Hi the problem is:

Prove that 6 is a factor of n^3+3n^2+2n for all natural numbers n.

I don't know where to start with these kinds of problems, can anyone help?

THANKS SO MUCH.

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- November 22nd 2008, 12:41 PMxoluosoxMathematical induction
Hi the problem is:

Prove that 6 is a factor of n^3+3n^2+2n for all natural numbers n.

I don't know where to start with these kinds of problems, can anyone help?

THANKS SO MUCH. - November 22nd 2008, 12:58 PMMathstud28
Ok so our hypothesis is that

So lets begin

Base Step:

When we have

So that is true

Restate the hypothesis:

Inductive step: Now we have to just using the hypothesis prove that

Now rexpanding this we get

Rewriting this we get

Now by our hypothesis the first term is divisible by six, so it follows that we must just prove that

Now if is odd then is even so

So there is a factor of two in so there is a factor of six in and consequently in

Now if is even then is even and the same argument follows.

So we have proved our iductive step and the result follows.

EDIT: Moo has informed me that I have the notation backwards, for my sake just pretend that means that the left hand side is divides the right hand site. - November 22nd 2008, 01:01 PMxoluosox
Thank you so much. I didn't know how to set up the equation at all.

- November 22nd 2008, 01:10 PMo_O
A note on notation.

When you say is divisible by 6 (or 6 divides ), we write it like this .

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Also in the inductive step, you can simply factor out the 3 in that second term and use the fact that if you add two numbers that are divisible by 3, then their sum is also divisible by 3:

- November 22nd 2008, 01:58 PMSoroban
Hello, xoluosox!

The proof depends upon what methods you are allowed.

. . Direct algebraic proof? .Induction?

Quote:

Prove that 6 is a factor of for all natural numbers

Factor the polynomial: .

We have the product of*three consecutive integers.*

Since they must be: .

. . at least one of the factors is even.

Hence, the polynomial is divisible by 2.

It can be shown (you can devise your own proof) that:

. . the product of three consecutive integers is divisible by 3.

Got it?

- November 22nd 2008, 02:04 PMmr fantastic