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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
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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FoolForMathBest ResponseYou've already chosen the best response.0
EDIT: #8 th now :)
 one year ago

zzr0ck3rBest ResponseYou've already chosen the best response.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:)
 one year ago

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

FoolForMathBest ResponseYou've already chosen the best response.0
No, but for \(n=10000\) the answer is \(\frac 12 \)
 one year ago

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

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

shubhamsrgBest ResponseYou've already chosen the best response.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..
 one year ago

zzr0ck3rBest ResponseYou've already chosen the best response.0
damn how do you easily see that?
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.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 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
 one year ago

CarnielBest ResponseYou've already chosen the best response.0
My head hurt but isn't it somewhere near 1/2 and 1/3
 one year ago

shubhamsrgBest 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
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.3
but @FoolForMath says its not 1/2..i must be wrong somewhere..hmm
 one year ago

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

maheshmeghwal9Best ResponseYou've already chosen the best response.0
i think no he is not here
 one year ago

zzr0ck3rBest ResponseYou've already chosen the best response.0
im guessing he would not have said it if it was not true:)
 one year ago

FoolAroundMathBest ResponseYou've already chosen the best response.2
\[S_{n} = \frac{n(n+1)}{2(n^{2}+n+1)}\]
 one year ago

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

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

shubhamsrgBest ResponseYou've already chosen the best response.3
@FoolAroundMath @FoolForMath
 one year ago

FoolAroundMathBest ResponseYou've already chosen the best response.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 : \[S = \frac{1}{2}(1  \frac{1}{n^{2}+n+1})\]
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.3
ohh lol..yes,,damn i missed that..nice :)
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
looks 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
 one year ago
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