Sum of First n Odd and Even Natural Numbers: AP Proofs | Class 10 Maths

Quick answer: The sum of the first $n$ odd natural numbers is $n^2$, and the sum of the first $n$ even natural numbers is $n(n + 1)$. Both results come straight from the AP sum formula $S_n = \frac{n}{2}(a + l)$.

The ancient Greek Pythagoreans studied numbers by arranging pebbles in patterns. They noticed something striking: start with one pebble and keep adding the next odd number of pebbles ($1 + 3 = 4 = 2^2$, $1 + 3 + 5 = 9 = 3^2$, $1 + 3 + 5 + 7 = 16 = 4^2$), and the pebbles always form a perfect square.

In modern algebra, the odd numbers $1, 3, 5, 7, \dots$ form an Arithmetic Progression (AP), and their sum is exactly $S_n = n^2$.

In CBSE Class 10 Mathematics, Chapter 5 (Arithmetic Progressions), questions on the sum of the first $n$ odd numbers, the sum of the first $n$ even numbers and the sum of multiples of a number (such as the first 40 positive integers divisible by 6) are common 2- to 3-mark board questions.

In this guide, we prove both results using the AP formula and work through two classic exam problems step by step.

What You Will Learn

• Proof that the sum of the first $n$ odd natural numbers is $S_n = n^2$
• A visual "square of dots" proof of $1 + 3 + 5 + \dots + (2n - 1) = n^2$
• Proof that the sum of the first $n$ even natural numbers is $S_n = n(n + 1)$
• How these give the sum of the first $n$ natural numbers, $S_n = \frac{n(n + 1)}{2}$
• Solved problem: sum of odd numbers between $0$ and $50$ ($S_{25} = 625$)
• Solved problem: sum of the first $40$ positive integers divisible by $6$ ($S_{40} = 4920$)

1. Sum of the First n Odd Natural Numbers ($S_n = n^2$)

Theorem

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

$1 + 3 + 5 + 7 + \dots + (2n - 1) = n^2$

Visual Proof: Odd Numbers Build Squares

Each new odd number adds an L-shaped layer that turns an $(n-1) \times (n-1)$ square into an $n \times n$ square:

n = 1:  ●                     = 1² = 1
n = 2:  ●  ○      (add 3)     = 2² = 4
        ○  ○
n = 3:  ●  ○  *   (add 5)     = 3² = 9
        ○  ○  *
        *  *  *

Algebraic Proof Using the AP Sum Formula

Step 1: Identify the AP. The odd natural numbers are:

$1, 3, 5, 7, 9, \dots, (2n - 1)$

• First term: $a = 1$
• Common difference: $d = 3 - 1 = 2$
• Number of terms: $n$
• Last term: $l = a_n = 2n - 1$

Step 2: Apply the AP sum formula.

$S_n = \frac{n}{2}(a + l)$

Step 3: Substitute $a = 1$ and $l = 2n - 1$.

$S_n = \frac{n}{2}\left[1 + (2n - 1)\right] = \frac{n}{2} \times 2n = n^2$

Hence proved: the sum of the first $n$ odd natural numbers is $S_n = n^2$.

2. Sum of the First n Even Natural Numbers ($S_n = n(n + 1)$)

Theorem

The sum of the first $n$ even natural numbers is $n(n + 1)$.

$2 + 4 + 6 + 8 + \dots + 2n = n(n + 1)$

Algebraic Proof Using the AP Sum Formula

Step 1: Identify the AP. The even natural numbers are:

$2, 4, 6, 8, \dots, 2n$

• First term: $a = 2$
• Common difference: $d = 4 - 2 = 2$
• Last term: $l = 2n$

Step 2: Apply the AP sum formula.

$S_n = \frac{n}{2}(a + l) = \frac{n}{2}(2 + 2n)$

Step 3: Take $2$ common from the bracket.

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

Hence proved: the sum of the first $n$ even natural numbers is $S_n = n(n + 1)$.

Useful ratio: dividing the two results gives

$\frac{\text{Sum of first } n \text{ odd numbers}}{\text{Sum of first } n \text{ even numbers}} = \frac{n^2}{n(n + 1)} = \frac{n}{n + 1}$

3. Connection to the Sum of Natural Numbers

The natural numbers from $1$ to $2n$ are exactly the first $n$ odd numbers plus the first $n$ even numbers:

$\text{Sum of } 1 \text{ to } 2n = (\text{sum of first } n \text{ odd}) + (\text{sum of first } n \text{ even})$

$\sum_{k=1}^{2n} k = n^2 + n(n + 1) = 2n^2 + n = n(2n + 1) = \frac{(2n)(2n + 1)}{2}$

Replacing $2n$ by $n$ gives the familiar formula for the sum of the first $n$ natural numbers:

$S_n = 1 + 2 + 3 + \dots + n = \frac{n(n + 1)}{2}$

4. Solved Board Exam Problems

Problem 1: Sum of Odd Numbers Between 0 and 50

Question: Find the sum of the odd numbers between $0$ and $50$.

Solution:

Step 1: The odd numbers between $0$ and $50$ are $1, 3, 5, 7, \dots, 49$. This is an AP with $a = 1$, $d = 2$ and $a_n = 49$.

Step 2: Find the number of terms $n$.

$a_n = a + (n - 1)d \implies 49 = 1 + (n - 1) \times 2$

$48 = 2(n - 1) \implies n - 1 = 24 \implies n = 25$

Step 3: Apply $S_n = n^2$.

$S_{25} = 25^2 = 625$

Answer: The sum of the odd numbers between $0$ and $50$ is $625$.

Problem 2: Sum of the First 40 Positive Integers Divisible by 6

Question: Find the sum of the first $40$ positive integers divisible by $6$.

Solution:

Step 1: The positive integers divisible by $6$ form the AP $6, 12, 18, 24, \dots$ with $a = 6$, $d = 6$ and $n = 40$.

Step 2: Apply the AP sum formula.

$S_n = \frac{n}{2}\left[2a + (n - 1)d\right]$

$S_{40} = \frac{40}{2}\left[2(6) + (40 - 1) \times 6\right] = 20\left[12 + 234\right] = 20 \times 246 = 4920$

Shortcut: $6 + 12 + \dots + 240 = 6(1 + 2 + \dots + 40) = 6 \times \frac{40 \times 41}{2} = 6 \times 820 = 4920$.

Answer: The sum of the first $40$ positive integers divisible by $6$ is $4920$.

5. Formula Summary Table

6. Exam Tips and Common Mistakes

Exam Tip: When asked for the sum of odd numbers between $0$ and $50$, always find $n$ first using $a_n = a + (n - 1)d$ (here $a_n = 49 \implies n = 25$). Writing $S_{25} = 25^2 = 625$ without showing how you got $n = 25$ can cost you a mark.

Common Mistake: Do not confuse the $n^{\text{th}}$ term with the sum. The $n^{\text{th}}$ odd number is $2n - 1$, while the sum of the first $n$ odd numbers is $n^2$.

Frequently Asked Questions

What is the sum of the first n odd natural numbers?

The sum of the first $n$ odd natural numbers is $n^2$. For example, $1 + 3 + 5 + 7 + 9 = 25 = 5^2$.

What is the sum of the first n even natural numbers?

The sum of the first $n$ even natural numbers is $n(n + 1)$. For example, $2 + 4 + 6 + 8 + 10 = 30 = 5 \times 6$.

What is the sum of odd numbers from 1 to 100?

There are $50$ odd numbers from $1$ to $99$, so the sum is $50^2 = 2500$.

What is the sum of even numbers from 1 to 100?

There are $50$ even numbers from $2$ to $100$, so the sum is $50 \times 51 = 2550$.

Which formula is used to prove these results?

Both are proved with the AP sum formula $S_n = \frac{n}{2}(a + l)$, where $a$ is the first term and $l$ is the last term.