Given
n3 + (n + 1)3 + (n + 2)3 is divisible by 9 ∀ n ∈ N
Proof
We will prove the given statement by induction
STEP 1
n = 1
13 + (1 + 1)3 + (1 + 2)3= 13 + 23 + 33
= 1 + 8 + 27
= 36
36 is divisible by 9. Therefore, the statement is true for n = 1
STEP 2
Let the given statement be true for n = k
k3 + (k + 1)3 + (k+ 2)3 = 9x
Now, we need to prove that if the statement is true for n = k then it is also true for n = k + 1
(k + 1)3 + (k+ 2)3 + (k+ 3)3 = ( (k + 1)3 + (k+ 2)3 ) + (k+ 3)3
= 9x – k3 + (k+ 3)3
= 9x – k3 + k3 + 33 + 3*k2*3 + 3*k*32
= 9x + 27 + 9k2 + 27k
= 9( x + 3 + k2 + 3k )
9( x + 3 + k2 + 3k ) is divisible by 9.
Therefore, we can say that if the given statement is true for n = k, then it is also true for n = k + 1. Hence, by the principle of mathematical induction, the given statement is true ∀ n ∈ N.