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    (Original post by SeanFM)
    What kind of things throw you? There is time to learn them is it when r is not equal to 1?
    Particularly that, yes! I was planning on looking for some online resources (examsolutions etc) but didn't know if it'd be worth revising it a lot, seeing as from the past papers I've done it doesn't seem to come up that often... But Edexcel have seemed to quite like throwing curveballs recently...
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    (Original post by chloeabeki)
    How likely do you guys think it is that sigma notation will come up? I can do standard geometric sequences questions but for some reason that always throws me.
    Are you referring to this?

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    (Original post by chloeabeki)
    Particularly that, yes! I was planning on looking for some online resources (examsolutions etc) but didn't know if it'd be worth revising it a lot, seeing as from the past papers I've done it doesn't seem to come up that often... But Edexcel have seemed to quite like throwing curveballs recently...
    Right, to use the formula for the sum of a geometric series, you need i=0 at the bottom of the notation:
    \sum_{i=0}^n ar^i , where the r=0 and n mean that the first that the first i value you put in is 0, then the next is 1, then 2, ..all the way up to n (and up all the terms), so you end up with ar^0 + ar + ar^2 .. + ar^n.

    When you have a non-zero number at the bottom then you can't use the formula straight away - to do this, let's look at a simple example.

    If a=5, r=2, find ar^2 + ar^3 + ar^4.

    This can be expressed as
    \sum_{i=2}^4 5*2^i as ar^2 is the first term you want to compute (hence i=2) and ar^4 is the last. But you can't just use the formula with a=5, because the first term in our sequence is ar^2, not ar^0.

    So you can either:

    Define it as a sequence with a_{new}=a_{old}r^2, r_{new}=r_{old} and n = 3 (there are 3 terms, given by (4-2+1) because although the difference is n, there are n+1 numbers. This would give you \frac{(5*2^2(2^3-1)}{2-1}. It makes sense because if the first term is ar^2, then (ar^2)r = ar^3 and (ar^2)r^2 = ar^4.

    The alternative requires more steps, but you use

    \sum_{i=2}^4 ar^i  = \sum_{i=0}^4 ar^i -\sum_{i=0}^1 ar^i, which again makes sense as, to find the sum of the third,fourth and fifth term in the sum from i=0 to 4, you take away the other two terms, which are the first and second (a and ar), and since the are of the form \sum_{i=0}^n ar^i you can use the formulae directly and get the same answer as before.

    I really can't remember what they ask you in C2 about it but hopefully this is of some use.
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    (Original post by SeanFM)
    Right, to use the formula for the sum of a geometric series, you need i=0 at the bottom of the notation:
    \sum_{i=0}^n ar^i , where the r=0 and n mean that the first that the first i value you put in is 0, then the next is 1, then 2, ..all the way up to n (and up all the terms), so you end up with ar^0 + ar + ar^2 .. + ar^n.

    When you have a non-zero number at the bottom then you can't use the formula straight away - to do this, let's look at a simple example.

    If a=5, r=2, find ar^2 + ar^3 + ar^4.

    This can be expressed as
    \sum_{i=2}^4 5*2^i as ar^2 is the first term you want to compute (hence i=2) and ar^4 is the last. But you can't just use the formula with a=5, because the first term in our sequence is ar^2, not ar^0.

    So you can either:

    Define it as a sequence with a_{new}=a_{old}r^2, r_{new}=r_{old} and n = 3 (there are 3 terms, given by (4-2+1) because although the difference is n, there are n+1 numbers. This would give you \frac{(5*2^2(2^3-1)}{2-1}. It makes sense because if the first term is ar^2, then (ar^2)r = ar^3 and (ar^2)r^2 = ar^4.

    The alternative requires more steps, but you use

    \sum_{i=2}^4 ar^i  = \sum_{i=0}^4 ar^i -\sum_{i=0}^1 ar^i, which again makes sense as, to find the sum of the third,fourth and fifth term in the sum from i=0 to 4, you take away the other two terms, which are the first and second (a and ar), and since the are of the form \sum_{i=0}^n ar^i you can use the formulae directly and get the same answer as before.

    I really can't remember what they ask you in C2 about it but hopefully this is of some use.
    PRSOM
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    (Original post by SeanFM)
    Right, to use the formula for the sum of a geometric series, you need i=0 at the bottom of the notation:
    \sum_{i=0}^n ar^i , where the r=0 and n mean that the first that the first i value you put in is 0, then the next is 1, then 2, ..all the way up to n (and up all the terms), so you end up with ar^0 + ar + ar^2 .. + ar^n.

    When you have a non-zero number at the bottom then you can't use the formula straight away - to do this, let's look at a simple example.

    If a=5, r=2, find ar^2 + ar^3 + ar^4.

    This can be expressed as
    \sum_{i=2}^4 5*2^i as ar^2 is the first term you want to compute (hence i=2) and ar^4 is the last. But you can't just use the formula with a=5, because the first term in our sequence is ar^2, not ar^0.

    So you can either:

    Define it as a sequence with a_{new}=a_{old}r^2, r_{new}=r_{old} and n = 3 (there are 3 terms, given by (4-2+1) because although the difference is n, there are n+1 numbers. This would give you \frac{(5*2^2(2^3-1)}{2-1}. It makes sense because if the first term is ar^2, then (ar^2)r = ar^3 and (ar^2)r^2 = ar^4.

    The alternative requires more steps, but you use

    \sum_{i=2}^4 ar^i  = \sum_{i=0}^4 ar^i -\sum_{i=0}^1 ar^i, which again makes sense as, to find the sum of the third,fourth and fifth term in the sum from i=0 to 4, you take away the other two terms, which are the first and second (a and ar), and since the are of the form \sum_{i=0}^n ar^i you can use the formulae directly and get the same answer as before.

    I really can't remember what they ask you in C2 about it but hopefully this is of some use.
    damn SeanFM all that Maths
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    For the geometric series question, how hard can they get? I've been looking at the past papers and I believe that a wordy question related to compound interest is possible to come up, but after that C1 exam I wouldn't be surprised if they even make the easy chapter to be challenging as well.
    Also, for binomial expansion do you reckon that estimation question might appear again? They haven't really asked that question since Jan 2012.
    Coordinate geometry is the one I'm getting a bit panicked about. Last year it was a decent question, but I do not feel this year will be the same. All I know about this chapter is finding the midpoint, the length , equation of a circ le and tangents. Anything else that might appear under this chapter?
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    (Original post by oni176)
    For the geometric series question, how hard can they get? I've been looking at the past papers and I believe that a wordy question related to compound interest is possible to come up, but after that C1 exam I wouldn't be surprised if they even make the easy chapter to be challenging as well.
    Also, for binomial expansion do you reckon that estimation question might appear again? They haven't really asked that question since Jan 2012.
    Coordinate geometry is the one I'm getting a bit panicked about. Last year it was a decent question, but I do not feel this year will be the same. All I know about this chapter is finding the midpoint, the length , equation of a circ le and tangents. Anything else that might appear under this chapter?
    do madasmaths papers, you'll feel more prepared.
    These papers are designed to test every every part of the spec in depth. Theyre good practise
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    (Original post by SeanFM)
    Right, to use the formula for the sum of a geometric series, you need i=0 at the bottom of the notation:
    \sum_{i=0}^n ar^i , where the r=0 and n mean that the first that the first i value you put in is 0, then the next is 1, then 2, ..all the way up to n (and up all the terms), so you end up with ar^0 + ar + ar^2 .. + ar^n.

    When you have a non-zero number at the bottom then you can't use the formula straight away - to do this, let's look at a simple example.

    If a=5, r=2, find ar^2 + ar^3 + ar^4.

    This can be expressed as
    \sum_{i=2}^4 5*2^i as ar^2 is the first term you want to compute (hence i=2) and ar^4 is the last. But you can't just use the formula with a=5, because the first term in our sequence is ar^2, not ar^0.

    So you can either:

    Define it as a sequence with a_{new}=a_{old}r^2, r_{new}=r_{old} and n = 3 (there are 3 terms, given by (4-2+1) because although the difference is n, there are n+1 numbers. This would give you \frac{(5*2^2(2^3-1)}{2-1}. It makes sense because if the first term is ar^2, then (ar^2)r = ar^3 and (ar^2)r^2 = ar^4.

    The alternative requires more steps, but you use

    \sum_{i=2}^4 ar^i  = \sum_{i=0}^4 ar^i -\sum_{i=0}^1 ar^i, which again makes sense as, to find the sum of the third,fourth and fifth term in the sum from i=0 to 4, you take away the other two terms, which are the first and second (a and ar), and since the are of the form \sum_{i=0}^n ar^i you can use the formulae directly and get the same answer as before.

    I really can't remember what they ask you in C2 about it but hopefully this is of some use.
    I wish I could give you infinite rep. Thank you!
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    I got an A* on the gold 1 and 3 paper but i know in the exam ill just bottle it
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    (Original post by not_lucas1)
    I got an A* on the gold 1 and 3 paper but i know in the exam ill just bottle it
    If you got an A* for it, and you have NEVER seen those questions before then your're pretty much prepared.
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    (Original post by imran_)
    If you got an A* for it, and you have NEVER seen those questions before then your're pretty much prepared.
    I hope i am Ill probs do the other papers tuesday night, got other exams on monday and tuesday
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    (Original post by not_lucas1)
    I hope i am Ill probs do the other papers tuesday night, got other exams on monday and tuesday
    do a madasmaths paper and see what you get, these papers are a better indication of what can come up. Theyre slightly harder so I would suggest starting with Paper D and moving on. Best of luck buddy.
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    (Original post by not_lucas1)
    I hope i am Ill probs do the other papers tuesday night, got other exams on monday and tuesday
    Try IYGB papers S and T, if you can complete them, you can do anything
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    (Original post by rosemondtan)
    Try IYGB papers S and T, if you can complete them, you can do anything
    I thought U to Z were the harder ones :/ Or do you mean how representative they are of the actual paper?
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    (Original post by Someboady)
    I thought U to Z were the harder ones :/ Or do you mean how representative they are of the actual paper?
    I'm quite sure S and T are the harder ones! (Someone mentioned it before somewhere I can't remember) Well paper T was the only one that made me falter a little so that's just my personal view :woo: but all the IYGB papers are supposed to be much tougher than the average edexcel paper so if you can smash those, there's no reason why you shouldn't feel confident :excited:
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    (Original post by Someboady)
    I thought U to Z were the harder ones :/ Or do you mean how representative they are of the actual paper?
    (Original post by rosemondtan)
    I'm quite sure S and T are the harder ones! (Someone mentioned it before somewhere I can't remember) Well paper T was the only one that made me falter a little so that's just my personal view :woo: but all the IYGB papers are supposed to be much tougher than the average edexcel paper so if you can smash those, there's no reason why you shouldn't feel confident :excited:
    I'd say if you even get more than 60% on one of TeeEm's papers, you're in a very good position imo
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    How to you prove the Sn formula for geometric series?
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    sn=a+ar+ar^2+ar^3+.....ar^n-2 +ar^n-1 +ar^n
    rsn=ar+ar^2+ar^3+ar^4.....ar^n-2 +ar^n-1
    do Sn-rSn
    Sn-rSn = a+ar^n
    Sn(1-r)=a(1-r^n)
    then Sn =a(1-r^n) Over (1-r)


    r
    (Original post by neil20143)
    How to you prove the Sn formula for geometric series?
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    (Original post by neil20143)
    How to you prove the Sn formula for geometric series?
    S_n = a + ar + ... + ar^{n-1}

    r \times S_n = rS_n = ar + ... + ar^{n}

    S_n - rS_n = (1-r)S_n = a - ar^{n}

    Therefore:

    \displaystyle S_n = \frac{a(1-r^{n})}{1-r}

    (Original post by riyadhussain123)
    ...
    There are some mistakes in that
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    How do you prove the sum to infinity of a geometric sequence (if we need to know it?)
 
 
 
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