# Find the sum of first 50 natural numbers

Arithmetic is the most basic part of the number theory and number theory entirely deals with numbers and the operations done with numbers. The operations involve division, multiplication, addition, subtraction, etc. There are different types of numbers in the number theory based on different characteristics, for instance, whole numbers (starting from 0 to infinity), natural numbers (starting from 1 to infinity), and so on. Let’s learn about natural numbers in more detail,

Natural numbers are the numbers that are used for counting and are a part of real numbers. The set of natural numbers include only the positive integers, i.e., 1, 2, 3, 4, 5, 6,….∞. In simple words, it could be stated as all the positive whole numbers except zero. So let’s first analyze the first five natural numbers, which could be stated as ⇢ 1, 2, 3, 4, 5,… A common observation that could be seen in these numbers is that given any number, the difference between the preceding numbers is constant and unity.

So its can be stated that **for a given number from the natural number series the common difference between preceding numbers is 1 (neglecting the sign).**

**common difference =|a-b| =1=|b-a|**

**Where** **a and** **b are natural numbers preceding each other.**

The second observation to notice is if the common difference between any two numbers in succession is 1. Then it could also be written as a number in terms of other numbers such as,

**a-b =1**

** a=1+b**

Where **T(n-1)** and **T(n)** are in succession, so to know the number at a given say ith position we can also write.

**T(n) _{ }=T(n-1) +1**

** or**

**T(n+1)=T(n)+1**

For example, let T(n) =3

Therefore, 3 = 2+1

1 is the first natural number and 3 is the third natural number, deduced its Arithmetic Progression and therefore its general form from any natural number which is,

**T(n)= (n-1)d +T(1)**

Where d is the common difference between two numbers in succession, n represents the number at an nth place in the natural number series, and T(1) represents the first number in the sequence.

### Sum of 50 natural numbers

Since the problem statement is asking for the sum of the first 50 natural numbers and that is a large amount to calculate, and also the first 50 natural numbers is actually an AP with a common difference of 1, finding the generalized formula for this would be better. In order to find the sum of more than one natural number, let’s take the sum of the first n natural number. Using the formula earlier discussed in the article we will find the first n natural number

**T(n) = 1+ 2+ 3+ … + n **

add T(n) on both sides

⇒T(n)+T(n) = 1 + 2 + 3 + … + n +T(n)

⇒T(n)+T(n) = 1 + 2 + 3 + … + (n-1) + n + n + (n-1) + (n-2) + … + 2 + 1

Now, pair of the terms such that there sum equal to be (n+1)

⇒2T(n)= (1 + n) + (2 + n-1) + (3 + n-2) + … + (n-1 + 2) + (n + 1)

All the n pair-sums are equal to (n+1),

⇒2T(n)= (n+1) + (n+1) + (n+1) + … + (n+1) + (n+1)

⇒2T(n) = n (n+1)

⇒**T(n) = n (n+1) /2 **

So the formula to deduce the sum of the first n natural number. So let calculate the sum of the first 50 Natural Numbers is written as follows,

T(50) = 50(50+1)/2

T(50)=25×51

T(50)=1275

**And hence the sum of the first 50 natural numbers to be 1275.**

### Similar Problems

**Question 1: What is the difference between twenty** **and** **ten natural numbers?**

**Solution:**

Lets first calculate the sum of the first of ten natural numbers by using the formula

T(n) = n (n+1) /2

Therefore n=10,

⇒ T(10) = 10(10+1)/2

⇒ T(10) = (10×11)/2

⇒ T(10) = 110/2

⇒ T(10) = 55

Now for n = 20,

⇒ T(20) = 20(20+1)/2⇒ T(20) = (20×21)/2

⇒ T(20) = 420/2

⇒ T(20) = 210

Therefore, T(20) -T(10) = 210-55 = 155

**Question 2: Solve (1+2+3+4+5…25)×(30+29+27+28…1) **

**Solution:**

As the given question given requires the product of two summations, therefore, use the formula n(n+1)/2

Lets first calculate the sum of natural number by using the formula,

T(n) = n(n+1)/2

Therefore n=25,

⇒ T(25) = 25(25+1)/2

⇒ T(25) = (25×26)/2

⇒ T(25) = 650/2

⇒ T(25) = 325

Now for n = 30,

⇒ T(30) = 30(30+1)/2

⇒ T(30) = (30×31)/2

⇒ T(30) = 930/2

⇒ T(30) = 465

Therefore,

⇒ (1+2+3+4+5…25)(30+29+27+28…1)

⇒ T(30) ×T(25)

⇒ 325×465

⇒ 151125

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