May 10, 2018

$\text{c} . 211$

#### Explanation:

We have to add the values of $16 {\left(\frac{3}{2}\right)}^{k - 1}$, when k is 1, 2, 3, 4, and 5 to find our sum.

$16 {\left(\frac{3}{2}\right)}^{1 - 1} + 16 {\left(\frac{3}{2}\right)}^{2 - 1} + 16 {\left(\frac{3}{2}\right)}^{3 - 1} + 16 {\left(\frac{3}{2}\right)}^{4 - 1} + 16 {\left(\frac{3}{2}\right)}^{5 - 1}$

$16 {\left(\frac{3}{2}\right)}^{0} + 16 {\left(\frac{3}{2}\right)}^{1} + 16 {\left(\frac{3}{2}\right)}^{2} + 16 {\left(\frac{3}{2}\right)}^{3} + 16 {\left(\frac{3}{2}\right)}^{4}$

$\left(16 \cdot 1\right) + \left(16 \cdot \frac{3}{2}\right) + \left(16 \cdot \frac{9}{4}\right) + \left(16 \cdot \frac{27}{8}\right) + \left(16 \cdot \frac{81}{16}\right)$

$16 + \frac{48}{2} + \frac{144}{4} + \frac{432}{8} + \frac{1296}{16}$

$16 + 24 + 36 + 54 + 81 = 211$

$\text{c} . 211$