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    Workings out is in the attachment, the answer is m=6, what did I mess up on? ********FP1, not C2*********Attachment 617292617294
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    (Original post by ckfeister)
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    Workings out is in the attachment, the answer is m=6, what did I mess up on? ********FP1, not C2*********Attachment 617292617294
    Surely you are over complicating this?

    List a few terms to see the sequence.
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    (Original post by ckfeister)
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    Workings out is in the attachment, the answer is m=6, what did I mess up on? ********FP1, not C2*********
    Like Muttley79 said, you're massively overcomplicating this. But, even so, if you insisit on pursuing this formulaic way, you've made quite a few errors.

    First off: the third line, the sum of 2 from r=1 to r=m is most certainly not m(m+1)/2. It's just 2m

    Second off: the third line, the sum of 2 from r=1 to r=1 is not 1/2* (2).

    Third off: in third line, your expression is of the form (a) - (a). Surely you can see it's 0?

    Fourth off: your addition is off on the fourth line. Of course, it's all moot given you started off wrongly, but perhaps you should review some basic algebra before starting off on more complicated tasks.
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    (Original post by ckfeister)

    Workings out is in the attachment, the answer is m=6, what did I mess up on? ********FP1, not C2*********
    As said above, this question can be approached by simply listing a few terms as the summation is small.

    Otherwise, I agree with the second line, but not the third. \displaystyle \sum_{r=1}^m 2 = 2m \not= \frac{1}{2}m(m+1) = \sum_{r=1}^{m} r

    That error aside, also your step from the 4th to 5th line doesn't make sense... both brackets are the same quantity therefore your result would be 0 anyway when you take one away from the other.
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    (Original post by Zacken)
    Like Muttley79 said, you're massively overcomplicating this. But, even so, if you insisit on pursuing this formulaic way, you've made quite a few errors.

    First off: the third line, the sum of 2 from r=1 to r=m is most certainly not m(m+1)/2. It's just 2m

    Second off: the third line, the sum of 2 from r=1 to r=1 is not 1/2* (2).

    Third off: in third line, your expression is of the form (a) - (a). Surely you can see it's 0?

    Fourth off: your addition is off on the fourth line. Of course, it's all moot given you started off wrongly, but perhaps you should review some basic algebra before starting off on more complicated tasks.
    This must be the most British politeness way I've ever heard of calling me a stupid 14 year old.

    (Original post by RDKGames)
    As said above, this question can be approached by simply listing a few terms as the summation is small.

    Otherwise, I agree with the second line, but not the third. \displaystyle \sum_{r=1}^m 2 = 2m \not= \frac{1}{2}m(m+1) = \sum_{r=1}^{m} r

    That error aside, also your step from the 4th to 5th line doesn't make sense... both brackets are the same quantity therefore your result would be 0 anyway when you take one away from the other.
    Thx

    (Original post by Muttley79)
    Surely you are over complicating this?

    List a few terms to see the sequence.
    Thx
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    (Original post by ckfeister)
    This must be the most British politeness way I've ever heard of calling me a stupid 14 year old.
    Nah, I mean it, you really should go over reviewing some basic algebra, it'll help more than you can imagine later on. You'll struggle with a flimsy foundation, knowing how to do all the manipulations will free up your mind to focus on the concepts and actual understanding.
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    (Original post by Zacken)
    Nah, I mean it, you really should go over reviewing some basic algebra, it'll help more than you can imagine later on, you'll struggle with a flimsy foundation.
    I noticed the mistakes I did the second I uploaded it, I have dyslexic so sometimes I write things when I meant another... its my huge flaw.
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    Attachment 617312617314

    What have I done wrong here? I don't see how that answer is correct...
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    Your step after calculating \alpha+\beta and \alpha \beta doesn't make sense. Can you explain what you were trying to do when you wrote down x^2 + \frac{1}{2}x + \frac{4}{3}  = 0?
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    (Original post by DFranklin)
    Your step after calculating \alpha+\beta and \alpha \beta doesn't make sense. Can you explain what you were trying to do when you wrote down x^2 + \frac{1}{2}x + \frac{4}{3}  = 0?
     \alpha \beta = \frac{c}{a}

    \alpha + \beta = -\frac{b}{a}
    a = 6
    b = -3
    c = 8

    :.  \alpha \beta = \frac{8}{6}

         \alpha + \beta = -\frac{-3}{6}
    as
     x^2 + (sum)x + (product) = 0
    x^2 + \frac{1}{2}x + \frac{4}{3}  = 0
    times it by 2,3

     6x^2 + 3x + 8 = 0
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    (Original post by ckfeister)
    ...
    First off, it's x^2 - (sum of roots) x + (product of roots). In your case, that would get you x^2 - \frac{1}{2}x + \frac{4}{3} = 0\Rightarrow 6x^2 - 3x + 8 = 0 which is precisely the equation that you started off with it. It has roots \alpha and \beta.

    You want to find the equation whose roots are \alpha + \beta and \alpha\beta. The sum of those two roots is \alpha + \beta + \alpha \beta and the product is \alpha \beta (a\lpha + \beta).

    So the equation should be x^2 - (\alpha + \beta + \alpha \beta)x + \alpha \beta(\alpha + \beta) = 0. Can you see where you got confused now?
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    (Original post by Zacken)
    First off, it's x^2 - (sum of roots) x + (product of roots). In your case, that would get you x^2 - \frac{1}{2}x + \frac{4}{3} = 0\Rightarrow 6x^2 - 3x + 8 = 0 which is precisely the equation that you started off with it. It has roots \alpha and \beta.

    You want to find the equation whose roots are \alpha + \beta and \alpha\beta. The sum of those two roots is \alpha + \beta + \alpha \beta and the product is \alpha \beta (a\lpha + \beta).

    So the equation should be x^2 - (\alpha + \beta + \alpha \beta)x + \alpha \beta(\alpha + \beta) = 0. Can you see where you got confused now?
    Ohh, mis-read the question, thanks!
 
 
 
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