In this section, we will establish a few useful formulas related to the sum of natural numbers. These formulas are often used in various competitive exams.

Table of Contents

**Sum of first n natural numbers:**

**Question:** What is the sum of first n natural numbers?

**Answer:** The formula of the sum of first n natural numbers is n(n+1)/2.

The sum of consecutive natural numbers is given by the following formula:

Prove that: $1+2+3+\cdots+n$ $=\dfrac{n(n+1)}{2}$ |

Proof: Note that $1, 2, 3, …, n$ form an arithmetic progression (AP) with the first term $a=1$ and with the common difference $d=1.$

By the sum formula of an AP, we have

$1+2+3+…+n$ $=\dfrac{n}{2}[2a+(n-1)d]$

$=\dfrac{n}{2}[2.1+(n-1).1]$

$=\dfrac{n}{2}[2+n-1]$

$=\dfrac{n(n+1)}{2}$

**SOLVED PROBLEMS**

Problem 1: Find the sum of first 10 natural numbers.

Solution: The desired sum = 1+2+…+10

$=\dfrac{10(10+1)}{2}$ by the above formula.

= 5×11

=55

So the sum of first 10 natural numbers is 55.

Problem 2: Find the sum of the first 100 natural numbers.

Solution: We have to find the sum 1+2+…+100

In the above formula of the sum of first n natural numbers, we put n=100

$\therefore$ the sum $=\dfrac{100(100+1)}{2}$

= 50 × 101

=5050

Thus, the sum of first 100 natural numbers is 5050.

Problem 3: Find the sum of the first 50 natural numbers.

Solution: We have to find the sum 1+2+…+50

In the formula $1+2+3+\cdots+n$ $=\dfrac{n(n+1)}{2},$ we put n=50

$\therefore$ the sum $=\dfrac{50(50+1)}{2}$

= 25 × 51

= 1275

So the sum of first 50 natural numbers is 1275.

Problem 4: In a similar way, the sum of the first 20 natural numbers is

$1+2+3+\cdots +20$

$=\dfrac{20(20+1)}{2}$

= 10 × 21 = 210.

So the sum of first 20 natural numbers is 210.

**Sum of first n even natural numbers:**

Prove that: $2+4+6+\cdots+2n$ $=n(n+1)$ |

Proof: Note that $2+4+6+…+2n$ $=2(1+2+3+…+n)$

$=2 \times \dfrac{n(n+1)}{2}$

$=n(n+1)$

**SOLVED PROBLEMS**

Problem I: Find the sum of the first 100 even natural numbers.

Solution: We have to find the sum 2+4+…+98+100

In the above formula of the sum of consecutive even natural numbers, we put 2n=100. So n=50.

$\therefore$ the sum $=50(50+1)$

$=50 \times 51$

$=2550$

**Sum of first n odd natural numbers:**

Prove that: $1+3+5+\cdots+(2n+1)$ $=n^2$ |

Proof: Note that $1, 3, 5, \cdots,$$(2n-1)$ forms an AP with $n$ terms. The first term is $a=1$ and the common difference is $d=2.$

$\therefore$ the sum $=1+3+5+$ $\cdots+(2n-1)$ $=\dfrac{n}{2}[2a+(n-1)d]$

$=\dfrac{n}{2}[2.1+(n-1).2]$

$=\dfrac{n}{2} \times 2n$

$=n^2$

**SOLVED PROBLEMS**

Problem A: Find the sum of the first 100 odd natural numbers.

Solution: We have to find the sum 1+3+…+97+99

In the above formula of the sum of consecutive odd natural numbers, we put 2n-1=99. So 2n=100, that is, n=50.

$\therefore$ the sum of the first 100 odd natural numbers is $50^2$ $=2500.$

**FAQs on Sum of Natural Numbers Formula**

**Q1: What is the sum of the first n natural numbers?**

Answer: The formula for the sum of the first n natural numbers is n(n+1)/2.

**Q2: What is the formula for the sum of first n even numbers?**

Answer: The sum of the first n even numbers is n(n+1).

**Q3: Find the sum of the first n odd numbers.**

Answer: The sum of the first n odd numbers is $n^2.$