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    I've just done section B of last year's FP1 paper as part of my mock and there's a question I couldn't figure out.

    The question is on summation of series, which we haven't actually covered yet, but I get another shot at the question of Friday and I kind of just want to show that I can do it anyway.
    The question (more or less - I memorised it as well as I could):

    45/(39*41*43) + 46/(40*42*44) + 47/(41*43*45) + ... + 105/(99*101*103)

    What I worked out so far:
    A general formula of n/((n-6)(n-4)(n-2))

    I had a friend try and explain it to me and he did a fairly good job of it but, since I haven't learnt about this yet, it didn't make a whole lot of sense.

    Could someone just walk me through this? Thanks!
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    (Original post by JustJusty)
    I've just done section B of last year's FP1 paper as part of my mock and there's a question I couldn't figure out.

    The question is on summation of series, which we haven't actually covered yet, but I get another shot at the question of Friday and I kind of just want to show that I can do it anyway.
    The question (more or less - I memorised it as well as I could):

    45/(39*41*43) + 46/(40*42*44) + 47/(41*43*45) + ... + 105/(99*101*103)

    What I worked out so far:
    A general formula of n/((n-6)(n-4)(n-2))

    I had a friend try and explain it to me and he did a fairly good job of it but, since I haven't learnt about this yet, it didn't make a whole lot of sense.

    Could someone just walk me through this? Thanks!
    I think we actually need to see the question before we can help. Unfortunately I don't have access to last years paper.
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    Look at the "Method of Differences" for summation.
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    (Original post by JustJusty)
    I've just done section B of last year's FP1 paper as part of my mock and there's a question I couldn't figure out.

    The question is on summation of series, which we haven't actually covered yet, but I get another shot at the question of Friday and I kind of just want to show that I can do it anyway.
    The question (more or less - I memorised it as well as I could):

    45/(39*41*43) + 46/(40*42*44) + 47/(41*43*45) + ... + 105/(99*101*103)

    What I worked out so far:
    A general formula of n/((n-6)(n-4)(n-2))

    I had a friend try and explain it to me and he did a fairly good job of it but, since I haven't learnt about this yet, it didn't make a whole lot of sense.

    Could someone just walk me through this? Thanks!

    Please give slightly more info. Did it ask for you to evaluate the summation? If so it will have given you some general formula to apply.

    And yes method of differences is what you want.
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    (Original post by Welshstig)
    Look at the "Method of Differences" for summation.
    Is there any way you could just point me in the right direction with this? I know how to do the method of differences, but the only example's I've been able to find had some kind of formula. I tried it with the formula that I made up for it but doesn't work because there's only one fraction, so nothing to cancel out.
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    (Original post by carpetguy)
    Please give slightly more info. Did it ask for you to evaluate the summation? If so it will have given you some general formula to apply.

    And yes method of differences is what you want.
    Yes, I need to sum the values. There was no formula at all, only the series, which is why I have no idea how to do it. I've never seen method of differences without a given formula, and the one I made up for it doesn't work because it's only one fraction.
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    (Original post by Welshstig)
    Look at the "Method of Differences" for summation.
    I'll right a full answer a bit later if you are still stuck (a bit busy rn sorry bout that).

    Do you know how to split a fraction into partial fractions?
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    (Original post by JustJusty)
    Yes, I need to sum the values. There was no formula at all, only the series, which is why I have no idea how to do it. I've never seen method of differences without a given formula, and the one I made up for it doesn't work because it's only one fraction.
    I think you copied the question wrongly... The general expression for this sum is unusual to be in any FP1 paper, but I could be wrong.

    https://www.wolframalpha.com/input/?...))+from+n%3D45
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    (Original post by RDKGames)
    I think you copied the question wrongly... The general expression for this sum is unusual to be in any FP1 paper, but I could be wrong.

    https://www.wolframalpha.com/input/?...))+from+n%3D45
    The formula in terms of n is just my own thing that I worked out. It wasn't part of the question. All that was in the question was the series and the instruction to find its sum. No formula or anything.
    I'm almost certain that the question is copied correctly. The pattern was definitely this. The only thing I'm slightly unsure about are the bounds.
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    (Original post by JustJusty)
    The formula in terms of n is just my own thing that I worked out. It wasn't part of the question. All that was in the question was the series and the instruction to find its sum. No formula or anything.
    I'm almost certain that the question is copied correctly. The pattern was definitely this. The only thing I'm slightly unsure about are the bounds.
    The thing is, although "standard" FP1 methods work to sum this (and are the only reasonble method, I think), with the fraction you've descrbed you have to consider a lot more terms that usual before you get the cancellation that lets you sum the series. Which makes it a *lot* of work to get an answer. Hence the doubt that it's totally correct. (It's possible you get better cancellation if you're only summing over n-odd or n-even; is that a possibility?)
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    (Original post by JustJusty)
    Thanks!
    Right, so in the actual paper. The question was split into several parts. The first part was to express

    \dfrac{2r+5}{(2r-1)(2r+1)(2r+3)}=\dfrac{3}{4(2r-1)}-\dfrac{1}{2r+1}+\dfrac{1}{4(2r+3  )} as partial fractions.

    Second part was to find \displaystyle \sum_{r=1}^n\dfrac{2r+5}{(2r-1)(2r+1)(2r+3)} using the method of differences utilising the partial fraction expansion.

    Third part was to find the limit as n \to \infty of the sum.

    And fourth part was to compute\displaystyle \sum_{20}^{50} \dfrac{2r+5}{(2r-1)(2r+1)(2r+3)} by writing it as \displaystyle \sum_{r=1}^{50} \dfrac{2r+5}{(2r-1)(2r+1)(2r+3)} - \sum_{r=1}^{19} \dfrac{2r+5}{(2r-1)(2r+1)(2r+3)}

    and using evaluating those two by using the previous expression for n=50 and n=19.

    Now that it's in this more structured form, try having a go? Although you should look up "method of differences" before. The final answer should be ~0.00813 or something along those lines.
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    (Original post by Zacken)
    Right, so in the actual paper. The question was split into several parts. The first part was to express

    \dfrac{2r+5}{(2r-1)(2r+1)(2r+3)}=\dfrac{3}{4(2r-1)}-\dfrac{1}{2r+1}+\dfrac{1}{4(2r+3  )} as partial fractions.

    Second part was to find \displaystyle \sum_{r=1}^n\dfrac{2r+5}{(2r-1)(2r+1)(2r+3)} using the method of differences utilising the partial fraction expansion.

    Third part was to find the limit as n \to \infty of the sum.

    And fourth part was to compute\displaystyle \sum_{20}^{50} \dfrac{2r+5}{(2r-1)(2r+1)(2r+3)} by writing it as \displaystyle \sum_{r=1}^{50} \dfrac{2r+5}{(2r-1)(2r+1)(2r+3)} - \sum_{r=1}^{19} \dfrac{2r+5}{(2r-1)(2r+1)(2r+3)}

    and using evaluating those two by using the previous expression for n=50 and n=19.

    Now that it's in this more structured form, try having a go? Although you should look up "method of differences" before. The final answer should be ~0.00813 or something along those lines.
    Hmm interesting. I'll give it a go

    Is this old spec? Never come across this is fp1. Although i know it from c4.
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    (Original post by carpetguy)
    Hmm interesting. I'll give it a go

    Is this old spec? Never come across this is fp1. Although i know it from c4.
    OCR MEI 2016 FP1 paper.
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    (Original post by Zacken)
    OCR MEI 2016 FP1 paper.
    Ah MEI. That'd explain it.
    Kinda wish we were doing MEI now. Sounds really fun.
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    (Original post by carpetguy)
    Ah MEI. That'd explain it.
    Kinda wish we were doing MEI now. Sounds really fun.
    Seems like a tedious Edexcel FP1 question, not sure what's fun about it.
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    (Original post by carpetguy)
    Hmm interesting. I'll give it a go

    Is this old spec? Never come across this is fp1. Although i know it from c4.
    Dear comrade,

    I also know this content from a recent STEP examination paper.

    Best wishes,

    The Don
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    Hegarty Maths. Don't worry about it.


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