AS June 2025 Paper 1 Q8

EdexcelCurrent spec7 marksSeries

8. The first \(n\) triangular numbers are

\[1,\ 3,\ 6,\ 10,\ \ldots,\ \frac{1}{2}n(n+1)\]

where \(n\) is a positive integer.

(a) Use the standard results for \(\displaystyle\sum_{r=1}^{n} r^2\) and \(\displaystyle\sum_{r=1}^{n} r\) to show that the sum of the first \(n\) triangular numbers is\[\frac{1}{6}n(n+1)(n+2)\] (5)

(b) Hence determine the value of \(n\) for which the sum of the first \(n\) triangular numbers is \(22n\). (2)

Mark scheme (a)

Scheme	Marks	AO
\(\sum \dfrac{1}{2}r(r+1) = \dfrac{1}{2}\left[\sum r^2 + \sum r\right]\)	M1	3.1a
\(\dfrac{1}{2}\left[\sum r^2 + \sum r\right] = \dfrac{1}{2}\left[\dfrac{1}{6}n(n+1)(2n+1) + \dfrac{1}{2}n(n+1)\right]\)	M1 A1	1.1b 1.1b
\(= \dfrac{1}{2}n(n+1)\left[\dfrac{1}{6}(2n+1) + \dfrac{1}{2}\right]\) Or \(= \dfrac{1}{2}n\left[\dfrac{1}{6}(n+1)(2n+1) + \dfrac{1}{2}(n+1)\right]\) Or \(= \dfrac{1}{6}n(n+1)\left[\dfrac{1}{2}(2n+1) + \dfrac{3}{2}\right]\)	dM1	1.1b
\(= \dfrac{1}{6}n(n+1)(n+2)\) * cso	A1*	2.1
	(5)

Notes

M1: Starts the method by writing the correct sum and splitting into two sums, condone use of \(n\) instead of \(r\)

M1: Substitutes in at least one standard formula

A1: Fully correct unsimplified expression

dM1: Dependent on the previous method mark. Factorises out at least \(n\).

A1*: Achieves the printed answer with no omission or errors, factorisation of (\(n\) + 1) term needed cso

Mark scheme (b)

Scheme	Marks	AO
\(\dfrac{1}{6}n(n+1)(n+2) = 22n \Rightarrow n^2 + 3n - 130 = 0 \Rightarrow n = \ldots\) Or \(\dfrac{1}{6}n(n+1)(n+2) = 22n \Rightarrow n^3 + 3n^2 - 130n = 0 \Rightarrow n = \ldots\)	M1	3.1a
\(n = 10\) only	A1	3.2a
	(2)
	(7 marks)

Notes

M1: Sets the printed answer equal to \(22n\), forms a quadratic/cubic and solves to find a non zero value for \(n\). If an incorrect equation please check that their value for \(n\) is correct for their equation if no method shown.

A1: Selects the correct positive value of \(n\). If \(n = -13\) appears this must be rejected