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    10 days to go now
    Formula booklet - Link
    Calculator - Link
    I would recommend this calculator for any module tbh. It can help you with Matrices, Complex numbers and Numerical solutions.
    Specification - Link
    Scroll down to page 43 to find the content you'll need to know for FP1.

    Edexcel Questions by topic:

    Ch.1 Complex Numbers - Link
    Ch.2 Numerical Solutions - Link
    Ch.3 Coordinate Systems - Link
    Ch.4 Matrix Algebra - Link
    Ch.5 Series - Link
    Ch.6 Induction - Link

    Edexcel Past Papers - Link

    Edexcel had a specification change in 2009 and unfortunately this has left us with papers from 2009 - 2015. Make the most of these papers and I'd recommend you do them all twice over by exam day.

    Harder Questions by topic: (Thanks to TeeEm)

    Complex Numbers - Link
    Complex roots - Link
    Induction - Link
    Series - Link

    Extra Questions by topic: (Thanks to kingaaran)
    Answers will soon follow for these worksheets and you can find past papers here on CrashMaths

    Complex Numbers Worksheet - Link
    Numerical Solutions Worksheet - Link
    Conics Worksheet - Link
    Matrices - Soon to come
    Series Worksheet - Link
    Proof By Induction Worksheet - Link



    ExamSolutions - Link
    MathsWebsite - Link
    FP1 book PDF - Link
    Complex Numbers - Link

    I want to wish everyone the best of luck for this module. Looking back at discussions from a few years ago, I've noticed a pattern. The people who score low in this module regard this module as 'easy' and never practice it. Hence, they end up with this being the module they scored lowest on. Stupid right? Don't make that mistake.

    I have given you everything you need to score high on this module. Use it to your advantage, so you're not having to retake modules next year. Again, Good luck everyone!
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    Only exam that I'm looking forward to
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    Please could anyone explain question 10ii on this paper https://7cba9babeb0db0ff9468853e0b2d...%20Edexcel.pdf as I don't understand what to do with the 2n in the series? Many thanks
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    (Original post by economicss)
    Please could anyone explain question 10ii on this paper https://7cba9babeb0db0ff9468853e0b2d...%20Edexcel.pdf as I don't understand what to do with the 2n in the series? Many thanks
    Did you notice that the the summation starts from r = 0 rather than r = 1? It might help if you write out the series explicitly. There's also a new FP1 thread here - you might want to start posting there!
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    (Original post by aymanzayedmannan)
    Did you notice that the the summation starts from r = 0 rather than r = 1? It might help if you write out the series explicitly. There's also a new FP1 thread here - you might want to start posting there!
    Thanks, I've just written it out and I thought there would be n lots of (2n+1) but the mark scheme seems to show something different? Thank you
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    (Original post by aymanzayedmannan)
    Did you notice that the the summation starts from r = 0 rather than r = 1? It might help if you write out the series explicitly. There's also a new FP1 thread here - you might want to start posting there!
    I think what's confusing me is why the mark scheme has (n+1) as I thought being n was on top it would be n? thanks
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    (Original post by economicss)
    I think what's confusing me is why the mark scheme has (n+1) as I thought being n was on top it would be n? thanks
    Sorry for the late reply. We know that \displaystyle \sum_{r=0}^{n}2n = 2n\sum_{r=0}^{n}1 = 2n\left \left ( 1 + \cdots +1 \right ), right? But because we're starting from r=0 we're adding an extra 1, so the result of the summation is n+1 rather than n when r=1 because there's an extra term.
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    (Original post by aymanzayedmannan)
    There's also a new FP1 thread here - you might want to start posting there!
    Moved it all over for you. :-)
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    Nice thread!

    As if it wasn't bad enough that for there are no Solomon papers for FP1, to make matters worse we can only properly use papers from 2009 onwards!

    For my other modules ive started doing papers from 2001 onwards which should give me plently of past papers to do right up to the exam

    But for fp1 I hope I dont end up running out of past papers to do, if I do I guess I could just re do some of them idk
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    Aiming for 100 in this module. The Madasmaths solutions have been very helpful.It helps me to answer questions quickly in ways I wouldn't have thought of before. But I'm currently making that stupid mistake and I'm only practising once per week/ two weeks.
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    (Original post by NotNotBatman)
    Aiming for 100 in this module. The Madasmaths solutions have been very helpful.It helps me to answer questions quickly in ways I wouldn't have thought of before. But I'm currently making that stupid mistake and I'm only practising once per week/ two weeks.
    Good luck - it's quite a nice module to get 100 in, which I unfortunately missed out on. Edexcel are loving their conic sections in FP1 from what it seems, so try getting really good at those!
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    (Original post by aymanzayedmannan)
    Sorry for the late reply. We know that \displaystyle \sum_{r=0}^{n}2n = 2n\sum_{r=0}^{n}1 = 2n\left \left ( 1 + \cdots +1 \right ), right? But because we're starting from r=0 we're adding an extra 1, so the result of the summation is n+1 rather than n when r=1 because there's an extra term.
    Thanks so much!!
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    (Original post by aymanzayedmannan)
    Sorry for the late reply. We know that \displaystyle \sum_{r=0}^{n}2n = 2n\sum_{r=0}^{n}1 = 2n\left \left ( 1 + \cdots +1 \right ), right? But because we're starting from r=0 we're adding an extra 1, so the result of the summation is n+1 rather than n when r=1 because there's an extra term.
    Thank you, I think I'm starting to understand, does the 2n come out the front because it's almost a constant in effect? If a series starts with r=0 will it always be the case that the terms are then (n+1)? Thanks
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    (Original post by economicss)
    Thank you, I think I'm starting to understand, does the 2n come out the front because it's almost a constant in effect?
    Yes.

    If a series starts with r=0 will it always be the case that the terms are then (n+1)? Thanks
    If it starts from r=0 and ends at r=n then yes, there will be (n+1) terms.
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    (Original post by Zacken)
    Yes.



    If it starts from r=0 and ends at r=n then yes, there will be (n+1) terms.
    Thanks so much!
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    Hi, this is probably a stupid question but on 8b of this paper https://0025309b76bc88a0e4c3444acd03...%20kprime2.pdf could anyone explain please how you know that the area of triangle OBC= the area of triangles OBS and OCS? Thanks
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    (Original post by economicss)
    Hi, this is probably a stupid question but on 8b of this paper https://0025309b76bc88a0e4c3444acd03...%20kprime2.pdf could anyone explain please how you know that the area of triangle OBC= the area of triangles OBS and OCS? Thanks
    Look at the sketch, it's fairly obvious. This is why sketches and diagrams are vital.
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    (Original post by Zacken)
    Look at the sketch, it's fairly obvious. This is why sketches and diagrams are vital.
    Thanks, seen it now! On question 6ib of this paper https://699d3c34b250207412778630ab10...%20Answers.pdf why is the rotation 210 degrees because when you solve using the equation matrix in the formula book you get cos theta equals -root3/2 which would be 150 degrees? Many thanks
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    (Original post by economicss)
    Thanks, seen it now! On question 6ib of this paper https://699d3c34b250207412778630ab10...%20Answers.pdf why is the rotation 210 degrees because when you solve using the equation matrix in the formula book you get cos theta equals -root3/2 which would be 150 degrees? Many thanks
    Nah, you have the system: \cos \theta = -\frac{\sqrt{3}}{2}, \sin \theta = -\frac{1}{2} so solving this using your normal cast diagram method gets you:

    \theta = \pi + \cos^{-1} \frac{\sqrt{3}}{2}, \pi - \cos^{-1} \frac{\sqrt{3}}{2} since \cos is negative in the second and third quadrant. This gives you 150^{\circ} or 210^{\circ}, you now need to check which one fits the second equation \sin \theta = -\frac{1}{2}, we have \sin 150^{\circ} = \frac{1}{2} \neq -\frac{1}{2} so we throw that one away and we check \sin 210^{\circ} = -\frac{1}{2} so \theta = 210^{\circ} is what we want.
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    (Original post by Zacken)
    Nah, you have the system: \cos \theta = -\frac{\sqrt{3}}{2}, \sin \theta = -\frac{1}{2} so solving this using your normal cast diagram method gets you:

    \theta = \pi + \cos^{-1} \frac{\sqrt{3}}{2}, \pi - \cos^{-1} \frac{\sqrt{3}}{2} since \cos is negative in the second and third quadrant. This gives you 150^{\circ} or 210^{\circ}, you now need to check which one fits the second equation \sin \theta = -\frac{1}{2}, we have \sin 150^{\circ} = \frac{1}{2} \neq -\frac{1}{2} so we throw that one away and we check \sin 210^{\circ} = -\frac{1}{2} so \theta = 210^{\circ} is what we want.
    Thanks, really appreciate your help
 
 
 
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