To evaluate the sum $s_4 = \sum_{k=1}^{4} 2(3^{n-1})$, let's proceed step-by-step.

1. Identify the series definition: The sum notation $\sum_{k=1}^{4} 2(3^{n-1})$ suggests that we need to evaluate the expression $2(3^{n-1})$ for $n = 1, 2, 3,$ and $4$, and then add up those results.

2. Evaluate each term:

   • When $n = 1$: $2(3^{1-1}) = 2(3^0) = 2 \times 1 = 2$
   • When $n = 2$: $2(3^{2-1}) = 2(3^1) = 2 \times 3 = 6$
   • When $n = 3$: $2(3^{3-1}) = 2(3^2) = 2 \times 9 = 18$
   • When $n = 4$: $2(3^{4-1}) = 2(3^3) = 2 \times 27 = 54$

3. Sum the results: $s_4 = 2 + 6 + 18 + 54 = 80$

So, the evaluated result of the sum $s_4$ is 80.

Would you like further details or explanations on any specific step?

Here are five related questions you might find interesting:

1. How would the result change if the sum extended to $n = 5$?
2. What if the base of the exponent were different, such as $4$ instead of $3$?
3. How would we approach this if it were an infinite series rather than a finite one?
4. Can we generalize this approach for any upper limit $n$?
5. How would changing the coefficient from $2$ to another number affect the series?

Tip: When evaluating sums, always check if the expression inside the summation is constant or depends on the index; this affects how each term is computed.