The sum of the first n natural numbers is a fundamental mathematical concept that arises in many different areas, including algebra, geometry, and calculus. In this post, we will explore the formula for the sum of n natural numbers and some examples of its applications.

The formula for the sum of the first n natural numbers is given by:

S = 1 + 2 + 3 + … + n = n(n+1)/2

where S is the sum of the first n natural numbers.

To see how this formula works, let’s consider an example. Suppose we want to find the sum of the first 10 natural numbers. We can use the formula above to get:

S = 1 + 2 + 3 + … + 10 = 10(10+1)/2 = 55

Therefore, the sum of the first 10 natural numbers is 55.

This formula can be proven using mathematical induction, a powerful tool in mathematics that allows us to prove statements for all positive integers by starting with a base case and then showing that if the statement is true for any particular integer, it must also be true for the next integer.

Now, let’s explore some applications of this formula.

- Average of n natural numbers: The formula for the sum of n natural numbers can be used to find the average (or mean) of the first n natural numbers. The average is simply the sum divided by the number of terms, which is n in this case. So, we have:

Average = S/n = n(n+1)/(2n) = (n+1)/2

For example, the average of the first 10 natural numbers is (10+1)/2 = 5.5.

- Gauss’s addition trick: The famous mathematician Carl Friedrich Gauss discovered a clever trick for summing consecutive integers that are equally spaced. To use this trick, we pair the first and last terms, the second and second-to-last terms, and so on. Each pair has a sum equal to the same constant value, which is equal to the sum of the first and last terms. Therefore, the sum of n consecutive integers equally spaced can be found using the formula:

S = (n/2)(a + l)

where a is the first term, l is the last term, and n is the number of terms. Using this formula, we can find the sum of any consecutive integers, such as the sum of the integers from 1 to 100, which is (100/2)(1 + 100) = 5050.

- Sum of cubes: The formula for the sum of n natural numbers can be extended to find the sum of the cubes of the first n natural numbers. The formula is:

S = 1^3 + 2^3 + 3^3 + … + n^3 = [n(n+1)/2]^2

For example, the sum of the cubes of the first 10 natural numbers is [(10×11)/2]^2 = 3025.

Here are some examples related to the summation of n natural numbers:

- Sum of first 100 natural numbers: 1 + 2 + 3 + … + 98 + 99 + 100 = 5050
- Sum of first 50 even natural numbers: 2 + 4 + 6 + … + 96 + 98 + 100 = 2550
- Sum of first 25 odd natural numbers: 1 + 3 + 5 + … + 45 + 47 + 49 = 625
- Sum of first 10 prime numbers: 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29 = 129
- Sum of first 15 Fibonacci numbers: 0 + 1 + 1 + 2 + 3 + 5 + 8 + 13 + 21 + 34 + 55 + 89 + 144 + 233 + 377 = 986

These examples show how the summation of natural numbers can be used to find the sum of specific sets of numbers, such as even numbers, odd numbers, prime numbers, and Fibonacci numbers.

In conclusion, the formula for the sum of n natural numbers is a powerful tool in mathematics that can be used to find the sum of any consecutive integers, the average of the first n natural numbers, and the sum of the cubes of the first n natural numbers, among other applications.