Sum of Odd Numbers Formula

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

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.

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