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FoolForMath

Find the sum up to \( n\) terms of the series: \[ \frac{1}{3} + \frac 2{21} + \frac{3}{91} + \frac{4}{273} + \cdots \text{upto n terms} \] PS: I encountered this problem in spoj: http://www.spoj.pl/problems/SUMUP/). Not much later I devised a solution which is currenly #9 in rankings (http://www.spoj.pl/ranks/SUMUP/). Although there is much scope of further optimization. Good luck!

  • one year ago
  • one year ago

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  1. FoolForMath
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    EDIT: #8 th now :)

    • one year ago
  2. zzr0ck3r
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    is this beyond the scope of a standard calc with series class? I spent about 2 weeks with series so I dont know if I want to get into this:)

    • one year ago
  3. FoolForMath
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    The mathematical part doesn't use any sort of fancy things, any high-school students should be able to do it :)

    • one year ago
  4. zzr0ck3r
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    ok ok I'll try :P

    • one year ago
  5. shubhamsrg
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    is ans 1/2 ?

    • one year ago
  6. FoolForMath
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    No, but for \(n=10000\) the answer is \(\frac 12 \)

    • one year ago
  7. shubhamsrg
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    hmm,,i used hit and trial i.e. checked each term..sum approaches 0.5 only when it gets bigger..

    • one year ago
  8. FoolForMath
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    Nopes :) You have to derive a term of \(S_n \) summation upto \(n\) terms

    • one year ago
  9. shubhamsrg
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    i easily see the denominator proceeds as (n^2 +n +1)(n^2 -n +1).. dont know what comes next.. hold on,,leme try..

    • one year ago
  10. zzr0ck3r
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    damn how do you easily see that?

    • one year ago
  11. shubhamsrg
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    somehow or the other way,,i get 1/2 only..please correct me where am going wrong(if i am :P) the given terms are (1/3)+(2/21)+(3/91)+(4/273)......... 1/(1*3) + 2/(3*7) + 3/(7*13) + 4/ (13*21) i notice that 3-1 =2,,7-3 =4,, 13-7 =6 ,,and so on ,,i.e a table of 2 so i take 1/2 common ,,we have : 1/2( (1-1/3) + (1/3-1/7) + (1/7 -1/13).........) =1/2*1 =1/2 o.O

    • one year ago
  12. Carniel
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    My head hurt but isn't it somewhere near 1/2 and 1/3

    • one year ago
  13. shubhamsrg
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    @zzr0ck3r i somehow saw that it is (n^2 +1)^2 - (n^2) dont know how,,even that amazes me!! :D

    • one year ago
  14. zzr0ck3r
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    nice work

    • one year ago
  15. shubhamsrg
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    but @FoolForMath says its not 1/2..i must be wrong somewhere..hmm

    • one year ago
  16. shubhamsrg
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    and how do we know its 1/2 only upto n=10000?? sir?? you there?

    • one year ago
  17. maheshmeghwal9
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    i think no he is not here

    • one year ago
  18. zzr0ck3r
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    code it and check

    • one year ago
  19. zzr0ck3r
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    im guessing he would not have said it if it was not true:)

    • one year ago
  20. shubhamsrg
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    hmmm.. :)

    • one year ago
  21. FoolAroundMath
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    \[S_{n} = \frac{n(n+1)}{2(n^{2}+n+1)}\]

    • one year ago
  22. shubhamsrg
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    ohh..i see,,i calcuated for infinite terms i.e n tends to infinity..hmm..not the general formula

    • one year ago
  23. shubhamsrg
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    how do you find summation (n/n^4 + n^2 +1) ??

    • one year ago
  24. shubhamsrg
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    @FoolAroundMath @FoolForMath

    • one year ago
  25. FoolAroundMath
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    I will give my solution: \[T_{k} = \frac{1}{2}(\frac{1}{k^{2}-k+1}-\frac{1}{k^{2}+k+1})\] It is easy to see that \[(k+1)^{2} - (k+1) + 1 = k^{2} + k + 1\] Thus when summing to 'n' terms, successive terms cancel. What remains is : \[S = \frac{1}{2}(1 - \frac{1}{n^{2}+n+1})\]

    • one year ago
  26. shubhamsrg
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    ohh lol..yes,,damn i missed that..nice :)

    • one year ago
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