Question

What is the sum of all odd numbers between 0 and 100?

Answer 1

Hint: We need to find the sum of all odd numbers between 0 and 100. We start to solve the given question by finding the number of odd numbers between 0 and 100 using the arithmetic progression formula ${{a}_{n}}=a+\left( n-1 \right)d$ . Then, we find the sum of all odd numbers between 0 and 100 using the formula ${{S}_{n}}=\dfrac{n}{2}\left( a+l \right)$ to get the desired result.

Complete step by step solution:
We are asked to calculate the sum of all odd numbers between 0 and 100. The given question can be solved using the concept of arithmetic progression.
Odd numbers are the numbers that cannot be divided exactly into pairs. All the odd numbers are not exactly divisible by 2.
Example: 1, 3, 5.
Arithmetic progression is a progression in which the difference between the consecutive terms is a constant.
The formula for the ${{n}^{th}}$ term of the arithmetic progression is given as follows,
$\Rightarrow {{a}_{n}}=a+\left( n-1 \right)d$
Here,
${{a}_{n}}$ is the ${{n}^{th}}$ term of the sequence
a is the first term of the sequence
n is the number of terms
d is the common difference
We need to find the number of odd numbers between 0 and 100. It is the same as finding the number of odd numbers from 1 to 99.
The sequence of odd numbers between 0 and 100 is given as follows,
$\Rightarrow 1,3,5,7...........99$
From the above,
${{a}_{n}}$ = 99
a = 1
d = difference between any two consecutive terms = 3 - 1 = 2
Substituting the above values in the formula, we get,
$\Rightarrow 99=1+\left( n-1 \right)2$
Moving the term 1 to the other side of the equation, we get,
$\Rightarrow 99-1=\left( n-1 \right)2$
Simplifying the above equation, we get,
$\Rightarrow 98=\left( n-1 \right)2$
Dividing the equation by 2 on both sides, we get,
$\Rightarrow \dfrac{98}{2}=\dfrac{\left( n-1 \right)2}{2}$
Canceling the common factors, we get,
$\Rightarrow 49=\left( n-1 \right)$
Moving the term 1 to the other side of the equation, we get,
$\Rightarrow n=49+1$
$\therefore n=50$
Now,
The sum of n terms of an arithmetic progression is given as follows,
$\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( a+l \right)$
Here,
${{S}_{n}}$ is the sum of n terms
a is the first term
l is the last term
n is the number of terms
The sequence of the sum of odd numbers between 0 and 100 is given as follows,
$\Rightarrow 1+3+5+7+...........+99$
From the above,
a = 1
l = 99
n = 50
Substituting the values in the formula, we get,
$\Rightarrow {{S}_{50}}=\dfrac{50}{2}\left( 1+99 \right)$
Simplifying the above equation, we get,
$\Rightarrow {{S}_{50}}=\dfrac{50}{2}\left( 100 \right)$
Canceling the common factors, we get,
$\Rightarrow {{S}_{50}}=25\times 100$
$\therefore {{S}_{50}}=2500$

$\therefore$ The sum of all odd numbers between 0 and 100 is 2500.

Note: The given question can be solved alternatively as follows,
The sum of n consecutive odd numbers is given by the formula,
$\Rightarrow 1+3+5+....+\left( 2n-1 \right)={{n}^{2}}$
According to the question, we have to find the sum of the sequence
$\Rightarrow 1+3+5+7+...........+99$
From the above,
$\Rightarrow 2n-1=99$
Moving the term 1 to the other side of the equation, we get,
$\Rightarrow 2n=100$
Dividing the equation by 2 on both sides, we get,
$\Rightarrow \dfrac{2n}{2}=\dfrac{100}{2}$
$\therefore n=50$
Substituting the value of n in the above formula, we get,
$\Rightarrow {{\left( 50 \right)}^{2}}$
$\Rightarrow 2500$