Answer :
[tex]n(n^2+11)=\\n(n^2-1+12)=\\
n(n^2-1)+12n=\\
(n-1)n(n+1)+12n[/tex]
[tex](n-1)n(n+1)[/tex] is the product of 3 consecutive natural numbers, so it has be to divisible by 3.
[tex]12n[/tex] 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.
[tex](n-1)n(n+1)[/tex] is the product of 3 consecutive natural numbers, so it has be to divisible by 3.
[tex]12n[/tex] 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.
@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)
-------------
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
-------------------------
So:
A=(n-1)n(n+1)+(4*3)n
Therefore, A is divisible by 3.
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)
-------------
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
-------------------------
So:
A=(n-1)n(n+1)+(4*3)n
Therefore, A is divisible by 3.
