$\sum_{i=0}^{log_4 n} (\frac{3}{16})^{i} cn^{2}$
$\Rightarrow cn^2\Bigg [ 1 + \frac{3}{16} + \frac{3}{16}^2 + \dots + \frac{3}{16}^k \Bigg]$ ($\color{green}{k = log_4n}$)
$\Rightarrow \large cn^2 \Bigg [ \frac{1 - (\frac{3}{16})^{log_4n + 1}}{1 - \frac{3}{16}} \Bigg ]$
$\Rightarrow \frac{16}{13}cn^2 \Bigg [ 1 - (\frac{3}{16})^{log_4n}*\frac{3}{16} \Bigg ]$
$\Rightarrow \frac{16}{13}cn^2 \Bigg [ 1 - n^{log_4\frac{3}{16}}*\frac{3}{16} \Bigg ]$
$\Rightarrow \frac{16}{13}cn^2 \Bigg [1 - n^{log_43 - 2}*\frac{3}{16} \Bigg ]$
$\Rightarrow \large \frac{16}{13}cn^2*\Bigg [\frac{16-3*\frac{n^{log_43}}{n^2}}{16} \Bigg ]$
$\color{navy}{\large \Rightarrow \frac{c}{13} \Bigg [ 16n^2 - 3n^{log_43} \Bigg ]}$