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 Sn 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 Sk is equal, to
Sk = k(k+1) / 2
Our goal is to demonstrate its validity for k+1. This implies that the sum Sk+1 equals:
Sk+1 = (k+1)(k+2) / 2
Now lets expand Sk+1
Sk+1 = Sk + (k+1)
Substituting Sk = k(k+1) / 2 we get:
Sk+1 = k(k+1) / 2 + (k+1)
which simplifies, to:
Sk+1 = k(k+1) / 2 + 2(k+1) / 2
Therefore we have:
Sk+1 = [k(k+1) + 2(k+1)] / 2
leading to the result
Sk+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, Sn = n(n+1) / 2
S10 = 10(10+1) / 2
⇒ S10 = 10(11) / 2
⇒ S10 = 110 / 2
⇒ S10 = 55