Question

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

Answer 1

Suppose the positive integer is n.

∴ n = 2q or n = 2q + 1         where q∈Z.

CASE 1:-  
n = 2q
∴ n² - n = (2q)² - 2q
             = 4q² - 2q
             = 2(2q² - q)

CASE 2:-
n = 2q + 1
∴n² - n = (2q + 1)² - (2q + 1)
           = 4q² + 4q + 1 - 2q - 1
           = 4q² + 2q
           = 2(2q² + q)

Thus, in any case, n² - n is divisible by 2.

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

Answer 2

Answer:

Since, n^2 + n = n * ( n+1)

Since, n^2 + n = n * ( n+1)=> n^2 + n is the product of 2 consecutive positive integers

Since, n^2 + n = n * ( n+1)=> n^2 + n is the product of 2 consecutive positive integers& the product of any consecutive integers is always an even number.

Since, n^2 + n = n * ( n+1)=> n^2 + n is the product of 2 consecutive positive integers& the product of any consecutive integers is always an even number.& even numbers are always divisible by 2

Since, n^2 + n = n * ( n+1)=> n^2 + n is the product of 2 consecutive positive integers& the product of any consecutive integers is always an even number.& even numbers are always divisible by 2=> (n^2 + n ) is divisible by 2

Since, n^2 + n = n * ( n+1)=> n^2 + n is the product of 2 consecutive positive integers& the product of any consecutive integers is always an even number.& even numbers are always divisible by 2=> (n^2 + n ) is divisible by 2[Hence Proved]