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    4+ 2root6 + 6root2 + 2root3
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    (Original post by Zacken)
    On the mean side, but whatever - show that:

    \displaystyle

\begin{equation*}\frac{\sqrt{3} + \sqrt{2} - 1}{\sqrt{3} - \sqrt{6} + 1} = 2 + a\sqrt{2} + b\sqrt{3} + c\sqrt{6}\end{equation*}

    where a,b,c are integers to be determined.
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    never mind - completely wrong :lol:
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    (Original post by not_lucas1)
    4+ 2root6 + 6root2 + 2root3
    You're supposed to get 2root3-2
    Try this for the denominator: (root3(1-root2)+1)(root3(1+root2)+1) which gives 2root3-2. It's easier because it's in quadratic form. Then for the numerator do this (root3 + root2 -1)(root3(1+root2)+1)

    Basically factories the denominator to make it easier p.s.(quote me so I can see the notification)
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    (Original post by saffronlord)
    isnt this c2?
    You might be right I do not know.
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    (Original post by SeanFM)
    Here is a question for you to try


    A sequence of numbers  a_0, a_1, a_2, ... is given by  a_{n+1} = 2a_{n} + 6 - 4n.

    Given that a_2 = 6, find a_1 and a_3.

    Hence find the value of \sum_{i=1}^{10} a_{i}.

    Hence, or otherwise, find the value of \sum_{i=0}^{10} a_{i}.
    Would they ever ask you that?
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    (Original post by YeSand)
    Would they ever ask you that?
    the question is standard, the last question would only be worth 1 mark I think.
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    (Original post by imran_)
    the question is standard, the last question would only be worth 1 mark I think.
    For this:
    Hence find the value of .


    You need to do find a1 - 10 and add them up.

    That will take a lot of time in the exam. They will never ask you something this ridiculous. Going up to a6 is possibly the furthest they should go.
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    (Original post by imran_)
    the question is standard, the last question would only be worth 1 mark I think.
    The last question you find a0 and add it to the previous a1-10. Simple.
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    (Original post by YeSand)
    Would they ever ask you that?
    I don't think it'd be as complicated as that. Plus, some of it might be some C2 stuff.
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    (Original post by YeSand)
    For this:
    Hence find the value of .


    You need to do find a1 - 10 and add them up.

    That will take a lot of time in the exam. They will never ask you something this ridiculous. Going up to a6 is possibly the furthest they should go.
    you just use the sequence equation Sn=n/2(2a+(n-1)d) or Sn=n/2(a+l)
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    (Original post by EndOfTheTour)
    Has anyone got access to the C1 mymaths? Is it worth using? My school's password doesn't seem to be working, should I even bother?
    I find MyMaths rubbish for explaining concepts for A level. The best revision tool is Exam Solutions.
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    c1 has never been hard, hopefully they dont make it hard this year
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    (Original post by XOR_)
    you just use the sequence equation Sn=n/2(2a+(n-1)d) or Sn=n/2(a+l)
    That's a common error. Those formulas won't work on recurrent sequences.
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    (Original post by surina16)
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    never mind - completely wrong :lol:
    (Original post by not_lucas1)
    cant solve it atm but multiply the whole thing by the denominator with a flipped sign so - to + then simplify it? is that correct?
    (Original post by Someboady)
    Er are these the correct answers:I probably messed up somewhere :/
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    a = 1, b = 2 , c = 1.
    (Original post by odera_dim)
    a=1b=1c=1This was a nasty question and I'm not too sure about the answer but this is what I got 😩
    (Original post by techfan42)
    Same, I got that too
    (Original post by JN17)
    ^ I also got a,b,c all equal to 1
    (Original post by imran_)
    this question is taking me too long smh

    \displaystyle

\begin{equation*}\frac{\sqrt{3} + \sqrt{2} - 1}{1 + \sqrt{3} - \sqrt{6}} \times \frac{1 + \sqrt{3} + \sqrt{6}}{1 + \sqrt{3} + \sqrt{6}} = \frac{1 + 2\sqrt{2} + \sqrt{3}}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3}+1} = 2 + \sqrt{2} + \sqrt{3} + \sqrt{6}\end{equation*}
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    (Original post by YeSand)
    That's a common error. Those formulas won't work on recurrent sequences.
    thats why you need to work them out
    after working out a1 that gives you the first term a=2
    work out a2 and a3 you'll get the next terms a2=6 a3=10

    then work out the difference and that would be your (d) d=4

    N= 10 cos theres 10 terms

    now work from there
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    (Original post by Zacken)
    \displaystyle

\begin{equation*}\frac{\sqrt{3} + \sqrt{2} - 1}{1 + \sqrt{3} - \sqrt{6}} \times \frac{1 + \sqrt{3} + \sqrt{6}}{1 + \sqrt{3} + \sqrt{6}} = \frac{1 + 2\sqrt{2} + \sqrt{3}}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3}+1} = 2 + \sqrt{2} + \sqrt{3} + \sqrt{6}\end{equation*}
    eh I was only 5 root 2s off

    I was lost at the first step because of the 2 surds on the bottom and stupidly multiplied by (root6 - root3 + 1)
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    only in year 11 though so have an excuse?
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    (Original post by surina16)
    eh I was only 5 root 2s off

    I was lost at the first step because of the 2 surds on the bottom and stupidly multiplied by (root6 - root3 + 1)
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    only in year 11 though so have an excuse?
    Yes, that's the tricky bit.

    Stuff like this came up in my Year 11 IGCSE all the time.
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    (Original post by imran_)
    thats why you need to work them out
    after working out a1 that gives you the first term a=2
    work out a2 and a3 you'll get the next terms a2=6 a3=10

    then work out the difference and that would be your (d) d=4

    N= 10 cos theres 10 terms

    now work from there
    Be careful. Recurrence sequences do not have same common difference.
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    (Original post by Zacken)
    Yes, that's the tricky bit.

    Stuff like this came up in my Year 11 IGCSE all the time.
    :getmecoat:
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    Fam C1 is easy n sounds like you already done enough revision
 
 
 
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