anonymous
  • anonymous
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!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
EDIT: #8 th now :)
zzr0ck3r
  • zzr0ck3r
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:)
anonymous
  • anonymous
The mathematical part doesn't use any sort of fancy things, any high-school students should be able to do it :)

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More answers

zzr0ck3r
  • zzr0ck3r
ok ok I'll try :P
shubhamsrg
  • shubhamsrg
is ans 1/2 ?
anonymous
  • anonymous
No, but for \(n=10000\) the answer is \(\frac 12 \)
shubhamsrg
  • shubhamsrg
hmm,,i used hit and trial i.e. checked each term..sum approaches 0.5 only when it gets bigger..
anonymous
  • anonymous
Nopes :) You have to derive a term of \(S_n \) summation upto \(n\) terms
shubhamsrg
  • shubhamsrg
i easily see the denominator proceeds as (n^2 +n +1)(n^2 -n +1).. dont know what comes next.. hold on,,leme try..
zzr0ck3r
  • zzr0ck3r
damn how do you easily see that?
shubhamsrg
  • shubhamsrg
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
Carniel
  • Carniel
My head hurt but isn't it somewhere near 1/2 and 1/3
shubhamsrg
  • shubhamsrg
@zzr0ck3r i somehow saw that it is (n^2 +1)^2 - (n^2) dont know how,,even that amazes me!! :D
zzr0ck3r
  • zzr0ck3r
nice work
shubhamsrg
  • shubhamsrg
but @FoolForMath says its not 1/2..i must be wrong somewhere..hmm
shubhamsrg
  • shubhamsrg
and how do we know its 1/2 only upto n=10000?? sir?? you there?
maheshmeghwal9
  • maheshmeghwal9
i think no he is not here
zzr0ck3r
  • zzr0ck3r
code it and check
zzr0ck3r
  • zzr0ck3r
im guessing he would not have said it if it was not true:)
shubhamsrg
  • shubhamsrg
hmmm.. :)
FoolAroundMath
  • FoolAroundMath
\[S_{n} = \frac{n(n+1)}{2(n^{2}+n+1)}\]
shubhamsrg
  • shubhamsrg
ohh..i see,,i calcuated for infinite terms i.e n tends to infinity..hmm..not the general formula
shubhamsrg
  • shubhamsrg
how do you find summation (n/n^4 + n^2 +1) ??
shubhamsrg
  • shubhamsrg
@FoolAroundMath @FoolForMath
FoolAroundMath
  • FoolAroundMath
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 : \[S = \frac{1}{2}(1 - \frac{1}{n^{2}+n+1})\]
shubhamsrg
  • shubhamsrg
ohh lol..yes,,damn i missed that..nice :)
experimentX
  • experimentX
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

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