Answer
$2032$
Work Step by Step
$\sum ^{10}_{k=4}2^{k}=\dfrac {a_{1}\times \left( r^{n}-1\right) }{r-1};a_{1}=2^{4}=16;r=2;n=10-4+1=7\Rightarrow S_{n}=\dfrac {16\times \left( 2^{7}-1\right) }{2-1}=16\times 127=2032$
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