The problem asks for the evaluation of the sum:

$S = \sum_{n=1}^{4} 3(7)^{n-1}.$

Step-by-step solution:

1. Understand the sum structure:

   • The sum has 4 terms where $n$ goes from 1 to 4.
   • The general term is $3(7)^{n-1}$.

2. Write out the terms: Substitute $n = 1, 2, 3, 4$ into the general term $3(7)^{n-1}$:

   • When $n = 1$: $3(7)^{1-1} = 3(7)^0 = 3(1) = 3$,
   • When $n = 2$: $3(7)^{2-1} = 3(7)^1 = 3(7) = 21$,
   • When $n = 3$: $3(7)^{3-1} = 3(7)^2 = 3(49) = 147$,
   • When $n = 4$: $3(7)^{4-1} = 3(7)^3 = 3(343) = 1029$.

3. Add the terms:

   • $S = 3 + 21 + 147 + 1029$,
   • $S = 1200$.

Final Answer:

$S = 1200.$

Would you like a detailed explanation of any step?

Related Questions:

1. How would the sum change if the upper limit $n = 4$ was replaced with $n = 5$?
2. What happens if the base $7$ is replaced with $5$?
3. Can the formula for a geometric series be used here to verify the result?
4. How is the formula for the sum of a geometric series derived?
5. What are some real-world applications of geometric series?

Tip:

When solving sums like this, identify if the sequence forms a geometric progression to simplify the computation.