Find the Sum of First n Natural Numbers

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

The sum of the first n natural numbers can be found using the following formula:

[Tex]\bold{S_n = \frac{n \cdot (n + 1)}{2}}[/Tex]

Where S_n​ is the sum, and n is the number of natural numbers being summed.

The formula is derived by pairing the first and last numbers in the sequence, the second and second-to-last numbers, and so on, resulting in [Tex]\bold{\frac{n}{2}}[/Tex]​ pairs. Each pair has a constant sum of n + 1, and when these pairs are added together, the formula simplifies to [Tex]\bold{\frac{n \cdot (n + 1)}{2}}[/Tex]​.

Proof of the Formula

For n = 1,

S₁ = 1(1+1) / 2 =2/2 = 1

Which is true.

Lets assume the formula holds true for an integer k. The sum S_k is equal, to

S_k = k(k+1) / 2

Our goal is to demonstrate its validity for k+1. This implies that the sum Sk+1 equals:

S_k+1 = (k+1)(k+2) / 2

Now lets expand S_k+1

S_k+1 = S_k + (k+1)

Substituting Sk = k(k+1) / 2 we get:

S_k+1 = k(k+1) / 2 + (k+1)

which simplifies, to:

S_k+1 = k(k+1) / 2 + 2(k+1) / 2

Therefore we have:

S_k+1 = [k(k+1) + 2(k+1)] / 2

leading to the result

S_k+1 = (k+1) (k+2) / 2

Which is the required formula for n = k + 1.

Thus, from principle of mathematical induction we can conclude that the formula Sn = n(n + 1) / 2, for the sum of the n numbers holds true for all positive integers n.

Let’s consider an example for how to use the formula.

Example: Calculate Sum of first n natural number for n is equal to 10.

Solution:

Given, n = 10

formula, S_n = n(n+1) / 2

S₁₀ = 10(10+1) / 2

⇒ S₁₀ = 10(11) / 2

⇒ S₁₀ = 110 / 2

⇒ S₁₀ = 55