**Proof:** According to the division algorithm, every n is of the form 6q, 6q + 1, 6q + 2, 6q + 3, 6q + 4, 6q + 5.Assume the first of these cases.

**When n = 6q, then**

*n(n + 1) (2n + 1) / 6*

*= 6q (6q +1) (12q + 1) / 6*

*= q(6q + 1) (12q + 1)*

## which is clearly is an integer.

**When n = 6q + 1, then**

n(n + 1) (2n + 1) / 6

= (6q + 1)(6q + 2)(12q + 3)/6

= (6q + 1) (72q^{2} + 42q + 6)/6

= (6q + 1) (12q^{2} + 7q + 1)

### is an integer in this instant also.

**When n = 6q + 2, then**

n(n + 1) (2n + 1) / 6

= (6q + 2)(6q + 3)(12q + 5)/6

= (36q^{2} + 30q + 6)(12q + 5)/6

= (6q^{2} + 5q + 1)(12q + 5)

is an integer.

**When n = 6q + 3, then**

n(n + 1) (2n + 1) / 6

= (6q + 3)( 6q + 4)(12q + 7) / 6

= (36q^{2} + 42q + 12)(12q + 7) /6

= (6q^{2} + 7q + 2)(12q + 7)

is an integer.

**When n = 6q + 4, then**

n(n + 1) (2n + 1) / 6

= (6q + 4)(6q + 5)(12q + 9) / 6

= (72q^{2} + 102q + 36)(6q + 5)/6

= (12q^{2} + 17q + 6)(6q + 5)

is an integer.

**When n = 6q + 5, then**

n(n + 1) (2n + 1) / 6

= (6q + 5)(6q + 6)(12q + 11) / 6

= (36q^{2} + 66q + 30)(12q + 11) / 6

= (6q^{2} + 11q + 5) (12q + 11)