A Sum

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$2\sum_{n=0}^{\infty} \frac{3}{4^n} = 6\sum_{n=0}^{\infty} \frac{1}{4^n}$

$= 6\times\left(1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}..............\right)$

$\left(\frac{1}{4} + \frac{1}{4^2} .................\right)\text{is a G.P with common ratio} \frac{1}{4}$

$= 6\times \left(1+\dfrac{\frac{1}{4}}{1-\frac{1}{4}}\right )= 6\times\frac{4}{3}=\boxed{8}$