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.