Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

FoolForMath

  • 2 years ago

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!

  • This Question is Closed
  1. FoolForMath
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    EDIT: #8 th now :)

  2. zzr0ck3r
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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:)

  3. FoolForMath
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The mathematical part doesn't use any sort of fancy things, any high-school students should be able to do it :)

  4. zzr0ck3r
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok ok I'll try :P

  5. shubhamsrg
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    is ans 1/2 ?

  6. FoolForMath
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    No, but for \(n=10000\) the answer is \(\frac 12 \)

  7. shubhamsrg
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    hmm,,i used hit and trial i.e. checked each term..sum approaches 0.5 only when it gets bigger..

  8. FoolForMath
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Nopes :) You have to derive a term of \(S_n \) summation upto \(n\) terms

  9. shubhamsrg
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  10. zzr0ck3r
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    damn how do you easily see that?

  11. shubhamsrg
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    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

  12. Carniel
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    My head hurt but isn't it somewhere near 1/2 and 1/3

  13. shubhamsrg
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    @zzr0ck3r i somehow saw that it is (n^2 +1)^2 - (n^2) dont know how,,even that amazes me!! :D

  14. zzr0ck3r
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    nice work

  15. shubhamsrg
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    but @FoolForMath says its not 1/2..i must be wrong somewhere..hmm

  16. shubhamsrg
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    and how do we know its 1/2 only upto n=10000?? sir?? you there?

  17. maheshmeghwal9
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i think no he is not here

  18. zzr0ck3r
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    code it and check

  19. zzr0ck3r
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    im guessing he would not have said it if it was not true:)

  20. shubhamsrg
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    hmmm.. :)

  21. FoolAroundMath
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[S_{n} = \frac{n(n+1)}{2(n^{2}+n+1)}\]

  22. shubhamsrg
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    ohh..i see,,i calcuated for infinite terms i.e n tends to infinity..hmm..not the general formula

  23. shubhamsrg
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    how do you find summation (n/n^4 + n^2 +1) ??

  24. shubhamsrg
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    @FoolAroundMath @FoolForMath

  25. FoolAroundMath
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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 : \[S = \frac{1}{2}(1 - \frac{1}{n^{2}+n+1})\]

  26. shubhamsrg
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    ohh lol..yes,,damn i missed that..nice :)

  27. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.