[tex]D)\frac{21}{2}[/tex]
1) Since this is a series, a geometric we can expand this, note that the common ratio is in the parentheses.
[tex]\begin{gathered} \sum_{n=1}^38\cdot\left(\frac{1}{4}\right)^{k-1}=8\cdot\left(\frac{1}{4}\right)^{1-1}+8\cdot\left(\frac{1}{4}\right)^{2-1}+8\cdot\left(\frac{1}{4}\right)^{3-1} \\ \\ \placeholder{⬚}^1 \\ \end{gathered}[/tex]
Note We plugged into k the number of the term from 1 to 3.
2) So we can solve it this way:
[tex]8\cdot(\frac{1}{4})^0+8\cdot(\frac{1}{4})^1+8\cdot(\frac{1}{4})^2=\frac{21}{2}[/tex]
Thus the answer is 21/2
