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To solve the summation \( 9 + 10 + 11 + 12 + \dots + 600 \), we will use the formula for the sum of an arithmetic series:

\[
S_n = \frac{n}{2} \times (a + l)
\]

Where:
- \( n \) is the number of terms in the series,
- \( a \) is the first term,
- \( l \) is the last term.

### Step 1: Identify the parameters
- The first term \( a = 9 \),
- The last term \( l = 600 \),
- The common difference \( d = 1 \).

### Step 2: Find the number of terms (\( n \))
Using the formula for the \( n \)-th term of an arithmetic series:
\[
l = a + (n-1)d
\]
Substitute the known values:
\[
600 = 9 + (n-1)(1)
\]
\[
600 - 9 = n-1
\]
\[
n = 592
\]

### Step 3: Compute the sum
Now substitute \( n = 592 \), \( a = 9 \), and \( l = 600 \) into the summation formula:
\[
S_n = \frac{592}{2} \times (9 + 600)
\]
\[
S_n = 296 \times 609
\]

Performing the calculation:
\[
S_n = 180264
\]

### Final Answer:
The sum of the series is:
\[
\boxed{180264}
\]

Let me know if you'd like a detailed explanation of any step! Here are 5 related questions you might consider:

1. How do you derive the formula for the sum of an arithmetic series?
2. What happens if the common difference is not 1? How does the calculation change?
3. Can we compute the sum using a different method, such as pairwise addition?
4. How do we solve similar problems if the series has fractional or negative terms?
5. Could we express this result symbolically instead of numerically?

**Tip:** Always double-check calculations for \( n \) in arithmetic series problems to avoid errors in counting the number of terms.

For each of the summations given below, use the formula for the sum of the first n integers either to evaluate the sum or to express it in closed form. (a) 9 + 10 + 11 + 12 + ⋯ + 600

Calculate the sum of the arithmetic series from 9 to 600. This problem uses the formula for the sum of an arithmetic series, ideal for high school students studying sequences and series.

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