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    (Original post by tazza ma razza)
    I think simultaneous equations w/ matricies will come up - never seen it before and i have no idea how to do them lol
    I was thinking that! and me neither lmaoo
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    (Original post by tazza ma razza)
    I think simultaneous equations w/ matricies will come up - never seen it before and i have no idea how to do them lol
    Do you mean using simultaneous equations to solve matrices? I don't think that's on spec (although it was in my GCSE).
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    (Original post by Zacken)
    Do you mean using simultaneous equations to solve matrices? I don't think that's on spec (although it was in my GCSE).
    chapter 4 ex 4j and point 4.11 example 24

    seems easy enough just want to confirm the order of each element within the matrix
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    (Original post by tazza ma razza)
    chapter 4 ex 4j and point 4.11 example 24

    seems easy enough just want to confirm the order of each element within the matrix
    Don't have a textbook, but:

    If you have the simultaneous equations:

    ax+by = c

    and

    dx + ey = f

    Then you write down:

    \displaystyle \begin{pmatrix}a  &b \\ d & e\end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix}c \\ f \end{pmatrix}

    Which makes sense, because if you multiple out the LHS you end up with:

    \displaystyle \begin{pmatrix} ax + by \\ dx + ey \end{pmatrix} = \begin{pmatrix} c \\ f \end{pmatrix}

    and comparing components gets you the original equation.
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    (Original post by Zacken)
    Don't have a textbook, but:

    If you have the simultaneous equations:

    ax+by = c

    and

    dx + ey = f

    Then you write down:

    \displaystyle \begin{pmatrix}a  &b \\ d & e\end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix}c \\ f \end{pmatrix}
    yh i'm just getting that - tbh i would just use my graphical to solve it for me and then i know which values to get lol
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    (Original post by tazza ma razza)
    yh i'm just getting that - tbh i would just use my graphical to solve it for me and then i know which values to get lol
    Look at my edited post, it's not hard to understand why that works.
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    (Original post by Zacken)
    Look at my edited post, it's not hard to understand why that works.
    ah i just expanded it on paper and i get it - working through the example in the book.

    tbh i can't think of anything hard that will trip me up. it will be a silly mistake somewhere. conics most likely will be that area.
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    (Original post by economicss)
    Is it part a or b you're struggling with?
    I'm also stuck on the question (part b), i cant seem to do the last part f(k+1), would someone be able to write down the answer?
    Thanks
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    (Original post by Strom)
    I was thinking that! and me neither lmaoo
    ex 1g also
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    example 10 ch4
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    Can anyone refer me to a tough proof by divisibility question if any?

    Edit: I found something - no worries.
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    For the proof of divisibility qs- I always use the following method- can someone pls check for me if it is the right way to do it/would get me full marks:

    E.g. prove that f(n)=7^(2n)-48n-1 is divisible by 2304
    sub in 1, make assumption n=k etc... then

    f(k+1)-f(k)= 7^(2k+2) - 48(k+1) -1 - 7^(2k) +48k + 1
    = 48(7^(2k)) - 48

    then I rearrange f(k) to make 7^(2k) subject of the equation
    i.e. 7^(2k)= f(k) + 48k + 1
    Then sub it in to f(k+1)-f(k)

    Is this a correct way of doing it??
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    (Original post by thesmallman)
    For the proof of divisibility qs- I always use the following method- can someone pls check for me if it is the right way to do it/would get me full marks:

    E.g. prove that f(n)=7^(2n)-48n-1 is divisible by 2304
    sub in 1, make assumption n=k etc... then

    f(k+1)-f(k)= 7^(2k+2) - 48(k+1) -1 - 7^(2k) +48k + 1
    = 48(7^(2k)) - 48

    then I rearrange f(k) to make 7^(2k) subject of the equation
    i.e. 7^(2k)= f(k) + 48k + 1
    Then sub it in to f(k+1)-f(k)

    Is this a correct way of doing it??
    Yes.
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    (Original post by thesmallman)
    For the proof of divisibility qs- I always use the following method- can someone pls check for me if it is the right way to do it/would get me full marks:

    E.g. prove that f(n)=7^(2n)-48n-1 is divisible by 2304
    sub in 1, make assumption n=k etc... then

    f(k+1)-f(k)= 7^(2k+2) - 48(k+1) -1 - 7^(2k) +48k + 1
    = 48(7^(2k)) - 48

    then I rearrange f(k) to make 7^(2k) subject of the equation
    i.e. 7^(2k)= f(k) + 48k + 1
    Then sub it in to f(k+1)-f(k)

    Is this a correct way of doing it??
    If this is when n is more than or equal to 2 - you need to sub in 2 as someone pointed out on the forum. Then do the simple induction process - finally on to your conclusion. Done.
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    (Original post by NotNotBatman)
    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.
    Oh, wait - do you mean when you have find f(k) in f(k+1)? - I say since f(k) is div. by n then f(k+1) must be div. by n - then proceed with the conclusion.
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    (Original post by Marxist)
    If this is when n is more than or equal to 2 - you need to sub in 2 as someone pointed out on the forum. Then do the simple induction process - finally on to your conclusion. Done.
    (Original post by Zacken)
    Yes.

    Thanks guys!!!
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    How many got full marks on June 2012 FP1?
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    Is  ZZ^* =\left |Z  \right |^2 For all complex numbers?
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    For the June 2015 IAL paper Q5 (b), will my answer get full marks? As the mark scheme says the 'use of negative lengths scores M0'. What does that mean?

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    Any hard non-edexcel papers?
 
 
 
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