Answer
See below.
Work Step by Step
$\sum_{i=1}^{4} ((\frac{2}{3})^k-k)=((\frac{2}{3})^1-1)+((\frac{2}{3})^2-2)+((\frac{2}{3})^3-3)+((\frac{2}{3})^4-4)=(\frac{2}{3}-\frac{3}{3})+(\frac{4}{9}-\frac{18}{9})+(\frac{8}{27}-\frac{81}{27})+(\frac{16}{81}-\frac{324}{81})=-\frac{27}{81}-\frac{126}{81}-\frac{219}{81}-\frac{308}{81}=-\frac{680}{81}$
