Answer
$=\dfrac {64}{49}$
Work Step by Step
$\sum ^{\infty }_{k=3}\dfrac {3\times 4^{k}}{7^{k}}=\sum ^{\infty }_{n=3}3\times \left( \dfrac {4}{7}\right) ^{k}=\dfrac {a_{1}}{1-r};a_{1}=3\times \left( \dfrac {4}{7}\right) ^{3};r=\dfrac {4}{7}\Rightarrow S_{\infty }=\dfrac {a_{1}}{1-r}=\dfrac {3\times \left( \dfrac {4}{7}\right) ^{3}}{1-\dfrac {4}{7}}=\dfrac {64}{49}$
