The sum of all the whole numbers between 0 and 100 is 4950.
Understanding the Question: Whole Numbers and the Range
To provide an exact answer, it's crucial to understand what "whole numbers" are and what "between 0 and 100" signifies:
- Whole Numbers: These are non-negative integers, meaning they include 0, 1, 2, 3, and so on.
- "Between 0 and 100": In mathematical contexts, "between A and B" typically implies strict inequality (A < x < B). Therefore, the numbers included in this specific sum are those whole numbers that are greater than 0 and less than 100. This set of numbers is:
- 1, 2, 3, ..., 99.
Calculating the Sum Using Arithmetic Progression
To find the sum of this sequence of numbers (1 to 99), we can leverage information about the sum of natural numbers, which form an arithmetic progression.
Leveraging the Reference Information
According to Unacademy's explanation on finding the sum of natural numbers, "the sum of the first 100 natural numbers is equal to 5050." This refers to the sum of:
- 1 + 2 + 3 + ... + 100 = 5050
The reference clearly states that natural numbers can be written as "1, 2, 3, 4, 5, 6, 7, and 8 to 100", confirming the set considered in their calculation.
Deriving the Sum for Numbers Between 0 and 100
Since we need the sum of numbers from 1 to 99, we can easily derive this from the given sum of 1 to 100:
- Start with the sum of the first 100 natural numbers:
Sum (1 to 100) = 5050 - Identify the extra number in this sum that is not "between 0 and 100":
The number 100 is included in the sum of 1 to 100, but it is not between 0 and 100. - Subtract this number from the total sum:
Sum (1 to 99) = Sum (1 to 100) - 100
Sum (1 to 99) = 5050 - 100
Sum (1 to 99) = 4950
Summary of Sums
| Description | Numbers Included | Sum |
|---|---|---|
| Sum of natural numbers from 1 to 100 | 1, 2, ..., 100 | 5050 |
| Sum of whole numbers between 0 and 100 | 1, 2, ..., 99 | 4950 |
Practical Insight: Using the Arithmetic Series Formula
The sum of an arithmetic progression can also be calculated using the formula:
$S_n = \frac{n}{2}(a_1 + a_n)$
Where:
- $S_n$ is the sum of $n$ terms.
- $n$ is the number of terms.
- $a_1$ is the first term.
- $a_n$ is the last term.
For the whole numbers between 0 and 100 (i.e., 1 to 99):
- $n = 99$ (there are 99 numbers from 1 to 99)
- $a_1 = 1$ (the first number)
- $a_n = 99$ (the last number)
Applying the formula:
$S{99} = \frac{99}{2}(1 + 99)$
$S{99} = \frac{99}{2}(100)$
$S{99} = 99 \times 50$
$S{99} = \textbf{4950}$
This calculation confirms the result obtained by leveraging the provided reference.
