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    Sorry, a bit of a click bait title with the capitals. Can someone check if this explanation is correct? Maths question -80% of those who clicked lost at this point. Thanks

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    Wasn't expecting this level of interest lmao

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    (Original post by Moltenmo)
    Sorry, a bit of a click bait title with the capitals. Can someone check if this explanation is correct? Maths question -80% of those who clicked lost at this point. Thanks

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    That looks fine. However, the proof would have worked much more neatly if you used  2n and  2n+1 as your consecutive numbers. You'd end up with  4(2n^2+n) + 1 , which is more explicit in showing the required result.
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    (Original post by kingaaran)
    That looks fine. However, the proof would have worked much more neatly if you used  2n and  2n+1 as your consecutive numbers. You'd end up with  4(2n^2+n) + 1 , which is more explicit in showing the required result.
    Ahh thank you! One question I have. Is it better to use 2n and 2n+1 as the consecutive numbers as opposed to n and n+1 in proof questions? Or was it only better in this scenario?

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    (Original post by Moltenmo)
    Ahh thank you! One question I have. Is it better to use 2n and 2n+1 as the consecutive numbers as opposed to n and n+1 in proof questions? Or was it only better in this scenario?

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    Providing there are no other conditions to the consecutive numbers of course.

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    (Original post by Moltenmo)
    Ahh thank you! One question I have. Is it better to use 2n and 2n+1 as the consecutive numbers as opposed to n and n+1 in proof questions? Or was it only better in this scenario?

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    Hmm, it depends on what you're trying to show it is a multiple of. Here we wanted a multiple of 4, so I realised that because we have squares, we may as well use  2n , as when squared, that'll give us the factor of 4 we want.

    However, using  2n and  2n+1 will result in larger coefficients, which given the nature of questions you're likely to get, will probably make the proofs come out a lot easier / nicer.
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    (Original post by kingaaran)
    Hmm, it depends on what you're trying to show it is a multiple of. Here we wanted a multiple of 4, so I realised that because we have squares, we may as well use  2n , as when squared, that'll give us the factor of 4 we want.

    However, using  2n and  2n+1 will result in larger coefficients, which given the nature of questions you're likely to get, will probably make the proofs come out a lot easier / nicer.
    That makes sense. Thank you.

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