Answer
$\dfrac {189}{4}$
Work Step by Step
$\sum ^{6}_{j=1}48\left( \dfrac {1}{2}\right) ^{j}=\dfrac {a_{1}\left( r^{n}-1\right) }{r-1};a_{1}=48\times \left( \dfrac {1}{2}\right) ^{1}=24;r=\dfrac {1}{2}\Rightarrow S_{n}=\dfrac {24\times \left( \left( \dfrac {1}{2}\right) ^{6}-1\right) }{\dfrac {1}{2}-1}=\dfrac {24\times \left( \dfrac {-63}{64}\right) }{-\dfrac {1}{2}}=\dfrac {189}{4}$
