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To prove that the sum of the first 60 natural numbers is 1830, we use the well-known formula for the sum of the first \( n \) natural numbers:

\[
S_n = \frac{n(n+1)}{2}
\]

In this case, \( n = 60 \). Substituting \( n \) into the formula:

\[
S_{60} = \frac{60(60+1)}{2} = \frac{60 \times 61}{2} = \frac{3660}{2} = 1830
\]

Thus, the sum of the first 60 natural numbers is indeed 1830.

Would you like more details or have any questions? Here are 5 related questions to expand your understanding:

1. How can you prove this formula works for any \( n \)?
2. What is the sum of the first 100 natural numbers?
3. Can you derive a general formula for the sum of squares of the first \( n \) natural numbers?
4. How does this formula relate to the concept of arithmetic progressions?
5. How would you calculate the sum of natural numbers between two arbitrary values, say from 20 to 60?

**Tip:** The sum of an arithmetic series can be calculated quickly using its first term, last term, and the number of terms.

Prove that the sum of the first 60 natural numbers is 1830.

Explore how to prove the sum of the first 60 natural numbers using the arithmetic progression formula. This problem illustrates the application of the well-known sum formula for natural numbers. Suitable for students in Grades 6-8.

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