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    If your summation formula is in your formula book, I think this is fine. The examiner will know that you have just used the formula from your book and taken this legal short cut (as opposed to just stating it from memory).

    What I would have a problem with is that you've not shown me that n^3 + 2n^2 + 7n + 14 factorises to  (n+2)(n^2 + 7) . It looks like you've just assumed it because the question says it, its not really an easy factorisation to notice lol. I would say you would need to show how you have factorised this, but using the formula to get to this stage is fine.
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    (Original post by claret_n_blue)
    What I would have a problem with is that you've not shown me that n^3 + 2n^2 + 7n + 14 factorises to  (n+2)(n^2 + 7) . I would say you would need to show how you have factorised this, but using the formula to get to this stage is fine.
    Yeah that's what I was worried about. I'm not too confident when trying to factorise cubics, so would expanding out the step I am trying to get to and comparing with my unfactorised bracket suffice?
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    (Original post by Lunch_Box)
    I am asked to prove the following:


    Am I simply allowed to do this step after using the summation formulae and conclude that it is true?


    ... or does the examiner expect me to include the intermediate steps?
    Since it's a proof, you should really include all the steps you take.

    \frac{1}{4}n[n^3 + 2n^2 + 7n + 14] = \frac{1}{4}n[n(n^2+7) + 2(n^2 +7)]
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    (Original post by claret_n_blue)
    If your summation formula is in your formula book, I think this is fine. The examiner will know that you have just used the formula from your book and taken this legal short cut (as opposed to just stating it from memory).

    What I would have a problem with is that you've not shown me that n^3 + 2n^2 + 7n + 14 factorises to  (n+2)(n^2 + 7) . It looks like you've just assumed it because the question says it, its not really an easy factorisation to notice lol. I would say you would need to show how you have factorised this, but using the formula to get to this stage is fine.
    Would it be fine to look at the given expression, looking for linear factors, dividing getting the quadratic and then factor that from there?
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    (Original post by Indeterminate)
    Since it's a proof, you should really include all the steps you take.

    \frac{1}{4}n[n^3 + 2n^2 + 7n + 14] = \frac{1}{4}n[n(n^2+7) + 2(n^2 +7)]
    Thank you.
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    (Original post by Lunch_Box)
    Yeah that's what I was worried about. I'm not too confident when trying to factorise cubics, so would expanding out the step I am trying to get to and comparing with my unfactorised bracket suffice?
    Not quite. Usually what happens with proofs is that you start on one side and then when you end the question your answer is the other side. So for example, in your proof, you started on the left hand side and then you have to keep on doing your manipulations and whatever until you get your answer, which is the right hand side. You aren't allowed to touch the right hand side at all.

    Seeing as your doing FP1, I assume you know how to factorise cubics yeah?
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    (Original post by UKBrah)
    Would it be fine to look at the given expression, looking for linear factors, dividing getting the quadratic and then factor that from there?
    To factorise? Yeah, thats fine. You've shown that you've not just guessed what the answer is, but how to get to the answer.
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    (Original post by claret_n_blue)
    Seeing as your doing FP1, I assume you know how to factorise cubics yeah?
    Nope. Not part of my course (Edexcel) I don't believe. I'll learn how to now then.
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    (Original post by UKBrah)
    Would it be fine to look at the given expression, looking for linear factors, dividing getting the quadratic and then factor that from there?
    That would all be a bit unnecessary, but you could do it like that.

    In FP1 proofs by induction, you often need to simplify complex expressions, but all of this can easily be done without use of the factor theorem/long division.
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    (Original post by Lunch_Box)
    Nope. Not part of my course (Edexcel) I don't believe. I'll learn how to now then.
    Is factor theorem not in C2?
    If not you'll certainly come across it in C3
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    (Original post by claret_n_blue)
    To factorise? Yeah, thats fine. You've shown that you've not just guessed what the answer is, but how to get to the answer.
    Thanks.
    While im at it, if you get a question on the sum from r = 1 to 2n like \sum\limits_{r=1}^{2n} r would it simply be the sum to r = 1 to n but with 2n instead of n? If so why?
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    (Original post by UKBrah)
    Thanks.
    While im at it, if you get a question on the sum from r = 1 to 2n like \sum\limits_{r=1}^{2n} r would it simply be the sum to r = 1 to n but with 2n instead of n? If so why?
    Use the formulae in the data booklet and replace n with 2n

    (Original post by NiceToMeetYou)
    Is factor theorem not in C2?
    If not you'll certainly come across it in C3
    Yes factor theorem is in C2, but at a very basic level where they always give you a solution, unlike in this.

    And in C1 cubics show up but you can always take a factor of x outside.
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    (Original post by Indeterminate)
    That would all be a bit unnecessary, but you could do it like that.

    In FP1 proofs by induction, you often need to simplify complex expressions, but all of this can easily be done without use of the factor theorem/long division.
    oh ok, how would you simply the cubic?
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    (Original post by Lunch_Box)
    Use the formulae in the data booklet and replace n with 2n



    Yes factor theorem is in C2, but at a very basic level where they always give you a solution, unlike in this.

    And in C1 cubics show up but you can always take a factor of x outside.
    In exam questions 99% of the time you'll only need to try from -2 to +2 to find one factor then use long division or inspection
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    (Original post by Lunch_Box)
    Nope. Not part of my course (Edexcel) I don't believe. I'll learn how to now then.
    Easiest way I do it is using the following steps:

    1) By trial and error, you find a solution such that your function is 0. So let's say you have some function, f(x). You want to find a value for x, such that f(x) = 0.

    This is normally just done by guessing, but they tend to make the numbers easy for this. So you would really just try like x = ±1, ±2, ±3 and so on. You wouldn't need to go too far until you find your solution.

    2) Let's say that at some point 'a', you get f(a) = 0. What this means is that x = a is solution to your function (i.e a point where it crosses the 'x' axis) and so we can say that (x - a) = 0 is a root. With your example, notice how n = -2 is a solution and so n + 2 = 0 is a root.

    3) Now you have two options. You can either use the long division way and divide f(x) by (x - a) and this will give you a quadratic. Either that or you can compare coefficients but i don't use this method so don't know how to explain it. You can then factorise this quadratic in the normal way.

    If you tried dividing your cubic by n + 2, you should get n² + 7. This can't be factorised any more so your final answer is what you got.

    Another way to do it is the way Indeterminate has done it.

    You have a cubic which is

     n^3 + 2n^2 +7n + 14 .

    What we see here is that 3 terms contain an 'n' and two terms contain a '2'. So lets take the two terms containing a '2' and then take the other two terms left that have an 'n' and so we can re-write and factorise to get

     n^3 + 7n + 2n^2 + 14 = n(n^2 + 7) + 2(n^2 + 7) .

    Now clearly, we have one common factor here and so we can factorise this out and get

    (n^2 + 7)(n + 2) .
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    (Original post by claret_n_blue)
    Easiest way I do it is:

    1) By trial and error, you find a solution such that your function is 0. So let's say you have some function, f(x). You want to find a value for x, such that f(x) = 0.

    This is normally just done by guessing, but they tend to make the numbers easy for this. So you would really just try like x = ±1, ±2, ±3 and so on. You wouldn't need to go too far until you find your solution.

    2) Let's say that at some point 'a', you get f(a) = 0. What this means is that x = a is solution to your function (i.e a point where it crosses the 'x' axis) and so we can say that (x - a) = 0 is a root. With your example, notice how n = -2 is a solution and so n + 2 = 0 is a root.

    3) Now you have two options. You can either use the long division way and divide f(x) by (x - a) and this will give you a quadratic. Either that or you can compare coefficients but i don't use this method so don't know how to explain it. You can then factorise this quadratic in the normal way.

    If you tried dividing your cubic by n + 2, you should get n² + 7. This can't be factorised any more so your final answer is what you got.

    Another way to do it is the way Indeterminate has done it.

    You have a cubic which is

     n^3 + 2n^2 +7n + 14 .

    What we see here is that 3 terms contain an 'n' and two terms contain a '2'. So lets take the two terms containing a '2' and then take the other two terms left that have an 'n' and so we can re-write and factorise to get

     n^3 + 7n + 2n^2 + 14 = n(n^2 + 7) + 2(n^2 + 7) .

    Now clearly, we have one common factor here and so we can factorise this out and get

    (n^2 + 7)(n + 2) .
    Quality never seen this before.
    Appriciate it claret/indeterminate.
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    In edexcel fp1 when proving summation formula, it is proved by mathematical induction? Show that LHS = RHS is true for n=1 then assumption n = k followed by n = k + 1

    Im pretty sure there's no such shortcut to proofs in fp1? Have you learnt this?

    I think other replies in the thread may have done a different exam board
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    (Original post by raiden95)
    In edexcel fp1 when proving summation formula, it is proved by mathematical induction? Show that LHS = RHS is true for n=1 then assumption n = k followed by n = k + 1

    Im pretty sure there's no such shortcut to proofs in fp1? Have you learnt this?

    I think other replies in the thread may have done a different exam board
    This could be an induction question or it could be a proof using standard results. The question will clearly state if induction is required.
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    (Original post by Mr M)
    This could be an induction question or it could be a proof using standard results. The question will clearly state if induction is required.
    I thought mathematical induction is the only way to prove it works for all numbers, and for the other type it says show that the left side is equal to the right side?
 
 
 
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