If a *whole number* **is not divisible by 2** into whole numbers then those numbers are called as **odd numbers**. Example: **1, 3, 5, 7,** and so on… are odd numbers. But, what if you want to find the sum of first n odd numbers? What formula will you use to do so?

This article is going to explain you the method and formula to find that sum and also the **sum of first n odd numbers**.

Related Topic: Sum of Even Numbers Formula

Index

**General Formula for Odd Numbers**

As we know, the odd numbers are

1, 3, 5, 7, 9, 11, …

If we observe carefully, we can find an Arithmetic Progression Sequence(AP),

Here, the AP can be made into the formula as,

The formula for odd numbers = \(2n \pm 1\)

\(\mbox{Odd Number} =

\left\{

\begin{array}{ll}

2n+1 & \mbox{if $n = 0, 1, 2, 3,$ and so on…} \\

2n-1 & \mbox{if $n = 1, 2, 3,$ and so on…} \\

\end{array}

\right.

\)

## Sum of Odd Numbers

Let us say the sequence of the n odd numbers is

=> (2k-1) + (2(k+1)-1) + …… + (2(k+n-1)-1)

From Arithmetic Progression, we know that,

\(S_n = \frac{n}{2}[2a + (n-1)d] \mbox{ OR } \frac{n}{2}[a + l]\)

Where,

\(n\) = no. of terms

\(a\) = First term of the A.P.

\(d\) = Common difference

\(l\) = Last term

And we know,

Here, a = 2k-1, d = 2, n = n & l = 2(k+n-1) – 1

So, the sum would be \(S_n = \frac{n}{2}[2(2k-1) + (n-1)2]\)

= \(S_n = \frac{n}{2}[4k -2 + 2n -2 ]\) = \(S_n = (n)[2k -2 + n]\)

**Sum of First n Odd Numbers**

Let us find the sum of first n odd numbers: (\(S_n\)),

Therefore, \(S_n = 1 + 3 + 5 + 7 + 9 + 11 + …. + (2n – 1) … (i)\)

\(S_n = \sum_{i = 1} ^{n} (2i – 1)\)

From Arithmetic Progression, we know that,

\(S_n = \frac{n}{2}[2a + (n-1)d] \mbox{ OR } \frac{n}{2}[a + l]\)

Where,

\(n\) = no. of terms

\(a\) = First term of the A.P.

\(d\) = Common difference

\(l\) = Last term

Here, \(a = 1, d = 2 \mbox{ & } l = 2n – 1\)

Therefore, \(S_n = \frac{n}{2} [a + l]\)

\(= \frac{n}{2} [1 + (2n – 1)]\)

\(= \frac{n}{2} [2n] = n^2\)

Therefore, \(S_n = n^2\).

For,

- \(n = 1\)
- \(S_1 = 1\) (&)
- \(S_1 = \frac{1}{2} [1 + (2\times1 – 1)] = 1\)

- \(n = 2\)
- \(S_2 = 1 + 3 = 4\) (&)
- \(S_2 = \frac{2}{2} [1 + (2\times2 – 1)] = 4\)

- \(n = 3\)
- \(S_3 = 1 + 3 + 5 = 9\) (&)
- \(S_3 = \frac{3}{2} [1 + (2 \times 3 – 1)] = 9\)

- \(n = 4\)
- \(S_4 = 1 + 3 + 5 + 7 = 16\) (&)
- \(S_4 = \frac{4}{2} [1 + (2\times4 – 1)] = 16\)

And so on… .

**Solved Examples**

**Question 1.** Find sum of first 50 odd number.

**Solution.** \(S_n = n^2\)

Therefore, \(S_{50} = 50^2 = 2500\)

**Question 2.** Find the sum of odd number between 100 & 200.

**Solution.** \(S_n = n^2\)

Therefore, sum of odd number between 100 & 200 = \(S_{100} – S_{50}\).

[Because there are 50 odd number in 1 to 100 & 100 odd numbers in between 1 to 200]

\(S_{50} = 50^2 = 2500\)

\(S_{100} = 100^2 = 10000\)

Therefore, sum of odd number between 100 & 200 = 10000 – 2500 = 7500.

**FAQs**

**What is the sum of first n odd numbers?**Sum of ‘n’ odd numbers: \(S_n = n^2\)

**What is the sum of n odd numbers?**The formula for sum between 1 and n is \((\frac{1}{2}(n+1))^2\).

**What is the sum of the first 25 odd numbers?**\(S_{25} = n^2 = 625\)

**What are Odd numbers?**Odd numbers are a set of numbers from Real Numbers where the digit cannot be divided by 2.