Sum of Odd Numbers

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Sum of Odd Natural Numbers

Odd numbers are those which give fractional form when divided by 2. Suppose we take the case of natural numbers, then the odd numbers among them will be given by 1, 3, 5, 7, …….We can say those numbers which are ending with 1, 3, 5, 7, and 9 are called odd numbers. After every odd number, there comes an even number. These are placed alternatively in mathematics. By using any arithmetic progression, we can easily find the sum of odd natural numbers.


Sum of Odd Numbers Formula

The numbers that have 1, 3, 5, 7, and 9 at the end are odd numbers. We are providing you with the explanation of the sum of odd numbers using Arithmetic Progression. However, this case can be defined as general for first n odd numbers or the sum of odd natural numbers to 10 or 100. We will look into each of the cases separately. 


Sum of First n Odd Numbers

Suppose we denote the sum of first n odd numbers as Sn. Then Sn is given by

Sn = 1+ 3+ 5+ 7……+ (2n-1)  - (i)

This is the case of Arithmetic Progression where d i.e. common difference is given by 2.

The formula used to find the sum of first n natural numbers is given by 

Sn = ½{n[2a + (n -1) d]}  - (ii)

In the above equation,

n is the total odd numbers that we want to add

a is the first term of the series i.e. 1 for the sum of odds

d is the common difference between two terms i.e. 2 for the sum of odd numbers. 

Hence, according to the equation, a =1 and d= 2

The last term of the odd number sequence is given by l = 2n -1

Put all these values in (ii)

Sn = ½{n [ a +l] }

Sn = n/2 [1 + 2n -1]

Sn = (n/2) x (2n) 

Sn = n2


The sum of first n odd natural numbers = n2

 

Now from the above formula, we can define the sum of total odd numbers in the given range.  

If n = 1 then sum of numbers is 1

   n =2 sum is 4

   n = 3 sum is 9

   …..

   …..

   …..

   …..

   . 

   n= 10 sum is 100

Here n is the consecutive odd number starting from 1


Example of Sum of Odd Numbers from 1 to 100

The case of finding the sum of odd numbers from 1 to 100 is quite different from that of finding the sum of even numbers. We can get it in two ways.


Case 1:

We know that the total number of odd natural numbers from 1 to 100 is 50. The other 50 are even numbers. 

Sum of odd natural numbers is given by

Sn = n2 

Hence, we give a sum of the first 50 odd natural numbers by:

S50 = (50)2

S50 = 2500


Case 2:

Alternatively, we can subtract the sum of even natural numbers from 1 to 100 from the total sum of numbers from 1 to 100.

We give this case as,

Sum of odds = Total - Sum of even numbers from 1 to 100

Se gives the sum of even numbers.  

Se  = 2+ 4+ 6+ 8……...100

Se  = 2 x ( 1+ 2+ 3+ ……...50)

Se  = 2 x [ 50/2 (1 +50)]

Se  = 2 x 1275

Se  = 2550

Now the sum of odds is given by So

So = Sum of first 100 natural numbers - sum of even numbers from 1 to 100

So = (50 x 101) - (2550)

So = 5050 - 2550

So = 2500

In both the cases discussed above, the sum of odd numbers from 1 to 100 is the same.


Sum of Three Consecutive Odd Integers

It becomes easy for you to solve the word problem for finding a generic term to get a sum of three consecutive odd integers. The case is represented as follows:

n = 2k

Here k is any integer

The general form for an odd term can be given by 

n = 2k =1 or n = 2k -1

Here k is an integer

We know that between two odd integers there is an even integer.


Case 1: If we use 2k +1 as the first odd integer

The next two consecutive odd integers will be 2k +1 +2 and 2k + 1 + 4

I.e. 2k +1, 2k + 3, 2k +5 are three consecutive odd integers.


Case 2: If we use 2k -1 as a first odd integer

The next two odd integers would be 2k - 1 +2 and 2k - 1 + 4

I.e. 2k -1, 2k+1, and 2k +3 are three consecutive odd integers.

FAQ (Frequently Asked Questions)

Q1. How will You Give a Sum of Odd Numbers from 11 to 60?

We have to give the sum of odds from 11 to 60. So, first, we will find the sum of odds from 1 to 60 and subtract the sum of odds from 1 to 10. This will give our solution to the range problem as follows:

S11-60 = S1-60 - S1-10

i.e.  S11-60 = (30)2- (5)2

S11-60 = 900 - 25

S11-60 = 875

In the above case, there is the only second case that we can follow to find the sum of natural numbers in the given range. To define the sum in the given range, we will use case 2.

Q2. Prove n2 Gives the Sum of Odd Natural Numbers.

Let us first understand the entire pattern to find the sum of odd numbers. Starting from 1, we have the odd natural numbers in the following sequence:

1, 3, 5, ……….(2n-1)

The sum of the first natural number is 1

Sum of first two natural numbers is 1 + 3 = 4 = 2*2

Sum of first three natural numbers is 1 + 3 + 5 = 9 = 3*3

Sum of first four natural numbers is 16 = 4*4

Hence proved, the sum of odd natural numbers is given by n2 where n is the number of odd terms that you are going to add.