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    http://mathsathawthorn.pbworks.com/f/FP%21Jan06.pdf
    How do I do q6b?
    Thanks
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    (Original post by English-help)
    http://mathsathawthorn.pbworks.com/f/FP%21Jan06.pdf
    How do I do q6b?
    Thanks
    Remember, you have to get the induction started!
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    (Original post by English-help)
    http://mathsathawthorn.pbworks.com/f/FP%21Jan06.pdf
    How do I do q6b?
    Thanks
    The proposition deals with the case where n is a positive integer, meaning it belongs to the set of natural numbers \mathbb{N}={1,2,3,4,...}

    Now mathematical induction requires you to have a base case, which in this proposition would be n=1, and showing that P works for it before moving onto the inductive step which you've done in part a.

    So, you need to cover the first step of proof by induction and show how P is actually false.
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    (Original post by RDKGames)
    The proposition deals with the case where n is a positive integer, meaning it belongs to the set of natural numbers \mathbb{N}={1,2,3,4,...}

    Now mathematical induction requires you to have a base case, which in this proposition would be n=1, and showing that P works for it before moving onto the inductive step which you've done in part a.

    So, you need to cover the first step of proof by induction and show how P is actually false.
    Okay so how would i do that? The thing is when i put n=1 and other numbers it shows it doesnt work? So do i put 2 numbers inside and show they dont work?
    & In q8 how would i do it and would i know what the roots are equal to?
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    (Original post by English-help)
    Okay so how would i do that? The thing is when i put n=1 and other numbers it shows it doesnt work? So do i put 2 numbers inside and show they dont work?
    & In q8 how would i do it and would i know what the roots are equal to?
    Yes, ifit isn;t true for n = 1 then it is false.

    If the roots form an AP then we can call them a, a + 1 and ....
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    see if it is true for n = 1

    :teehee:
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    (Original post by English-help)
    Okay so how would i do that? The thing is when i put n=1 and other numbers it shows it doesnt work? So do i put 2 numbers inside and show they dont work?
    & In q8 how would i do it and would i know what the roots are equal to?
    Well yes that's the point. The base case fails the initial check so the proposition collapses.

    For Q8 consider the fact that you have roots a,a+1,a+2 and what restriction must a have in order for all roots to be negative?

    Then just consider \displaystyle \sum \alpha \beta = 47 in terms of a.
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    (Original post by Muttley79)
    Yes, ifit isn;t true for n = 1 then it is false.

    If the roots form an AP then we can call them a, a + 1 and ....
    Ahh okay thanks!
    (Original post by the bear)
    see if it is true for n = 1

    :teehee:
    I did that tbh, i just never understood the question too well tbh

    (Original post by RDKGames)
    Well yes that's the point. The base case fails the initial check so the proposition collapses.

    For Q8 consider the fact that you have roots a,a+1,a+2 and what restriction must a have in order for all roots to be negative?

    Then just consider \displaystyle \sum \alpha \beta = 47 in terms of a.
    Ahhh okay and whats Beta? Why do we equal it to the coefficient of x though?
    When I have the roots a,a+1 and a+2 , do i times them all with each other and could i have any roots like a-1,a, a+1?
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    (Original post by English-help)

    Ahhh okay and whats Beta? Why do we equal it to the coefficient of x though?
    When I have the roots a,a+1 and a+2 , do i times them all with each other and could i have any roots like a-1,a, a+1?
    I assume you have learnt about roots of cubic polynomials?

    For an arbitrary cubic in the form ax^3+bx^2+cx+d=0 such that a,b,c,d \in \mathbb{Q} it has roots \alpha, \beta, \gamma such that:

    \alpha + \beta + \gamma=-\frac{b}{a}

    \alpha \beta + \beta \gamma + \alpha \gamma = \frac{c}{a}

    \alpha \beta \gamma = -\frac{d}{a}
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    (Original post by RDKGames)
    I assume you have learnt about roots of cubic polynomials?

    For an arbitrary cubic in the form ax^3+bx^2+cx+d=0 such that a,b,c,d \in \mathbb{Q} it has roots \alpha, \beta, \gamma such that:

    \alpha + \beta + \gamma=-\frac{b}{a}

    \alpha \beta + \beta \gamma + \alpha \gamma = \frac{c}{a}

    \alpha \beta \gamma = -\frac{d}{a}
    Brilliant! Thanks , now i understand it
    Much appreciated!
    Soo i have this paper but i cant find the MS and ive done the first 3 questions , could you tell me if they are right?
    http://mathsathawthorn.pbworks.com/f/FP1Jun06.pdf
    For q1, i got 1/4n (n+7)(n+1)(n+2)
    For q2, i got -2/(2x-3)^2
    For q3, i got z=1+3i
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    (Original post by English-help)
    Brilliant! Thanks , now i understand it
    Much appreciated!
    Soo i have this paper but i cant find the MS and ive done the first 3 questions , could you tell me if they are right?
    http://mathsathawthorn.pbworks.com/f/FP1Jun06.pdf
    For q1, i got 1/4n (n+7)(n+1)(n+2)
    For q2, i got -2/(2x-3)^2
    For q3, i got z=1+3i
    First two are fine but not the third one.
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    (Original post by RDKGames)
    First two are fine but not the third one.
    Shall i tell you how i did the 3rd one so you can tell me where I have gone wrong?
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    (Original post by English-help)
    Shall i tell you how i did the 3rd one so you can tell me where I have gone wrong?
    Go for it.
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    (Original post by RDKGames)
    Go for it.
    So the main question is z/z+1 =2+3i

    I cancelled the z's and got
    1+z=2+3i
    1+(x+iy)=2+3i
    1+x+iy=2+3i
    x+iy=1+3i
    Therefore z=1+3i?
    Ohh sugar, i realised i cant cancel the z's right?
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    (Original post by English-help)
    Brilliant! Thanks , now i understand it
    Much appreciated!
    Soo i have this paper but i cant find the MS and ive done the first 3 questions , could you tell me if they are right?
    http://mathsathawthorn.pbworks.com/f/FP1Jun06.pdf
    For q1, i got 1/4n (n+7)(n+1)(n+2)
    For q2, i got -2/(2x-3)^2
    For q3, i got z=1+3i
    Q1: Correct, but your notation on here is ambiguous: 1/4n (n+7)(n+1)(n+2) [i]could[/b] be interpreted as \dfrac{1}{4n(n+7)(n+1)(n+2)} (or more likely, [latex]\dfrac{1}{4n} (n+7)(n+1)(n+2)). It's better to over bracket and remove any ambiguity.

    Q2: That's right but with a question like this It's not whether you got the right answer, it's what your method looks like.

    Q3: No. If z/(1+z) = 2+3i then z = (2+3i)(1+z). If z = 1+3i, the RHS = (2+3i)(2+3i) = -5 + 12i, which definitely isn't the same as z (the LHS).
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    (Original post by English-help)
    So the main question is z/z+1 =2+3i

    I cancelled the z's and got
    1+z=2+3i
    1+(x+iy)=2+3i
    1+x+iy=2+3i
    x+iy=1+3i
    Therefore z=1+3i?
    Ohh sugar, i realised i cant cancel the z's right?
    Not quite sure how you cancelled the z's.

    Just express z=x+iy and then you have \frac{x+iy}{(x+1)+iy}=2+3i and multiply both sides by the denominator of the LHS and proceed from there.
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    (Original post by RDKGames)
    Not quite sure how you cancelled the z's.

    Just express z=x+iy and then you have \frac{x+iy}{(x+1)+iy}=2+3i and multiply both sides by the denominator of the LHS and proceed from there.
    Okay ill redo that now thanks
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    (Original post by DFranklin)
    Q1: Correct, but your notation on here is ambiguous: 1/4n (n+7)(n+1)(n+2) [i]could[/b] be interpreted as \dfrac{1}{4n(n+7)(n+1)(n+2)} (or more likely, [latex]\dfrac{1}{4n} (n+7)(n+1)(n+2)). It's better to over bracket and remove any ambiguity.

    Q2: That's right but with a question like this It's not whether you got the right answer, it's what your method looks like.

    Q3: No. If z/(1+z) = 2+3i then z = (2+3i)(1+z). If z = 1+3i, the RHS = (2+3i)(2+3i) = -5 + 12i, which definitely isn't the same as z (the LHS).
    Okayy thanks , ill redo number 3 now
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    (Original post by RDKGames)
    Not quite sure how you cancelled the z's.

    Just express z=x+iy and then you have \frac{x+iy}{(x+1)+iy}=2+3i and multiply both sides by the denominator of the LHS and proceed from there.
    Okay so i have got x+iy=2x+3ix+2iy+3i-3y+2
    Which can give-(x+iy)=3ix+3i-3y+2
    What do i do next?
    DFranklin
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    (Original post by English-help)
    Okay so i have got x+iy=2x+3ix+2iy+3i-3y+2
    Which can give-(x+iy)=3ix+3i-3y+2
    What do i do next?
    DFranklin
    Cant check if thats correct right now, but you essentially need to equate the real and imaginary parts of both sides of the equation.


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