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.
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.
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:
Hello, xoluosox!
The proof depends upon what methods you are allowed.
. . Direct algebraic proof? .Induction?
Here's one approach . . .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?