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    I really need help with summation series where r=0. How does this affect the formulas? Also, what do you do when 'n' is featured in the summation series (mixed in with the r-values).

    To be more specific, question 10ii in the Edexcel June 2013 (R) paper is where I'm really baffled, I don't understand the mark scheme either

    If anyone could please explain this, I would be eternally grateful!

    Paper: https://7cba9babeb0db0ff9468853e0b2d...%20Edexcel.pdf

    Mark Scheme: https://7cba9babeb0db0ff9468853e0b2d...%20Edexcel.pdf

    p.s. I really this question was asked previously, but I didn't understand the explanation at all and it's quite old now
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    Hi

    r=0 problem:

    The standard formulae only apply when r=1.

    Therefore, if the summation given starts with r=0, work out the first term 'manually' by subbing in r=0 and then add the sum from r=1 to n.


    This is just like how you might do a proof for a summation: when you have a sum from r=1 to n=k+1, you do the sum from r=1 to n=k and ADD the r=k+1 term 'manually',

    n in the formula problem:

    just like how you might get a 'show that' question for a summation and you separate out each component in the formula, do the same. by this i mean, you currently have sum (r^2 -2r +2n +1). Separate this to give: sum (r^2) - sum (2r) + sum(+1) + sum (2n)

    For sum (2n), it's just effectively 2n + 2n + 2n +...+ 2n, n times, so sum(2n), is 2n^2,

    Hope this helps. if you need clarification, feel free to ask
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    (Original post by Horagontus)
    I really need help with summation series where r=0. How does this affect the formulas? Also, what do you do when 'n' is featured in the summation series (mixed in with the r-values).

    To be more specific, question 10ii in the Edexcel June 2013 (R) paper is where I'm really baffled, I don't understand the mark scheme either

    If anyone could please explain this, I would be eternally grateful!

    Paper: https://7cba9babeb0db0ff9468853e0b2d...%20Edexcel.pdf

    Mark Scheme: https://7cba9babeb0db0ff9468853e0b2d...%20Edexcel.pdf

    p.s. I really this question was asked previously, but I didn't understand the explanation at all and it's quite old now
    when 'n' is featured in the summation series (mixed in with the r-values).
    just multiply the formula by n
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    (Original post by Salzy)
    when 'n' is featured in the summation series (mixed in with the r-values).
    just multiply the formula by n
    how does this help?
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    Yeah what Amar says, only thing I do slightly differently is instead of taking out the 2n term from the (r^2 - 2r +2n + 1), you can treat the (2n+1) term the same as you'd treat any other constant. Normally, the sum from r=1 to n of a constant k would be kn, so the sum from r=1 to n of (2n+1) is n(2n+1)

    In this question, it is the sum from r=0 to n. As Amar said, you can just treat it like it is from r=1, then put 0 into the r equation to get the term when r=0 then add it on to the sum from r=1 to n, which would give you the the sum from r=0 to r=n

    When you put in 0 to any equation, the value you will have will always equal the constant on the end of the r equation, here out constant is (2n+1). This means you can use the formula as normal to get your sum of the r^2 and -2r terms, then for the (2n+1) you have n(2n+1) as shown in the 1st paragraph, then you have an extra constant added on from when r=0, as shown in the 2nd paragraph, meaning that for the constant you have n(2n+1) + (2n+1) = (n+1)(2n+1)

    This can be used anytime it is the sum from r=0, you use the formula as normal but instead of the sum of any constant k being kn, it will be k(n+1) to account for the addition of the 0 term.

    Hope this helps, did that question just now!
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    (Original post by AmarPatel98)
    Hi

    r=0 problem:

    The standard formulae only apply when r=1.

    Therefore, if the summation given starts with r=0, work out the first term 'manually' by subbing in r=0 and then add the sum from r=1 to n.


    This is just like how you might do a proof for a summation: when you have a sum from r=1 to n=k+1, you do the sum from r=1 to n=k and ADD the r=k+1 term 'manually',


    Hope this helps. if you need clarification, feel free to ask
    When you "manually" obtain the value of r=0, does anything happen to the n at the top? i.e. would I have to use (n-1) rather than n as the sequence is reduced by one component?

    thanks in advance
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    (Original post by Horagontus)
    I really need help with summation series where r=0. How does this affect the formulas? Also, what do you do when 'n' is featured in the summation series (mixed in with the r-values).

    To be more specific, question 10ii in the Edexcel June 2013 (R) paper is where I'm really baffled, I don't understand the mark scheme either

    If anyone could please explain this, I would be eternally grateful!

    Paper: https://7cba9babeb0db0ff9468853e0b2d...%20Edexcel.pdf

    Mark Scheme: https://7cba9babeb0db0ff9468853e0b2d...%20Edexcel.pdf

    p.s. I really this question was asked previously, but I didn't understand the explanation at all and it's quite old now
    I think someone's method on here is the same as mine - which I got from Zacken, where you take the 2n+1 term out. You can refer to the post here. http://www.thestudentroom.co.uk/show...955045&page=19
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    (Original post by TommyBrecon)
    When you "manually" obtain the value of r=0, does anything happen to the n at the top? i.e. would I have to use (n-1) rather than n as the sequence is reduced by one component?

    thanks in advance
    Nope you wont.

    Let me put it this way to make it clear:

    When the sum is from r=0 to r=n, the series is: (r=0 term) + (r=1 term) + (r=2 term) + ... + (r=n term)

    When we do it 'manually', we do: (r=0 term manually) + (sum from r=1 to n) = (r=0 term) + (r=1 term) + (r=2 term) + ... + (r=n term)

    We break it up so we can use the standard formulae.

    Hope this makes sense
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    (Original post by Nikhilm)
    Can you not be so passive aggressive.
    stop trolling
 
 
 
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