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    (Original post by RichE)
    Do you mean 2011 Q5 (ii)? That is not asking for a specific answer. Part (iii) is, but the given answer of 2^(n-1) is very different from yours.
    Yeah iii I meant. I know its different but my reasoning is that there is only 1 way to get to the first shaded and last shaded square. so that's 2 then for each of the left over squares there are n-1 ways to get to it, therefore (n-1)(n-2) + 2 ?
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    (Original post by Mystery.)
    Yeah iii I meant. I know its different but my reasoning is that there is only 1 way to get to the first shaded and last shaded square. so that's 2 then for each of the left over squares there are n-1 ways to get to it, therefore (n-1)(n-2) + 2 ?
    Yours can't be the right answer as it's a different function.

    If you wanted to argue it your way you would be summing binomial coefficients

    1 + (n-1)C1 + (n-2)C2 + ... + (n-1)C(n-1) + 1

    which would again give their answer of 2^(n-1).
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    (Original post by RichE)
    Yours can't be the right answer as it's a different function.

    If you wanted to argue it your way you would be summing binomial coefficients

    1 + (n-1)C1 + (n-2)C2 + ... + (n-1)C(n-1) + 1

    which would again give their answer of 2^(n-1).
    I know its not the right answer obviously but it works. I am not saying its the same
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    (Original post by Mystery.)
    I know its not the right answer obviously but it works. I am not saying its the same
    In what sense does it work? Your reasoning is wrong (see my previous post) and your formula gives the wrong answers.
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    (Original post by RichE)
    In what sense does it work? Your reasoning is wrong (see my previous post) and your formula gives the wrong answers.
    well when I tried it with different n x n grids it gave the same answer as their function.
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    (Original post by Mystery.)
    well when I tried it with different n x n grids it gave the same answer as their function.
    Unless I'm misreading your posts you're suggesting the answer is

    (n-1)(n-2) + 2

    and their answer is

    2^(n-1).

    The second function grows exponentially and the first is a polynomial, so there's no way they could be the same - for large n the second function would be colossally bigger.

    But you only have to try n = 5 to get different answers from the two formulae - 14 from the first and 16 from the second. (Try n=100 on your calculator to see how different the formulae really are.)
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    Hi
    In MAT 2007, Q2, part (ii) where we find an expression in terms of k....

    How would you come up with (4k - 1) / 3 as the last part of that expression? Is that just pattern spotting or is there a method to develop a relationship between k and the final term of each expression in the sequence?
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    (Original post by Someboady)
    Hi
    In MAT 2007, Q2, part (ii) where we find an expression in terms of k....

    How would you come up with (4k - 1) / 3 as the last part of that expression? Is that just pattern spotting or is there a method to develop a relationship between k and the final term of each expression in the sequence?
    It's a geometric sum.
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    Can anyone explain Part V, question 5 of the 2007 paper?
    Thanks
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    (Original post by RichE)
    It's a geometric sum.
    Ah yes, can't believe I didn't spot that...
    But how do we take into account the "+1" for each term? As in for the geometric series... r is 4 but don't we add 1 to each term to get the next term also?
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    (Original post by Someboady)
    Ah yes, can't believe I didn't spot that...
    But how do we take into account the "+1" for each term? As in for the geometric series... r is 4 but don't we add 1 to each term to get the next term also?
    I'm not really sure what you're asking. The (4^k-1)/3 expression appears as it's the sum of the geometric series given.

    That multiplying by 4 and adding 1 each time leads to a geometric sum at all is just a matter of spotting the pattern

    1
    4+1
    4^2 + 4 + 1
    4^3 + 4^2 + 4 + 1
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    What are people generally getting for the 2009 and 2011 papers?
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    What are the best resources for improving at MAT papers, I feel like Im not really understanding how to approach most of these questions
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    Do you guys normally do all the questions or do like 2/3 from Q2-3 and try to secure all the marks?
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    (Original post by Mystery.)
    Do you guys normally do all the questions or do like 2/3 from Q2-3 and try to secure all the marks?
    I'm not really sure what you mean. Generally, I do questions in the order of the booklet (for me that's 1,2,3,5 and 6), and where I can I try to do a whole question at once, because of how they progress. But if I'm really struggling I'll usually either skip the last part or a "show that" part (because I can still progress in the question with the given result) and come back to it later. It's kind of a dangerous game, though - the part you're skipping could be 1-2 marks, or it could be 4-6.

    If you're really struggling for time, then it's probably better to do the first few parts from all your questions rather than try and complete fewer questions, as earlier marks in a question tend to be easier. But really it's personal preference.
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    (Original post by Mystery.)
    Do you guys normally do all the questions or do like 2/3 from Q2-3 and try to secure all the marks?
    you should do all of them
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    I don't understand how they go about the long division in 2013 question 1, G?

    Posted from TSR Mobile
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    Hi can someone please explain MAT 2007, Question 7d...

    Is there an alternative method to the one provided in the solutions?
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    (Original post by Someboady)
    Hi can someone please explain MAT 2007, Question 7d...

    Is there an alternative method to the one provided in the solutions?
    This is pretty similar to the given solution, but may help

    Assume we have N of the first cards and M of the second cards to make a match.

    Therefore 7N + M = N + 10M (on the first tile, we have N lots of 7 Xs on the top, and M lots of one X on the bottom; on the second tile, we have N lots of one X on the top and M lots of 10 Xs on the bottom - to make a match).

    We can rewrite this as 2N = 3M.

    Trialing we find N = 3 and M = 2, and get 2(3) = 3(2), which is correct.

    Therefore we have 3 + 2 = 5 tiles here.

    Seeing the working in bold clarifies that the only way of making this work on any occasion will be by using N as a multiple of 3 and M as a multiple of 2.

    Therefore N = 3k, M = 2k. 3k + 2k = 5k. Therefore we always have a multiple of 5.
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    2009, 1(f)

    How have they got the turning points to be x=0,1,2 in the worked solutions

    I've got x=-1,0, 3 through differentiating, equating to zero and then factorising
 
 
 
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