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    (Original post by iMacJack)
    Any predictions then?
    I think divisibility will (unfortunately) come up, the other induction proofs are quite a bit easier. I think edexcel might throw in an r=0 series but that's been covered so many times in this thread it's simple now. Also I think we'll get a harder than usual /obscure complex numbers question and probably a parabola question with a lot of letters and algebra.
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    If I carried on that Iast Iine with:
     k^2 -2k + 3 -(\frac{1}{2})^k^+^1 + k^2 + 2k + 1
     = 2k^2 +4 - (\frac{1}{2})^k^+^1
     = 2(k+1)^2 - 4(k+1) + 6 -(\frac{1}{2})^k^+^1
    WouId I get aII the marks for that inductive step?

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    (Original post by Yunique)
    If I carried on that Iast Iine with:
     k^2 -2k + 3 -(\frac{1}{2})^k^+^1 + k^2 + 2k + 1
     = 2k^2 +4 - (\frac{1}{2})^k^+^1
     = 2(k+1)^2 - 4(k+1) + 6 -(\frac{1}{2})^k^+^1
    WouId I get aII the marks for that inductive step?

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    Yes.
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    (Original post by yesyesyesno)
    I think edexcel might throw in an r=0 series but that's been covered so many times in this thread it's simple now.
    Oh god, I've answered the same question like 7/8 times on this thread. :lol:
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    For divisibility tests, is the conclusion different from true for n=k, k+1,n =1, so all n. Do you have to state since f(k) divisible by n, so f(k+1) is divisible by n etc.
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    If we were to find a point on a parabola that is equidistant from a point on the directrix and the focus, would the line joining the points always be perpendicular to the directrix?
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    (Original post by Zacken)
    Oh god, I've answered the same question like 7/8 times on this thread. :lol:
    Ahh want to make it 8/9 Pleeeease! Just incase I remember seeing this in a past paper last question just cant find it anymore! :O Didn't understand it fully at the time

    If I had to predict something coming up I think a proof by induction of Un+2=... type of question may come up!
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    Why do some mark schemes say 'all n and nEZ+'. Surely it's just the latter? (induction)
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    (Original post by SeanFM)
    :dontknow: it'd be a bit nasty in that the sum of n terms would be going from r=0 to r=n-1. I wouldn't put it past them but it seems needlessly complicated.
    In this case you would prove n = 0 (instead of n = 1) right?
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    (Original post by Nikhilm)
    Why do some mark schemes say 'all n and nEZ+'. Surely it's just the latter? (induction)
    How you state it using set theory is  \forall n, n \in \mathbb{Z}^+ .

    This reads as "for all n where n is an element of  \mathbb{Z}^+ ".
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    Mathematical induction Q:

    say if you have a summation formula, 1 to n, for (2n + r^2 + 4r). I.e. the n is within the equation. It would be impossible to prove an equality by mathematical induction right? Because the actual equation would change when you have k or k+1?

    Zacken
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    So, matrix multiplication is not commutative, but for the induction questions it is, because (where \mathbf A is a matrix)

     \mathbf{A}^k^+^1 = \mathbf{A}^k\cdot \mathbf{A}^1 and  \mathbf{A}^k^+^1 = \mathbf{A}^1\cdot \mathbf{A}^k due to index laws, so does it matter what way around we multiply it in the in inductive proof?
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    Anyone know any good tips on how to work out the transformations of a matrix?
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    (Original post by Eshanth)
    Anyone know any good tips on how to work out the transformations of a matrix?
    for rotations find the inverse of it using your calculator and the formula in the booklet
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    Not sure if this has been brought up before but I noticed there was one proof by induction question in the textbook that had a factorial in it and I was wondering the best way to go about these problems if it were to come up in the exam tomorrow.
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    (Original post by Don John)
    The determinant of the product of two square and equal dimension matrices is equal to the product of both matrix's determinants. Also, the determinant of the inverse of a matrix is equal to the inverse of the determinant of the original matrix.

    So if the determinant of the inverse is a third, what does that tell you about \det(\mathbf{A})? Then can you find \det(\mathbf{B}) after that?
    Great, got it now, thanks so much
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    (Original post by NotNotBatman)
    So, matrix multiplication is not commutative, but for the induction questions it is, because (where \mathbf A is a matrix)

     \mathbf{A}^k^+^1 = \mathbf{A}^k\cdot \mathbf{A}^1 and  \mathbf{A}^k^+^1 = \mathbf{A}^1\cdot \mathbf{A}^k due to index laws, so does it matter what way around we multiply it in the in inductive proof?
    No it doesn't matter.
    Although matrix multiplication isn't commutative, it is associative. So, \mathbf{(AA)A} is the same as \mathbf{A(AA)} ... therefore if you extend it further you will see that\mathbf{A}^k\cdot \mathbf{A}^1 is the same as \mathbf{A}^1\cdot \mathbf{A}^k.
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    (Original post by LukeB98)
    Not sure if this has been brought up before but I noticed there was one proof by induction question in the textbook that had a factorial in it and I was wondering the best way to go about these problems if it were to come up in the exam tomorrow.
    treat it as a normal induction question and try to collect like terms
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    (Original post by rashid.mubasher)
    No it doesn't matter.
    Although matrix multiplication isn't commutative, it is associative. So, \mathbf{(AA)A} is the same as \mathbf{A(AA)} ... therefore if you extend it further you will see that\mathbf{A}^k\cdot \mathbf{A}^1 is the same as \mathbf{A}^1\cdot \mathbf{A}^k.
    I see, thanks for the reply.
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    Just did the Jan 15 IAL paper - probably one of the worst things ever


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