Question

Prove that n2 - n is divisible by 2 for every positive integer ' n' ?

Answer 1

Given :

•n²-n is divisible by 2 for every positive integer 'n'

To Prove :

•Prove that n²-n is divisible by 2 for every positive integer 'n'=?

Solution :

• We know that,

• Every positive integer is in the form of 2q or 2q+1 ,for some integer q

Case (1)

• When $\boxed{\bf{n=2q}}$

$\implies \sf \: {n}^{2} - n = {(2q)}^{2} - 2q$

$\implies \sf \: 4 {q}^{2} - 2q$

$\implies \sf \: 2q(2q - 1)$

Let,

• n²-n=2m

Case (2)

• When $\boxed{\bf{n=2q+1}}$

$\\ \implies \sf \: {n}^{2} - n = {(2q + 1)}^{2} - (2q + 1)$

$\implies \sf \: {4q}^{2} + 1 + 4q - 2q - 1$

$\implies \sf \: 4 {q}^{2} + 2q$

$\implies \sf \: 2q(2q + 1)$

Let,

2q+1=m

➡ n²-n=2m

Hence, n²-n is divisible by 2 for every positive integer