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    $\begingroup$ damn I made super stupid mistake in asking question. Now corrected. However now am not able to get how the stuff cancel each other in the series: $2^0\log_2\frac{n}{2^0}+2^1\log_2\frac{n}{2^1}+2^2\log_2\frac{n}{2^2}+...+2^{ \log_2{n}}\log_2\frac{n}{2^{ \log_2{n}}}=\log_2{n}+2\log_2{\frac{n}{2}}+4\log_2{\frac{n}{4}}+...+n\log_2{1}$.. This is the series that you come up with right? Trying with $n=8$, I got $\log_2{8}+2\log_2{4}+4\log_2{2}+8\log_2{1}=3+4+4=11$. Whats wrong here? $\endgroup$ Commented May 14, 2016 at 12:05
  • $\begingroup$ @anir - Use equality $\log_2(\frac{n}{2^k}) = \log_2 n - k$ $\endgroup$ Commented May 14, 2016 at 15:36
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    $\begingroup$ I still didn't get how that series collapses to $\Theta(n)$ :'¬( Math-noob-here... $\endgroup$ Commented May 19, 2016 at 18:42
  • $\begingroup$ @anir - I'll expand my answer $\endgroup$ Commented May 19, 2016 at 22:43
  • $\begingroup$ @HEKTO If you solve above equ of comment, you still get nlog(n) ?? I have tried a lot. would you please help me here ? $\endgroup$ Commented Jul 15, 2017 at 17:15