1. **State the problem:** We want to evaluate the infinite series $$\sum_{n=2}^{\infty} \left(\frac{3}{4}\right)^n \cos(180n^\circ).$$ 2. **Recall the cosine values:** Since $$\cos(180n^\circ) = \cos(\pi n) = (-1)^n,$$ the series becomes $$\sum_{n=2}^{\infty} \left(\frac{3}{4}\right)^n (-1)^n = \sum_{n=2}^{\infty} \left(-\frac{3}{4}\right)^n.$$ 3. **Recognize the series type:** This is a geometric series with first term $$a = \left(-\frac{3}{4}\right)^2 = \frac{9}{16}$$ and common ratio $$r = -\frac{3}{4}.$$ 4. **Check convergence:** Since $$|r| = \frac{3}{4} < 1,$$ the series converges. 5. **Use the geometric series sum formula:** For $$|r|<1,$$ $$\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}.$$ Here, our series starts at $$n=2,$$ so rewrite as $$\sum_{n=2}^{\infty} r^n = \sum_{n=0}^{\infty} r^n - r^0 - r^1 = \frac{1}{1-r} - 1 - r.$$ 6. **Calculate the sum:** Substitute $$r = -\frac{3}{4}$$ $$\sum_{n=2}^{\infty} \left(-\frac{3}{4}\right)^n = \frac{1}{1 - \left(-\frac{3}{4}\right)} - 1 - \left(-\frac{3}{4}\right) = \frac{1}{1 + \frac{3}{4}} - 1 + \frac{3}{4} = \frac{1}{\frac{7}{4}} - \frac{1}{1} + \frac{3}{4} = \frac{4}{7} - 1 + \frac{3}{4}.$$ 7. **Simplify the expression:** $$\frac{4}{7} - 1 + \frac{3}{4} = \frac{4}{7} - \frac{7}{7} + \frac{3}{4} = -\frac{3}{7} + \frac{3}{4} = \frac{-12}{28} + \frac{21}{28} = \frac{9}{28}.$$ **Final answer:** $$\sum_{n=2}^{\infty} \left(\frac{3}{4}\right)^n \cos(180n^\circ) = \frac{9}{28}.$$