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

- jamiebookeater

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- anonymous

EDIT: #8 th now :)

- 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

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

ok ok I'll try :P

- shubhamsrg

is ans 1/2 ?

- anonymous

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

- shubhamsrg

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

- anonymous

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

- 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

damn how do you easily see that?

- 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

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

- shubhamsrg

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

- zzr0ck3r

nice work

- shubhamsrg

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

- shubhamsrg

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

- maheshmeghwal9

i think no he is not here

- zzr0ck3r

code it and check

- zzr0ck3r

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

- shubhamsrg

hmmm.. :)

- FoolAroundMath

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

- shubhamsrg

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

- shubhamsrg

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

- shubhamsrg

@FoolAroundMath @FoolForMath

- 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

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

- 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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