Question

Prove that n^2 - n is divisible by 2 for every positive integer n.

Answer 1

$given \: that \: n {}^{2} - n \\ \\ = n(n - 1) \\ \\ for \: n = 0 \\ \\ n(n - 1) = 0 \times (0 - 1) = 0 \: which \: is \: divisible \: by \: 2 \\ \\ for \: n = 1 \\ \\ n(n - 1) = 1 \times (1 - 1) = 0 \: which \: is \: divisble \: by \: 2 \\ \\ for \: n = 2 \\ \\ n(n - 1) = 2 \times (2 - 1) = 2 \: which \: is \: divisible \: by \: 2 \\ \\ for \: n = 3 \\ \\ n(n - 1) = 3 \times (3 - 1) = 6 \: which \: is \: divisible \: by \: 2 \\ \\ for \: n = 4 \\ \\ n(n - 1) = 4 \times (4 - 1) = 12 \: which \: is \:divisible \: by \: 2$
Hence for n =0, 1, 2, 3, 4,.......................inf.

$n {}^{2} - n \: is \: divisible \: by \: 2 \\ \\ \\ \: \: \: \: proved \\ \\ hope \: it \: helps...$