Answer
$4$
Work Step by Step
$\sum ^{\infty }_{i=1}3\left( \dfrac {1}{4}\right) ^{i-1}=\dfrac {a_{1}}{1-r};a_{1}=3\times \left( \dfrac {1}{4}\right) ^{1-1}=3;r=\dfrac {1}{4}\Rightarrow S_{\infty }=\dfrac {3}{1-\dfrac {1}{4}}=4$
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