Logarithmic series

Question

$T(n) = \log k + \log (k/2) + \log (k/4) + \log(k/8) + \ldots+ 1$

What is the sum of this series?

Answer 1

$\log \left(k.k/2.k/4\ldots 1\right)= \log \left(k^n \left(1.1/2.1/2^2.1/2^3\ldots 1\right)\right)=\log\left(k^n \frac{1}{2^0.2^1.2^2 \ldots 2^n}\right)$

$=\log\left(k^n\left(\frac{1}{2^{0+1+2+\ldots +n}}\right)\right)$

$=\log\left(k^n\frac{1}{2^{n(n+1)/2)}})\right)$

$=n\log k-\frac{n(n+1)}{2}\log 2$

Comments

yes done correction

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@srestha You can continue reducing- it is $\log ab$ form. You can convert back to $\log a + \log b$.

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Thank u :)