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anonymous
 4 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!
anonymous
 4 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!

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zzr0ck3r
 4 years ago
Best ResponseYou've already chosen the best response.0is 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:)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The mathematical part doesn't use any sort of fancy things, any highschool students should be able to do it :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No, but for \(n=10000\) the answer is \(\frac 12 \)

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.3hmm,,i used hit and trial i.e. checked each term..sum approaches 0.5 only when it gets bigger..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Nopes :) You have to derive a term of \(S_n \) summation upto \(n\) terms

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.3i easily see the denominator proceeds as (n^2 +n +1)(n^2 n +1).. dont know what comes next.. hold on,,leme try..

zzr0ck3r
 4 years ago
Best ResponseYou've already chosen the best response.0damn how do you easily see that?

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.3somehow 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 31 =2,,73 =4,, 137 =6 ,,and so on ,,i.e a table of 2 so i take 1/2 common ,,we have : 1/2( (11/3) + (1/31/7) + (1/7 1/13).........) =1/2*1 =1/2 o.O

Carniel
 4 years ago
Best ResponseYou've already chosen the best response.0My head hurt but isn't it somewhere near 1/2 and 1/3

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.3@zzr0ck3r i somehow saw that it is (n^2 +1)^2  (n^2) dont know how,,even that amazes me!! :D

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.3but @FoolForMath says its not 1/2..i must be wrong somewhere..hmm

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.3and how do we know its 1/2 only upto n=10000?? sir?? you there?

maheshmeghwal9
 4 years ago
Best ResponseYou've already chosen the best response.0i think no he is not here

zzr0ck3r
 4 years ago
Best ResponseYou've already chosen the best response.0im guessing he would not have said it if it was not true:)

FoolAroundMath
 4 years ago
Best ResponseYou've already chosen the best response.2\[S_{n} = \frac{n(n+1)}{2(n^{2}+n+1)}\]

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.3ohh..i see,,i calcuated for infinite terms i.e n tends to infinity..hmm..not the general formula

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.3how do you find summation (n/n^4 + n^2 +1) ??

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.3@FoolAroundMath @FoolForMath

FoolAroundMath
 4 years ago
Best ResponseYou've already chosen the best response.2I 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 : \[S = \frac{1}{2}(1  \frac{1}{n^{2}+n+1})\]

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.3ohh lol..yes,,damn i missed that..nice :)

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0looks like that worked http://www.wolframalpha.com/input/?i=sum%5Bn%2F%28%28%28n%2B1%29%5E2++%28n%2B1%29+%2B+1%29%28%28n%29%5E2++%28n%29+%2B+1%29%29%2C+1%2C+infinity%5D
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