What is the sum of the first n natural numbers?

The sum of the first n natural numbers is given by the formula: n(n+1)/2.

Here's a breakdown:

• Natural Numbers: These are positive integers starting from 1 (i.e., 1, 2, 3, 4, ...).

• Formula: The formula n(n+1)/2 provides a quick way to calculate the sum without having to add each number individually.

Explanation

The formula can be derived using the concept of arithmetic progression. An arithmetic progression is a sequence of numbers such that the difference between any two consecutive terms is constant. The sequence of natural numbers (1, 2, 3, ..., n) is an arithmetic progression with a common difference of 1.

The sum (S) of an arithmetic progression can be calculated as:

S = (n/2) * (first term + last term)

In the case of the first n natural numbers:

• First term = 1
• Last term = n

Therefore, S = (n/2) * (1 + n) = n(n+1)/2

Examples

• Sum of the first 5 natural numbers (1+2+3+4+5):
  Using the formula: 5(5+1)/2 = 5(6)/2 = 30/2 = 15

• Sum of the first 10 natural numbers (1+2+3+...+10):
  Using the formula: 10(10+1)/2 = 10(11)/2 = 110/2 = 55

• Sum of the first 100 natural numbers (1+2+3+...+100):
  Using the formula: 100(100+1)/2 = 100(101)/2 = 10100/2 = 5050

Summary

The formula n(n+1)/2 provides an efficient and accurate method for calculating the sum of the first n natural numbers. This formula is derived from the properties of arithmetic progressions.