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n ∈ lN
We set A= n(n²+11)
Show that A is divisible by 3
( n is natural number )
please try to help me in this i need it tomorrow plz...

Sagot :

konrad509
[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.
@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.
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