Answer
$$-\frac{680}{81}$$
Work Step by Step
We simplify the summation by plugging in the values from $1$ to $4$ as follows:
$$\displaystyle \sum_{k=1}^{4} [(\frac{2}{3})^k-k]\\=[(\frac{2}{3})^1-1]+[(\frac{2}{3})^2-2]+[(\frac{2}{3})^3-3]+[(\frac{2}{3})^4-4]\\=[(\frac{2}{3})-1]+[(\frac{4}{9})-2]+[(\frac{8}{27})-3]+[(\frac{16}{81})-4]\\=\frac{2}{3}+\frac{4}{9}+\frac{8}{27}+\frac{16}{81}-10\\=-\frac{680}{81}$$
