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.
Thank you for your guidance.