1. 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

2. We can find the sum of first n natural number using (n/2(a+l))

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

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

Where Sn​ is the sum, n is the number of natural numbers being summed, a is the first term (1), l is the last term.

Proof of this Formula:

1. List the series:

[Tex]S_n = a + (a+d) + (a+2d) + … + (a+(n-2)d) + (a+(n-1)d)[/Tex]

Here, a is the first term, d is the common difference, and n is the number of terms in the series.

2. Express the nth term:

The nth term l can be expressed as:

[Tex]l = a+(n-1)d[/Tex]

This is the last term of the series.

3. Write the series is reverse:

[Tex]S_n = l + (l-d) + (l-2d) + … + (l-(n-2)d) + (l-(n-1)d)[/Tex]

Reversing the series helps in pairing the terms to simplify the calculation.

4. Add the two series equations:

[Tex]S_n = a+(a+d)+(a+2d)+ … + (a+(n-2)d)+(a+(n-1)d)[/Tex]

[Tex]S_n = l+(l-d) + (l-2d) + … + (l-(n-2)d)+(l-(n-1)d)[/Tex]

Adding both equation term by term gives:

[Tex]2S_n = (a+l)+(a+l)+(a+l)+ … + (a+l) [/Tex]

[Tex]2S_n = n(a+l)[/Tex]

Each term in the series sums to (a+l) and there are n such terms:

5. Solve for S_n :

The sum of the first n terms of an arithmetic series is :

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

This formula simplifies the process of finding the sum by using the first term a and the last term l.

Example – How to Use the Formula

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

Given, n = 10, a = 1, l = 10

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

=> S₁₀ = 10/2(1+10)

=> S₁₀ = 10/2(11)

=> S₁₀ = 5*11

=> S₁₀ = 55