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    (Original post by coefficient)
    In the first line they are splitting up that power to get 1^ x / 3^x and 1^x = 1 so it's just 1/3^x. They then multiply both sides by 3^x.
    This gives 1 = 2 * (3^x) * (3*x)
    We then divide by two to get (3^x)^2 = 1/2
    we then take log of both sides and bring down the power 2x to get 2xlog(3)=log(1/2).
    We then divide by log3 to get 2x = log(1/2) / log(3) and then divide by 2 to get the final answer
    Thankss
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    Good luck guys.

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    (Original post by jshep000)
    Hopefully, will be bless if we don't get one of those crappy weird questions at the end haha.
    probably will unfortunately haha
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    (Original post by jshep000)
    It still works, it just gives you a negative value. Our teacher taught it to us, that if r>1 than we had to swap it and do r-1.
    It won't give a negative value (unless the terms are negative) as the denominator is swapped as well. As you can see in the proof, the two formulae give the exact same answer and they are purely conjugates for ease of use:

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    Just out of curiosity, when a given range for trigonometric questions, includes >= or <=, do we include the values they're equal to as angles? Like 0 <= x <= 180, is 180 included as an answer?
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    What is an A for the June 2013 R paper?
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    (Original post by Skygon)
    Just out of curiosity, when a given range for trigonometric questions, includes >= or <=, do we include the values they're equal to as angles? Like 0 <= x <= 180, is 180 included as an answer?
    Yes, because it means "less than or equal to 180" and not "less than 180".

    0 \leq x \leq 180 \equiv \text{values of }x\text{ from 0 to 180 (inclusive)}
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    (Original post by coefficient)
    Because sometimes if we start in year 2000, it's easier to call that year 0 than year 1. Really it's just people who are more confident saving time. There's no real benefit of it, it's just being fancy.
    Or just do what I do and count the value of n on your finger if you are confused as to what n might be and use ar^(n-1). Haha I know we went past finger counting at primary school but it still helps to avoid confusion.
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    (Original post by TrentL)
    thank you very much Abbey and yes that would be amazing!
    took me a while :/
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    Someone please help me on question 7?! https://c4a3f001dcd45afe69d0ceec8300...%20Edexcel.pdf
    Thanks! (also please could you show all your working as to how you got the answer if possible)
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    (Original post by Da Di Doo)
    What is an A for the June 2013 R paper?
    64 for an A
    57 for a B
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    goodnight people. best of luck tomorrow!
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    we get 1 , n, nc2, nc3, for the ncr coefficients and then accending powers of kx.
    

(1+kx)^n = 1 

1   n    nc2     nc3

1   kx   k^2 x^2 k^3 x^3



so overall  1 + nkx + (nc2)k^2 x^2 + (nc3)k^3 x^3



ncr = n!/(n-1)!

nc2 = n(n-1)/2

nc3= n(n-1)(n-2)/3x2
    If the coefficient of x^2 and x^3 are equal then:

    

n(n-1)/2 * k^2 = n(n-1)(n-2)/6 *k^3
    1/2 = (n-2)/6 * k
    3 = k(n-2)
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    Any tips on circles at all


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    I reckon the paper will be quite chill tbh, C1 had some strange worded questions for some people. So they probably were nice and gave us a good C2 paper <3
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    (Original post by Shanahey)
    Someone please help me on question 7?! https://c4a3f001dcd45afe69d0ceec8300...%20Edexcel.pdf
    Thanks! (also please could you show all your working as to how you got the answer if possible)
    Use the other method for binomial expansion the n(n-1)/2! Method from formula booklet and see if you can work it out


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    (Original post by Shanahey)
    Someone please help me on question 7?! https://c4a3f001dcd45afe69d0ceec8300...%20Edexcel.pdf
    Thanks! (also please could you show all your working as to how you got the answer if possible)
    when A=4 then
     nk=4 and 3 = k(n-2) = nk-2k = 4-2k

2k = 1 

k = 1/2
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    Besides sequences in c2, what else do we have to be able to prove the formula for?
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    Somebody needs to help me with optimisation
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    (Original post by Skygon)
    Besides sequences in c2, what else do we have to be able to prove the formula for?
    I might be wrong but I rarely ever see a question asking to prove a formula?


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