0 votes 0 votes Could anyone pls help me to solve this series ? Raghav Khajuria asked Sep 6, 2018 Raghav Khajuria 269 views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments air1ankit commented Sep 6, 2018 reply Follow Share yes! you are in the correct way, by using the extended master method the time complexity will be T(n)= $\Theta$(n_logn) and in your image, you got the series na?? just because of, we are dealing with asymptotic complexity we need not to solve whole equation just take the first term of that series i.e T(n) = n logn and left all the term..bczz all term will certainly be less bczz all term is dividing by 2, so take first term and left all other 0 votes 0 votes Shubhanshu commented Sep 6, 2018 reply Follow Share The actual value will be nlogn + n $ * \frac{k-1}{2^k}$ which is asymtotically equal to O(nlogn). 0 votes 0 votes Shaik Masthan commented Sep 7, 2018 reply Follow Share @Raghav Khajuria convert that series as 2.log 2 + 22 . log (22) + 23 . log(23) + ..... + 2k . log(2k) = (1.2 + 2.22 +3.23 + .... + k . 2k) log2 ====> it is AGP series, solve it 1 votes 1 votes Please log in or register to add a comment.