Proof question

Can somebody help with part ii) thanks!

Look at your answer to part (a). What sort of numbers must n be in order for you to take a factor of 2 out completely??

Can somebody help with part ii) thanks!

could I see your part 1?

Look at your answer to part (a). What sort of numbers must n be in order for you to take a factor of 2 out completely??

erm odd?

But how can I give a really good explanation to get the marks for this?

Correct.

One way you could begin by saying since $n$ is odd then $n^2$ is odd, hence $3n^2$ is odd. Then logically show that $3n^2+6n$ is odd, etc...

Alternatively, if $n$ is odd then it can be written in the form $n=2k+1$ so substitute that in, expand, and factor out a 2 to show that the result is even.

what if I said;
let n+1 be an even number where n is an odd number.....so for any odd values of n in the expression 3n^2 + 6n + 5...Will give an even answer... (I also use some odd values in the expression and show my answer).

I know this is very stupid but would it work. I haven't done proofs yet but just curious thanks

No this is does not constitute a proof.

In a general proof like this you need to show logically step-by-step how you arrive at the conclusion that the expression is even when $n$ is odd. Also 'testing out' a few values of $n$ does not add validity to the proof itself, but it can help you understand it better and generalise on them.

Correct.

One way you could begin by saying since $n$ is odd then $n^2$ is odd, hence $3n^2$ is odd. Then logically show that $3n^2+6n$ is odd, etc...

Alternatively, if $n$ is odd then it can be written in the form $n=2k+1$ so substitute that in, expand, and factor out a 2 to show that the result is even.

If n is odd, then how does n+1 being even relate to 3n^2+6n+5 being even? A proof should be written to convince someone who doesn't believe what you're saying, so you have to be very clear in your assertions.

Also, giving some examples is not valid because it only shows that the statement is true for certain values of n, not all values of n.

Thanks a lot so I got the second way : 2(6k^2 +12k +7)
But am stuck on the first way just when you mentioned the logically part since n is odd, n^2 is odd cause odd x odd=odd so 3n^2 is odd cause odd x odd x odd =odd and then ..I got stuck

So, is $6n$ odd or even? Hence is $3n^2 + 6n$ odd or even? Hence is $3n^2+6n+5$ odd or even?

Oh so you don't have to explain why 6n is even?

Not really, it should be obvious. But you can if you want, it doesn't take much to rewrite it as $2 \times 3n$.

But then how do you prove 3n?

Apologies if I annoy you

Prove what?

All you need to have is that $6n$ is an even number because it two lots of some integer $3n$.