**Given**

n(n+1)(2n+1) is divisible by 6 ∀ n ∈ N

**Proof**

**STEP 1 : We will check if the statement is true for n = 1**

n*(n + 1)*(2n + 1) = 1 * (1 + 1) * ( 2*1 + 1 )

= 1 * 2 * 3

= 6

The statement is true for n = 1.

**STEP 2** **: We need to prove that if the statement is true for n = k, then it is also true for ****n = k + 1**

Let the given statement be true for n = k. Then k*(k + 1)*(2k + 1) is a multiple of 6. Let,

k*(k + 1)*(2k + 1) = 6m ∀ m ∈ N …… (equation 1)

We need to prove that the statement ii also true for n = k + 1. We need to show that

(k + 1)*(k + 2)*(2*(k+1) + 1) is a multiple of 6.

(k + 1)*(k + 2)*(2*(k+1) + 1) = (k + 1)*(k + 2)*(2*(k + 1) + 1)

= (k + 1)*(k + 2)*(2k + 2 + 1)

= (k + 2)*(k + 1)*( (2k + 1) + 2)

= k * (k + 1) * ( (2k + 1) + 2 ) + 2 * (k + 1) * ( (2k + 1) + 2 )

= k * (k + 1) * (2k + 1) + k * (k + 1) * 2 + 2 * (k + 1) * (2k + 3)

Substituting form equation 1

= 6m + 2 * k * (k + 1) + 2 * (k + 1) * (2k + 3)

= 6m + 2 * (k + 1) * ( k + 2k + 3 )

= 6m + 2 * (k + 1) * (3k + 3)

= 6m + 6 * (k + 1) * (k + 1)

From the above equation, we can say that (k + 1)*(k + 2)*(2*(k+1) + 1) is divisible by 6 if k*(k + 1)*(2k + 1) is divisible by 6.

Thus, if the statement is true for n = k, then it is also true for n = k + 1. Therefore, by the principle of mathematical induction, the given statement is true ∀ n ∈ N.