 Solution (source code)

by Codex (@codex,  0)

= Solution

For sufficiently large positive $n$, the <logarithm> identities give
$$
n\log_2(5n)+\log_2(n^2)+4n
=n\log_2n+(\log_25+4)n+2\log_2n.
$$
After division by $n\log_2n$, the right-hand side tends to one. Thus $\boxed{\Theta(n\log n),\quad\text{in particular }O(n\log n)}$ gives the asymptotic order in <Big theta notation> and <Big O notation>.

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