SPM.1.SL.TZ0.3

pestleMathematicsAASLPaper 1SPM· sl-1-6-simple-proofsource ↗

Show that ${\left( {2n - 1} \right)^2} + {\left( {2n + 1} \right)^2} = 8{n^2} + 2$ , where $n \in \mathbb{Z}$ .

[2]

Hence, or otherwise, prove that the sum of the squares of any two consecutive odd integers is even.

[3]

Markscheme / solution

attempting to expand the LHS (M1)

LHS $= \left( {4{n^2} - 4n + 1} \right) + \left( {4{n^2} + 4n + 1} \right)$ A1

$= 8{n^2} + 2$ (= RHS) AG

[2 marks]

METHOD 1

recognition that ${2n - 1}$ and ${2n + 1}$ represent two consecutive odd integers (for $n \in \mathbb{Z}$ ) R1

$8{n^2} + 2 = 2\left( {4{n^2} + 1} \right)$ A1

valid reason eg divisible by 2 (2 is a factor) R1

so the sum of the squares of any two consecutive odd integers is even AG

METHOD 2

recognition, eg that $_n$ and ${n + 2}$ represent two consecutive odd integers (for $n \in \mathbb{Z}$ ) R1

${n^2} + {\left( {n + 2} \right)^2} = 2\left( {{n^2} + 2n + 2} \right)$ A1

valid reason eg divisible by 2 (2 is a factor) R1

so the sum of the squares of any two consecutive odd integers is even AG

[3 marks]

Examiners’ report

[N/A]