Yet another question 6

It is given that $f(z) = \Re \{ z\}$ . Therefore,

$\begin{aligned} S & = \sum_{n=1}^{6a} \log_2 \left(\left| f \left((1+\sqrt{3}i)^n \right) \right| \right) \\ & = \sum_{n=1}^{6a} \log_2 \left(\left| \Re \left \{2^n \left(\frac 12 +\frac{\sqrt{3}}{2} i \right)^n \right \} \right| \right) \\ & = \sum_{n=1}^{6a} \log_2 \left(\left| \Re \left \{2^n e^{\frac{n\pi}{3}i} \right \} \right| \right) \\ & = \sum_{n=1}^{6a} \log_2 \left(\left| 2^n \cos \left( \frac{n\pi}{3} \right) \right| \right) \\ & = \small \log_2 \left(\left| 2 \left( \frac{1}{2}\right) \right| \right) + \log_2 \left(\left| 2^2 \left( -\frac{1}{2}\right) \right| \right) + \log_2 \left(\left| 2^3 \left(-1\right) \right| \right) + \log_2 \left(\left| 2^4 \left( - \frac{1}{2}\right) \right| \right) + \cdots + \log_2 \left(\left| 2^{6a} \left(1\right) \right| \right) \\ & = \small \log_2 \left(2 \left( \frac{1}{2}\right) \right) + \log_2 \left(2^2 \left( \frac{1}{2}\right) \right) + \log_2 \left(2^3 \right) + \log_2 \left( 2^4 \left( \frac{1}{2}\right) \right) + \log_2 \left( 2^5 \left( \frac{1}{2}\right) \right) + \cdots + \log_2 \left( 2^{6a} \right) \\ & = \log_2 \left(2 \left( 1 \right) \right) + \log_2 \left(2 \right) + \log_2 \left(2^3 \right) + \log_2 \left( 2^3\right) + \log_2 \left( 2^4 \right) + \log_2 \left( 2^6 \right) + \cdots + \log_2 \left( 2^{6a} \right) \\ & = 0 + 1 + \color{#3D99F6}{3} + 3 + 4 + \color{#3D99F6}{6} + 6 + 7 + 9 + 9 + 10 + \color{#3D99F6}{12} + 12 + 13 + \color{#3D99F6}{15} + 15 + 16 + \cdots + \color{#3D99F6}{6a} \\ & = 1 + \color{#3D99F6}{(2+1)} + 3 + 4 + \color{#3D99F6}{(5+1)} + 6 + 7 + 9 + 9 + 10 + \color{#3D99F6}{(11+1)} + 12 + \cdots + \color{#3D99F6}{(6a-1+1)} \\ & = \sum_{n=1}^{6a-1} n + 2a = \frac{(6a-1)6a}{2} + 2a = 18a^2-3a+2a = \boxed{18a^2-a} \end{aligned}$