https://file.aiquickdraw.com/mathbot/file/1737433017656jp922wy5.jpg

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?  

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### 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.

Evaluate the sum S = ∑ (from n = 1 to 4) 3(7)^(n-1).

Learn how to evaluate the finite geometric series S = ∑ (from n = 1 to 4) 3(7)^(n-1) step by step. This problem covers concepts like sequences, series, and properties of geometric progressions. Suitable for students in Grades 10-12.

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