The formula of the partial sum of the first n terms of a geometric series is:
![Sn=(a1(1-r^n))/(1-r)](https://img.qammunity.org/2023/formulas/mathematics/college/3232t2z4jw3n6m3lsehk7fg76mftkf9uu1.png)
In this case, a1 is 6 (the first term of the series) and r is 1/4 (the base). By replacing 16 for n and the rest of the values we should get:
![Sn=(6(1-((1)/(4))^(16)))/(1-((1)/(4)))\approx8](https://img.qammunity.org/2023/formulas/mathematics/college/5c51edzywxxewj3goeqwso6i84axhucbvu.png)
Then, the sum of the given geometric series is 8