Question

N ∈ lN

Answer 1

@Konrad509's answer was fantastic. 

What you first need to know is this. 3 consecutive natural numbers multiplied by each other, for instance: (1*2*3) or (2*3*4) or (3*4*5) or (4*5*(2*3)) can be described using the abstract expression:

(n-1)n(n+1)

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Now:

A=n(n²+11),

and:

n(n²+11)

=n(n²-1+12)

=n³-n+12n

=n(n²-1)+12n

=n(n+1)(n-1)+12n

=(n-1)n(n+1)+(4*3)n

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So:

A=(n-1)n(n+1)+(4*3)n

Therefore, A is divisible by 3.

Answer 2

$n(n^2+11)=\\n(n^2-1+12)=\\ n(n^2-1)+12n=\\ (n-1)n(n+1)+12n$

$(n-1)n(n+1)$ is the product of 3 consecutive natural numbers, so it has be to divisible by 3.

$12n$ is also divisible by 3 because 12 is divisible by 3.

If both elements of the sum are divisible by 3 so the sum itself is divisible by 3 as well.