@Siva: Work from the inside out, using the identities sum from i = 1 to n (i) = n*(n+1)/2
sum from i = 1 to n (i^2) = n*(n+1)*(2*n+1)/6 sum from i = 1 to n (i^3) = n^2*(n+1)^2/4 sum from i = 1 to n (i^4) = n^5/5 + n^4/2 + n^3/3 - n/30 Dave On Sep 21, 10:03 pm, siva viknesh <[email protected]> wrote: > somebody plz reply... > > On Sep 21, 10:53 pm, sivaviknesh s <[email protected]> wrote: > > > > > for(i=0;i<n;i++) > > for(j=0;j<i*i;j++) > > for(k=0;k<j;k++) > > sum++; > > > Is it n^5 log n ..... n * (n^2 log n) * (n^2 log n) ??? > > > correct me if i m wrong? also anybody can tell some easy approach to tackle > > these ques ?? I worked out for some values of n and arrived at the ans..... > > .... > > > -- > > Regards, > > $iva- Hide quoted text - > > - Show quoted text - -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
