermediate line of working. cso

Note If uses \(\sum 1 = 1\) scores B1 M1 A0 M0 A0

An attempt at proof by induction may score B1 only

Mark scheme (b)

Scheme	Marks	AO
\(\displaystyle\sum_{r=51}^{500} (2r - 1)^2\)	B1	3.1a
\[\begin{aligned}\sum_{r=51}^{500} (2r - 1)^2 &= \sum_{r=1}^{500} (2r - 1)^2 - \sum_{r=1}^{50} (2r - 1)^2\\ &= \frac{500}{3}\left(4(500)^2 - 1\right) - \frac{50}{3}\left(4(50)^2 - 1\right)\\ &\{= 166666500 - 166650\}\end{aligned}\]	M1	1.1b
166 499 850	A1	1.1b
	(3)
	(8 marks)

Notes

B1: Correct summation formula for the sum of the squares of all positive odd three-digit integers including limits. This can be implied by later work.

M1: Uses the answer to part (a) and \(\displaystyle\sum_{r=p}^{q} (2r - 1)^2 = \sum_{r=1}^{q} (2r - 1)^2 - \sum_{r=1}^{p-1} (2r - 1)^2\) where \(p\), \(q\) are numerical and \(q > p\), to find a value. There must be some indication of the sum that they are finding or the correct values for \(p\) and \(q\).
States \(\displaystyle\sum_{r=1}^{500} (2r - 1)^2 - \sum_{r=1}^{50} (2r - 1)^2\) implies B1

States \(\dfrac{500}{3}\left(4(500)^2 - 1\right) - \dfrac{50}{3}\left(4(50)^2 - 1\right)\) this scores B1 (implied) and M1

A1: Correct value

Note \(\displaystyle\sum_{r=51}^{500} (2r - 1)^2 = 166499850\) or correct answer only scores B1 M0 A0, must be evidence of using the answer to (a)