## FoolForMath Group Title 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! 2 years ago 2 years ago

1. FoolForMath Group Title

EDIT: #8 th now :)

2. zzr0ck3r Group Title

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 Group Title

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

4. zzr0ck3r Group Title

ok ok I'll try :P

5. shubhamsrg Group Title

is ans 1/2 ?

6. FoolForMath Group Title

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

7. shubhamsrg Group Title

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

8. FoolForMath Group Title

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

9. shubhamsrg Group Title

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 Group Title

damn how do you easily see that?

11. shubhamsrg Group Title

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 Group Title

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

13. shubhamsrg Group Title

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

14. zzr0ck3r Group Title

nice work

15. shubhamsrg Group Title

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

16. shubhamsrg Group Title

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

17. maheshmeghwal9 Group Title

i think no he is not here

18. zzr0ck3r Group Title

code it and check

19. zzr0ck3r Group Title

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

20. shubhamsrg Group Title

hmmm.. :)

21. FoolAroundMath Group Title

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

22. shubhamsrg Group Title

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

23. shubhamsrg Group Title

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

24. shubhamsrg Group Title

@FoolAroundMath @FoolForMath

25. FoolAroundMath Group Title

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 Group Title

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

27. experimentX Group Title