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    Hi guys, I am just doing a past paper for FP1 and have stumbled across this sort of question a few times and I am unable to answer it! Could you please give me a little bit of assistance as to what to do please?

    Thank you!
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    (Original post by iMacJack)
    Hi guys, I am just doing a past paper for FP1 and have stumbled across this sort of question a few times and I am unable to answer it! Could you please give me a little bit of assistance as to what to do please?

    Thank you!
    You'll need to use the result that you proved in the first part, so it'd be helpful if you could show us that.
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    (Original post by iMacJack)
    Hi guys, I am just doing a past paper for FP1 and have stumbled across this sort of question a few times and I am unable to answer it! Could you please give me a little bit of assistance as to what to do please?

    Thank you!
    I got the answer to be 2213/29, could you check if this is right then I can explain my working if you would like
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    (Original post by Dapperblook22)
    I got the answer to be 2213/29, could you check if this is right then I can explain my working if you would like
    It's wrong.
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    Oh, I remember this question... it was in my FP1 exam.
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    (Original post by Zacken)
    It's wrong.
    Oops, that's why I asked if it was right . I would assume an arithmetic error, as I split the problem up to 8 times the sum of r^3 from 5 to 8, minus 3 times the sum of r^2 from 5 to 8 plus K times the sum of r from 5 to 8
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    This is pretty easy, just use the standard series formulae, and basic algebra, to find k
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    (Original post by Dapperblook22)
    Oops, that's why I asked if it was right . I would assume an arithmetic error, as I split the problem up to 8 times the sum of r^3 from 5 to 8, minus 3 times the sum of r^2 from 5 to 8 plus K times the sum of r from 5 to 8
    The sums are to 10 not 8. The correct answer is
    Spoiler:
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    -7/5.


    The way you did it is good, but the first part of this question makes this part a trivial 2 liner or such instead of having to manually re-derive everything again.
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    Hey guys - sorry!
    Here's the full question
    https://gyazo.com/8024cf8825c10420ba3c3a62b256dbee

    Appreciate the help you guys are going to give
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    (Original post by Zacken)
    The sums are to 10 not 8. The correct answer is
    Spoiler:
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    -7/5.

    The way you did it is good, but the first part of this question makes this part a trivial 2 liner or such instead of having to manually re-derive everything again.
    Ah, I might need glasses, honestly thought it was an 8 . At least the method was correct so OP may follow that if they want.
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    (Original post by iMacJack)
    Hey guys - sorry!
    Here's the full question
    https://gyazo.com/8024cf8825c10420ba3c3a62b256dbee

    Appreciate the help you guys are going to give
    So right away you can split your sum as \displaystyle \sum_{r=5}^{10} (8r^3  - 3r) + \sum_{r=5}^{10} kr^2

    You should be able to say that, right away:

    \displaystyle \sum_{r=5}^{10} (8r^3 - 3r) = \sum_{r=1}^{10} (8r^3 - 3r) - \sum_{r=1}^{4} (8r^3 - 3r)

    which lets you compute that by plugging in the relevant values of n in the formula derived in the first part.

    Similarly: k\sum_{r=5}^{10} kr^2 = k\left(\sum_{r=1}^{10} r^2 - \sum_{r=1}^{4} r^2 \right) where standard formula booklet plugging in evalutes the bracket.

    This gets you 355k + \text{some number} = \text{that number I don't remember} and you solve for k.
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    (Original post by Zacken)
    So right away you can split your sum as \displaystyle \sum_{r=5}^{10} (8r^3  - 3r) + \sum_{r=5}^{10} kr^2

    You should be able to say that, right away:

    \displaystyle \sum_{r=5}^{10} (8r^3 - 3r) = \sum_{r=1}^{10} (8r^3 - 3r) - \sum_{r=1}^{4} (8r^3 - 3r)

    which lets you compute that by plugging in the relevant values of n in the formula derived in the first part.

    Similarly: k\sum_{r=5}^{10} kr^2 = k\left(\sum_{r=1}^{10} r^2 - \sum_{r=1}^{4} r^2 \right) where standard formula booklet plugging in evalutes the bracket.

    This gets you 355k + \text{some number} = \text{that number I don't remember} and you solve for k.
    Thank you mate. I will give this a go now and let you know how it goes!
    Thanks <3
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    (Original post by iMacJack)
    Thank you mate. I will give this a go now and let you know how it goes!
    Thanks <3
    No problem, this method of splitting sums from one number to another by writing them as sums from 1 to a number - sum from 1 to another number is in pretty much every FP1 paper ever.
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    (Original post by iMacJack)
    Thank you mate. I will give this a go now and let you know how it goes!
    Thanks <3
    If you need the working, please find it attached. Zacken explained it pretty clearly so you may not even need this

    Also sorry if it is a bit choppy at places.

    Name:  IMG_1359[1].jpg
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Size:  360.5 KB
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    I got K = -497/355 which = -7/5??
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    (Original post by Zacken)
    No problem, this method of splitting sums from one number to another by writing them as sums from 1 to a number - sum from 1 to another number is in pretty much every FP1 paper ever.
    yeah I can do the normal ones but when there's a constant involved it seems to throw me off a bit! thanks for the help!
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    (Original post by iMacJack)
    yeah I can do the normal ones but when there's a constant involved it seems to throw me off a bit! thanks for the help!
    Constants are constants, pull them out and do your thang with the series.
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    (Original post by Dapperblook22)
    If you need the working, please find it attached. Zacken explained it pretty clearly so you may not even need this

    Also sorry if it is a bit choppy at places.

    Name:  IMG_1359[1].jpg
Views: 42
Size:  360.5 KB
    Yeah thank you, appreciated
 
 
 
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