Sum of n Natural Numbers Formula

The sum of natural numbers formula is obtained by using the arithmetic progression formula where the common difference between the preceding and succeeding numbers is 1. Natural numbers are also called counting numbers start from the number 1 until infinity such as 1,2,3,4,5,6,7, and so on. Let us learn about the sum of n natural numbers, how the formula is derived, and solve a few examples.

What is the Sum of Natural Numbers Formula?

The sum of n natural numbers formula is used to find 1 + 2 + 3 + 4 +..... up to n terms. This is arranged in an arithmetic sequence. Hence we use the formula of the sum of n terms in the arithmetic progression for deriving the formula for the sum of natural numbers.

Sum of Natural Numbers Formula: \(\sum_{1}^{n}\) = [n(n+1)]/2, where n is the natural number.

Definition of Sum of n Natural Numbers

Sum of n natural numbers can be defined as a form of arithmetic progression where the sum of n terms are arranged in a sequence with the first term being 1, n being the number of terms along with the n^th term. The sum of n natural numbers is represented as [n(n+1)]/2. Natural numbers are the numbers that start from 1 and end at infinity. Natural numbers include whole numbers in them except the number 0.

Derivation of Sum of Natural Numbers Formula

Let us derive the sum of natural numbers using the sum of n terms in an AP. In an AP, 'a' is the first term, 'd' is a common difference, 'l' is the last term i.e. n^th term, l = a+(n-1)d

In the arithmetic sequence of natural numbers, the common difference between the numbers is 1.

The sum of n terms of arithmetic progression will be:

Sum = a + (a+d) + (a+2d) …… + (l-2d) + (l-d) + l——————– (1)

When the order is reversed, the sum remains the same, hence,

Sum = l+(l-d)+(l-2d)..…+(a+2d)+(a+d)+a——————- (2)

Adding equations 1 and 2, we get

2 × Sum = (a+l)+[(a+d)+(l-d)]………+[(l-d)+(a+d)]+(l+a)]

2× Sum = (a+l)+(a+l)………+(a+l)+(a+l)

2 × Sum = n×(a+l)

⇒ Sum = n/2(a+l)

Substituting the value of l from the previous equation, we get

Sum of n terms of arithmetic progression = n/2[2a + (n – 1)d]

For natural numbers, a = 1 and d = 1, therefore,

S = n/2[2×1+(n-1)1]

S = [n(n+1)]/2

Hence, Sum of natural numbers formula = [n(n+1)]/2

Examples on Sum of Natural Numbers Formula

Example 1: Find the sum of the first 35 natural numbers.

Solution: Given, n = 35

The sum of natural numbers formula is:

S = [n(n+1)]/2

S = [35(35+1)]/2

S = 630

Therefore, the sum of the first 35 natural numbers is 630

Example 2: Find the sum of the natural numbers from 1 to 100.

Solution: We can use the arithmetic progression formula to find the sum of the natural numbers from 1 to 100. Where a = 1, n = 100, and d = 1

Sum of n terms of arithmetic progression = n/2[2a + (n – 1)d]

S = 100/2[2×1 + (100 - 1)1]

S = 5050

Therefore, the sum of the natural numbers from 1 to 100 is 5050

Example 3: Find the sum of the first 5 natural numbers.

Solution: Given, n = 5

Using the sum of natural numbers formula we get,

Sum of Natural Numbers Formula = [n(n+1)]/2

S = 5(5+1)/2

S = 15

Therefore, the sum of the first 5 natural numbers is 15