$\displaystyle a_1 = 3(2^{-1}) = \frac{3}{2}$

$\displaystyle a_2 = 3(2^{-2}) = \frac{3}{2^2} = \frac{3}{4}$

$\displaystyle a_3 = 3(2^{-3}) = \frac{3}{2^3} = \frac{3}{8}$

$\displaystyle a_4 = 3(2^{-4}) = \frac{3}{2^4} = \frac{3}{16}$