The total of the integers 1–100 is the number of pairs (50) multiplied by the sum of **each pair** (101), or 50 x 101 = 5,050. Karl was able to handle what appeared to be a difficult job in a short amount of time by using what he understood about numbers. The fact that his estimate was very close to the actual answer shows that he used his knowledge effectively.

In mathematics, an estimate is a judgment made about a certain quantity or set of quantities, usually without relying on all the information available. For example, when asked how much milk there is in the refrigerator, one might say "There is almost exactly enough for one cup of coffee." Even though one knows from past experience that more than one cup of coffee is never served at once, one makes an estimate based on current needs. Similarly, when asked to calculate the sum of all the integers from 1 to 20, one would probably start with a rough guess and then use mathematical methods to refine this initial estimate.

Estimates can be important in mathematics, especially in areas where exact solutions are not possible or hard to find. For example, when faced with the task of finding a formula for computing mortgage payments, one might begin by estimating how long it will take to pay off the loan if it were paid in full every month. Using this estimate as a guide, one could then calculate the correct monthly payment without having to actually perform **the arduous task** of paying off the debt!

- What is the sum of the first 100 numbers?
- What is the sum of the consecutive numbers from 1 to 100?
- What is the addition of 1 to 100?
- How many whole numbers have a sum of 110?
- What is the sum of the first 100 natural numbers?
- What is the sum of all natural numbers from 1 to 100?
- What are the first 10 multiples of 105?
- What are the first 100 square numbers?

As a result, the total of the successive numbers from 1 to 100 is 5,050. Since 2 × 250 + 3 × 200 = 5,000, this number may be reduced by 2 times the difference between them, or 500.

Thus, the answer is 5,050. This can be verified by using a calculator. However, without using a calculator, it can be proved in **two ways**:

1 It can be proved by **mathematical induction**. The induction hypothesis will be that the number is equal to the sum of the first n terms of the sequence. Thus, the initial case will be for n = 1, which implies that the number must be equal to the first term, i.e., 1. Therefore, assuming that the statement is true up to n-1, we can say that the sum of the first n terms of the sequence is equal to n-1 + (n-1) + (n-1) +... + 1 = (n-1)2 + n = 5(n-1).

2 Another way to prove it is by noting that the sequence is an arithmetic progression with **common difference** 500.

Gauss discovered that if he divided the numbers into two groups (1 to 50 and 51 to 100), he could add them vertically to achieve a total of 101. Gauss knew that his ultimate total would be 50 (101) = 5050 at **that point**. He began the process by adding 1 to each number in the first group.

This added up to 51. He then did the same for **the second group**, adding 52 until he got to 100. The final sum was 102.

He realized that he could have stopped adding numbers after 50, but wanted to show that it wasn't necessary to add all the numbers from 1 to 100. In other words, there are many ways to calculate the sum of integers from 1 to 100, and they might give **different results**. However, one must always end with the same final total as Gauss did here: 5050.

The moral of this story is that there are many ways to calculate something, but no matter which way you take, you will always get the same result. In mathematics, it is important to find multiple proofs of concepts or techniques because some people may find those ideas more appealing than others. For example, some people may want to see how many times the sum of integers from 1 to 100 can be divided by 10 before continuing on while others might stop when they reach 100!

Starting with 0 and counting up to 55, there are 56 pairs of whole numbers that add up to 110. As a result, there are 56 pairs of whole numbers with a total of 110.

Each time you count forward or backward by 10, you get to a multiple of 11 or a number between 0 and 10. These are the only numbers that can be the sum of **two whole numbers**.

So, the first pair of **whole numbers** is 0 and 10. They add up to 20, which is less than 30. So, the next pair is greater than or equal to 30. The only number that is greater than or equal to 30 and less than 40 is 36. It's not possible to say more about **this number** since it's not possible to say whether it's greater than or equal to 30. We know only that it is less than 40. Thus, the only pair of whole numbers that adds up to 70 is 36 and 46.

Since we're only considering pairs of whole numbers, we don't need to worry about numbers like 35 or 39. They can't be the sum of two whole numbers because no matter what numbers we choose, they will always be larger than the first number and smaller than the second number.

[2a+(n-1)] S100=n2 We obtain S100 = 1002 [2x1+(100-1)x1] by swapping the variables. [2+99] S100=50 50x101 = S100 5050 = S100 5050 is the sum of **the first 100 natural numbers**.

Clearly, it is an arithmetic progression with a start term of 1, a last term of 100, and a total of 100 terms. As a result, the sum of **the first 100 natural numbers** equals 5050.

105, 210, 315, 420, 525, 630, 735, 840, 945, and 1050 are the first ten multiples of 105. As a result, the total of the first ten multiples is 5775: 105 + 210 + 315 + 420 + 525 + 630 + 735 + 840 + 945 + 1050. This number can be divided by 15, which reveals that it is equal to 40. Therefore, the sum of **all the multiples** of 105 up to but not including 1050 is 40 * 15 = 600.

The first 100 square numbers are listed below.