$\displaystyle f(n) = \sum_{i = 1}^{n} log_2(i)$ Having a hard time figuring this one out.
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My guess is. $\displaystyle \log_2(1)+\log_2(2)+\log_2(3)+\dots+\log_2(n-1)+\log_2(n)$ $\displaystyle \log_2(1\times 2\times3\times\dots\times (n-1)\times n)$ $\displaystyle \log_2(n!)$
Applying the basic properties of the logarithms You obtain... $\displaystyle \sum_{i=1}^{n}\log_{2} i = \log_{2} \prod_{i=1}^{n} i = \log_{2} n!$ Kind regards $\displaystyle \chi$ $\displaystyle \sigma$
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