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    Can somebody help with part ii) thanks!
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    (Original post by Kalabamboo)
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    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??
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    (Original post by RDKGames)
    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?
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    (Original post by Kalabamboo)
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    Can somebody help with part ii) thanks!
    could I see your part 1?
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    (Original post by brainmaster)
    could I see your part 1?
    3n^2 +6n+5
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    (Original post by RDKGames)
    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??
    So Ik it's odd.

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

    The ms version is a bit all over the place
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    (Original post by Kalabamboo)
    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.
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    (Original post by RDKGames)
    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
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    (Original post by brainmaster)
    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. What you've said is basically "So n is odd hence 3n^2+6n+5 is even" but where's the convincing part?
    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.
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    (Original post by RDKGames)
    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.
    aha thanks alot
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    (Original post by RDKGames)
    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.
    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:p: 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
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    (Original post by MR1999)
    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.
    Yeah I just saw that... thanks buddy
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    (Original post by Kalabamboo)
    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:p: 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?
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    (Original post by RDKGames)
    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?

    I thought you had to do some algebra to prove 6n is even in the answer-so it's not necessary?
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    (Original post by Kalabamboo)
    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.
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    (Original post by RDKGames)
    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

    I guess you can't prove 3n right? so 3(odd) x even = even
    3(odd) x odd=odd
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    (Original post by Kalabamboo)
    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.
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    (Original post by RDKGames)
    Prove what?

    All you need to have is that 6n is an even number because it two lots of some integer 3n.
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    Can also note that you can factorize this as

    3*(n+1)2 + 2

    which makes the arguments about being odd or even a bit simpler.
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    I think it's equally important to show that the sum is odd when n is even if you want the full marks. (But at the same time I think one line explanations are sufficent).
 
 
 
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