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Math Help - Geometric Series [Problem]

  1. #1
    Newbie bybuti's Avatar
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    Geometric Series [Problem]

    Hi everyone, this is my first post here

    I would be glad if someone help me with this problem

    1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,6,6,6,6,6,6,7....

    a1000000= ?

    a its geometric serie.

    thanks in advance
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  2. #2
    Moo
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    Hello,

    You can notice that :

    if a_n=k, then \sum_{i=1}^k i<n<\sum_{i=1}^{k+1} i

    So the problem is equivalent to finding k such as :

    \underbrace{\sum_{i=1}^k i}_{\dfrac{k(k+1)}{2}} < 1000000 < \underbrace{\sum_{i=1}^{k+1} i}_{\dfrac{(k+1)(k+2)}{2}}

    You can find k by the calculus, but i can give you an approximation :

    if k=1413, k(k+1)/2=998991 and (k+1)(k+2)/2=1000405

    So k=1413 is solution.

    Hence a_{1000000}=1413




    Err...actually, i don't know why you are told a is geometric series though the series they gave you is kinda arithmetical oO
    Perhaps i made a mistake, so if it seems wrong to you, don't hesitate to tell me :s
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  3. #3
    Newbie bybuti's Avatar
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    hmm, it's looking small number that solve that you post

    hmm, i think its like this 1+2+2+3+3+3+4+4+4+4+5+5+5+5+5+6+6+6+6+6+6+7+7+7+...+

    as you can see this goes like this 1 is one time, number 2 its 2 times, number 3 three times , ... number 10 its repeated ten times and continues like this.

    so from this up we have a_{10}=30

    a_{20}=85

    a_{30}=156

    so it's going like this, so a_{1000000}= to me looks to be bigger number

    anyway thanks very much for your help and your time

    Merci
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  4. #4
    Moo
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    Ha !

    I thought that the series of a_n was beginning with 1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,6,6,6,6,6,6,7....

    I'm sorry, but i can't tell you for the sum of all these terms (need to think more seriously about it)


    (you're welcome )
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  5. #5
    Moo
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    Oh, actually, it's simple...

    1+2+2+3+3+3+4+4+4+4+5+5+5+5+5+6+6+6+6+6+6+7+7+7+.. .+

    This is \sum_{k=1}^n k*k = \sum_{k=1}^n k^2

    Well, after two lines, i manage to remove reason from my message... I don't remember the value for this sum...
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  6. #6
    Newbie bybuti's Avatar
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    hmm, now looks more logical

    more near to solution
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  7. #7
    Moo
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    I have a way to find it... Perhaps you can think a little about it (it's very hard to write in latex when ideas just come out...)

    (k+1)=k+2k+1

    => \sum_{k=1}^n (k+1)^2 = \sum_{k=1}^n k^2 +2\sum_{k=1}^n k + \sum_{k=1}^n 1

    Assuming that \sum_{k=1}^n k^2 is in the form P(n) with P a polynomial of degree 2, you can try to take a recurrence formula from it, as you know 2\sum_{k=1}^n k + \sum_{k=1}^n 1


    Ok it seems that P is of degree 3.

    Maxima gave me this :

    (%i12) sum(k^2, k, 1, n), simpsum;
    (%o12) (2*n^3+3*n^2+n)/6



    & Good luck, because it's possible to find k when you have this formula, but it's very long to do it without calculator.
    I'm pretty sure there is a more direct way for this problem, but i'm sorry i can't find any...
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  8. #8
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    Quote Originally Posted by bybuti View Post
    so it's going like this, so a_{1000000}= to me looks to be bigger number.
    It is a larger number. But, only by one: 1414.
    Using the ceiling function, a_n  = \left\lceil {\frac{{ - 1 + \sqrt {1 + 8n} }}{2}} \right\rceil .
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  9. #9
    Moo
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    Gargl..

    I hesitated between 1413 and 1414, i didn't remember why i took 1413 although my first result was 1414 :'( Can you help me please ? Now i'm wondering ~
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  10. #10
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    Quote Originally Posted by Moo View Post
    a_n=k, then \sum_{i=1}^k i<n<\sum_{i=1}^{k+1} i
    Your mistake is right there. Using that scheme we get a_n=4 implies that 10 \le n \le 15 but a_{10}  = 4,\,a_{11}  = 5,\,a_{12}  = 5,\,a_{13}  = 5,\,a_{14}  = 5,\,a_{15}  = 5,\,a_{16}  = 5.
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  11. #11
    Moo
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    Hm so the solution is k+1, not k ^^
    Thanks
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  12. #12
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    Quote Originally Posted by Moo View Post
    Hm so the solution is k+1, not k
    Just change it to
    a_n=k>1, then \sum_{i=1}^{k-1} i < n \le \sum_{i=1}^{k} i
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  13. #13
    Moo
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    Yeah this sounds better :-)
    Now i have to find back why i changed my mind ~
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  14. #14
    Newbie bybuti's Avatar
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    Thanks guys, i will see if this is the wanted result
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