1. **State the problem:** We need to find the sum of the series $$\sum_{n=4}^{89} \left(\frac{1}{4}\right)^n + 7$$. 2. **Rewrite the series:** The series can be split into two parts: the sum of the geometric series $$\sum_{n=4}^{89} \left(\frac{1}{4}\right)^n$$ plus the constant term 7 added for each term from 4 to 89. 3. **Number of terms:** The number of terms from 4 to 89 inclusive is $$89 - 4 + 1 = 86$$. 4. **Sum of geometric series formula:** For a geometric series $$\sum_{k=0}^m ar^k = a \frac{1-r^{m+1}}{1-r}$$ where $$a$$ is the first term and $$r$$ is the common ratio. 5. **Apply formula to our series:** Here, the first term $$a = \left(\frac{1}{4}\right)^4 = \frac{1}{4^4} = \frac{1}{256}$$, the ratio $$r = \frac{1}{4}$$, and the number of terms $$m+1 = 86$$. 6. **Calculate the sum of the geometric part:** $$ S = \frac{1}{256} \cdot \frac{1 - \left(\frac{1}{4}\right)^{86}}{1 - \frac{1}{4}} = \frac{1}{256} \cdot \frac{1 - \left(\frac{1}{4}\right)^{86}}{\frac{3}{4}} = \frac{1}{256} \cdot \frac{4}{3} \left(1 - \left(\frac{1}{4}\right)^{86}\right) $$ 7. **Simplify:** $$ S = \frac{4}{768} \left(1 - \left(\frac{1}{4}\right)^{86}\right) = \frac{1}{192} \left(1 - \left(\frac{1}{4}\right)^{86}\right) $$ 8. **Sum of the constant term:** Since 7 is added for each of the 86 terms, the total is $$7 \times 86 = 602$$. 9. **Total sum:** $$ \text{Total} = S + 602 = \frac{1}{192} \left(1 - \left(\frac{1}{4}\right)^{86}\right) + 602 $$ 10. **Approximate:** Since $$\left(\frac{1}{4}\right)^{86}$$ is extremely small, it can be approximated as 0. Thus, $$ \text{Total} \approx \frac{1}{192} + 602 = 602 + 0.0052083 = 602.0052083 $$ **Final answer:** $$\boxed{602.0052}$$ (approximately)