## FoolForMath 3 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!

1. FoolForMath

EDIT: #8 th now :)

2. 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:)

3. FoolForMath

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

4. zzr0ck3r

ok ok I'll try :P

5. shubhamsrg

is ans 1/2 ?

6. FoolForMath

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

7. shubhamsrg

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

8. FoolForMath

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

9. 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..

10. zzr0ck3r

damn how do you easily see that?

11. 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

12. Carniel

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

13. shubhamsrg

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

14. zzr0ck3r

nice work

15. shubhamsrg

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

16. shubhamsrg

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

17. maheshmeghwal9

i think no he is not here

18. zzr0ck3r

code it and check

19. zzr0ck3r

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

20. shubhamsrg

hmmm.. :)

21. FoolAroundMath

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

22. shubhamsrg

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

23. shubhamsrg

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

24. shubhamsrg

@FoolAroundMath @FoolForMath

25. 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 : $S = \frac{1}{2}(1 - \frac{1}{n^{2}+n+1})$

26. shubhamsrg

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

27. experimentX