f(z) = the real part of z. If a∈N,n∈N then the value of ∑n=16a log2f[1+i3)n has the value equal to

Re(1+i3)n=2ncosnπ3⇒∑n=16a log2f[1+i3)n=log22ncosnπ3=n+log2cosnπ3⇒∑16a n+ln2cosnπ3=6a(6a+1)2=18a2+3a

f(z) = the real part of z. If a∈N,n∈N then the value of ∑n=16a log2f[1+i3)n has the value equal to

Re(1+i3)n=2ncosnπ3⇒∑n=16a log2f[1+i3)n=log22ncosnπ3=n+log2cosnπ3⇒∑16a n+ln2cosnπ3=6a(6a+1)2=18a2+3a

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f(z) = the real part of z. If $a \in N, n \in N$ then the value of $\sum_{n = 1}^{6 a} \log_{2} |[f [1 + i \sqrt{3})^{n}]|$ has the value equal to

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18a² – 3a

18a² – a

18a² + 3a

18a² + a

answer is A.

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Detailed Solution

$\begin{array}{l} Re (1 + i \sqrt{3})^{n} = 2^{n} \cos \frac{nπ}{3} \\ \Rightarrow \sum_{n = 1}^{6 a} \log_{2} |[f [1 + i \sqrt{3})^{n}]| = \log_{2} |2^{n} \cos \frac{nπ}{3}| = n + \log_{2} \cos \frac{nπ}{3} \\ \Rightarrow \sum_{1}^{6 a} n + \ln_{2} \cos \frac{nπ}{3} = \frac{6 a (6 a + 1)}{2} = 18 a^{2} + 3 a \end{array}$