Question

Question was provided via image upload.

Explanation:

Identify series parameters

$$ \sum_{n=1}^{5} 6(3)^{n-1} $$

$$ a_1 = 6(3)^{1-1} = 6 $$

$$ r = 3,\quad k = 5 $$

Apply geometric sum formula

$$ S_k = \frac{a_1(1 - r^k)}{1 - r} $$

$$ S_5 = \frac{6(1 - 3^5)}{1 - 3} $$

Calculate final value

$$ S_5 = \frac{6(1 - 243)}{-2} = -3( -242 ) = 726 $$

Answer:

Evaluate:

$$\sum_{n=1}^{5} 6(3)^{n-1}$$

\(S_n =\) <blank>726</blank>