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The given problem asks to find the sum:

\[
\sum_{n=1}^{60} (3n - 9)
\]

We can break this sum down and solve it step by step.

### Step 1: Break the summation into two parts.
\[
\sum_{n=1}^{60} (3n - 9) = \sum_{n=1}^{60} 3n - \sum_{n=1}^{60} 9
\]

This separates into two simpler sums: one involving \(3n\) and the other a constant \(9\).

### Step 2: Calculate the first summation: \(\sum_{n=1}^{60} 3n\)
The constant \(3\) can be factored out:
\[
3 \sum_{n=1}^{60} n = 3 \cdot \frac{60(60 + 1)}{2} = 3 \cdot \frac{60 \times 61}{2} = 3 \cdot 1830 = 5490
\]

### Step 3: Calculate the second summation: \(\sum_{n=1}^{60} 9\)
Since \(9\) is constant, this is simply:
\[
9 \times 60 = 540
\]

### Step 4: Subtract the two results:
\[
5490 - 540 = 4950
\]

Thus, the sum is:

\[
\boxed{4950}
\]

Would you like more details or have any questions?

Here are 5 related questions:
1. How would the sum change if the upper limit of the summation were \(n = 100\)?
2. How can you use properties of arithmetic progressions to solve similar summation problems?
3. What is the general formula for the sum of the first \(n\) natural numbers?
4. How would you sum an expression with quadratic terms, such as \( \sum (n^2 - 5n)\)?
5. How does factoring constants out of summations simplify the process?

**Tip:** Always look for ways to split and simplify summation problems by separating constants and variable-dependent terms.

Find the sum from n=1 to n=60 of (3n - 9).

Learn how to solve the summation problem where n ranges from 1 to 60 for the expression (3n - 9). The problem involves algebraic concepts such as arithmetic progression and the summation formula. Suitable for students in Grades 9-12.

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