Techniques for Adding the Numbers 1 to 100
There’s a well-known story about how Gauss the mathematician extraordinaire was a teacher who was lazy. The teacher tried to keep the students entertained so that he could have his nap. He asked students to add numbers 1 through 100.
Gauss was approached and offered his response”5050. What happened so fast? The teacher was suspicious of cheating however there was none. Manual addition was a trick for suckers, and Gauss discovered an answer to the issue:
Let’s discuss a few examples of this equation to grasp it easily. In these instances, we’ll use 1-10 as a multiplier and then examine how this works for one up to 100 (or 1 to any number).
Technique 1: Pair Numbers
Combining numbers is a popular method to solve this issue. Instead of listing all numbers on one column, we can wrap the numbers around in this manner:
1 2 3 4 5 10 9 8 7 6
A fascinating pattern is revealed: the sum of each column is 11. The row that is at the top increases while the bottom row shrinks and the sum remains the same.
Since 1 is paired with 10 (our number) So, we can conclude that each column contains (n+1). How many pairs are there? Since we have two equally sized rows. We must have two pairs.
that is the formula above.
Wait , what’s with the an odd amount of things?
Hey, I’m glad that you have brought this up. What happens when we add the numbers 1 to 9? There isn’t an equal number of items to pair. Many explanations will simply explain the previous explanation and then leave the rest to chance. I’m not going to.
So let’s multiply the number 1 through 9. However, instead of starting with 1 let’s start counting starting at 0 instead.
0 1 2 3 4 9 8 7 6 5
When we count from 0 we can get the “extra item” (10 in total) therefore we will get an equal number of rows. But, our formula may appear slightly different.
Note that each column has an amount of one (not the previous n+1) because the numbers 0-9 are grouping. In addition, instead of having precisely two rows of n items (for two pairs total) There are 1 item for each of the two rows (for (n + 1)/2 pairs in total). If you put these numbers together you will get:
that is the same formula that was used before. It was always a problem for me that the formula works for odd as well as even numbers. will you not be able to get fractions? You have the same formula but with different results.
Technique 2: Use Two Rows
The method above works but you must deal with odd numbers and even ones in a different way. Do you think there is a better method? Yes.
In lieu of looping them around we can record the numbers in two columns:
1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1
Note that we have 10 pairs. each pair is worth 10+1.
The sum of all of the above numbers is
However, we are only interested in the total of one row but not both. We subtract the formula in the previous formula by two and then get:
This is pretty cool (as amazing as rows of number may be). This works for odd or even numbers of identical items!
Technique 3: Make a Rectangle
I have recently discovered a different explanation that is a new approach to the previous pairing explanation. Different explanations are better for different individuals I find that I prefer this explanation more.
Instead of writing numbers, let’s pretend that you have beans. We’d like to increase 1 bean from 2 beans and three beans… all until 5 beans.
x x x x x x x x x x x x x x x
We could certainly increase the number to 100 or 10 beans, but at 5 you’ll get the idea. How do we determine the amount of beans in our pyramid?
The sum is evidently one plus two + three + four and five. Let’s consider it in differently. Let’s imagine that we reflect our pyramid (I’ll make use of the “o” for the mirrored beans) and then we topple it:
x o x o o o o o x x o o x x o o o o x x x o o o => x x x o o o x x x x o o o o x x x x o o x x x x x o o o o o x x x x x o
Cool, huh? If you’re wondering if the pattern “really” lines up, it actually does. Look at the lower row of the regular pyramid. It’s 5’x (and 1 one). The next row contains one lower than x (4 to total) and 1 additional O (2 to total) to make up the gap. Similar to the pair the sides are increasing while the other side decreases.
Let’s look at the explanation. What number of beans are we able to count in total? That’s only the size of the rectangle.
We have the number of rows (we did not change how many rows of the pyramid) The assortment comprises (n plus 1) units wide, because the 1 “o” is paired up with all “x”s.
It’s important to note that this time we don’t really care about whether n is odd or even, the formula for total area does the job perfectly. If the number n is odd, then we’ll get one even amount of things (n+1) in every row.
Of course we don’t need the entire surface (the amount of x’s as well as O’s) We just want to know the amount of x’s. Because we’ve doubled the x’s in order to obtain the o’s alone are only one-half of what’s the area
We’re back to the formula we used in the beginning. In this case, the number of”x’s” in the Pyramid = 1 , 2, 3, 5 + 4 or the sum of 1 to the number n.
Technique 4: Average it out
We all know this.
average = sum/ amount of objects
We can modify it that we can rewrite
sum = the average * the amount of items
Let’s calculate the total. In the event that we’ve 100 numbers (1…100) Then we have clearly 100 items. This was simple.
For the purpose of calculating the average, note how the figures are equally dispersed. For every large number it’s a tiny number that’s on the other side. Let’s examine a tiny number:
1 2 3
The mean is 2. 2. The 2nd is now in the middle while 1 and 3 “cancel out” so their average is 2.
To have a uniform quantity of items
1 2 3 4
The average ranges in between 3 and 2 – which is 2.5. Even although we’re dealing with an average that is fractional it’s okay because there is the even number of items once we divide the total by number the ugly fraction will vanish.
In both cases that 1. is located on one end of the average , and N is similarly far to the left. We can therefore say that”the average” of the whole set is really just the sum of 1 and the number n: (1 + n)/2.
Incorporating this formula into our formula
And voilà! There is a fourth way to think concerning our mathematical formula.
Then why is this helpful?
1.) Making quick calculations is useful to estimate. Note that the formula can be expanded to:
Let’s suppose you add up the numbers from one to 1000: suppose that you receive one additional user to your site every day. How many visitors would you receive after a period of 1000 days? Since 1000 squared is 1 million We get millions / 2 + 1,000/2 equals 500,500.
2) The idea of adding numbers 1 and N is also seen in other areas, such as the process of determining the likelihood of that Birthday paradox. Being able to comprehend this formula can help you gain understanding of many different areas.
3.) In the end, this illustration shows that the many different ways to comprehend the formula. Perhaps you are a fan of the pairing technique, or maybe you prefer the rectangle method or there’s an alternative method that is more suitable for you. Don’t quit in the event that you don’t understand and try to find a new explanation that is more effective. Happy math.
And, by the way, there are more information regarding the origins of this tale and the technique Gauss could use.
In place of 1 and n what do you feel about 5 to 5?
Begin with the equation (1 + 2 + 3 … + n = (n + 1) + 2) and subtract the portion you don’t need (1 + 2 + 3 = 4 = (4 + 1)) + 2. = 10.).
Sum for 5 + 6 + 7 + 8 + … n = [n * (n + 1) / 2] – 10
And for any number starting with one of the following:
Sum from A to N = [n + (n + 1) + 2] [(a 1) * 2]
We’d like to rid ourselves of all numbers from 1 to 1.
What about even numbers such as two + four + six + eight … and n?
Simply double the formula. To multiply evens between 2 – 50 take 1 + 2 + 3 , 3 … plus 25. Then multiply it:
Sum of 2+4 +6 … + n = 2* (1 + 2+ 3 … + n/2) = 2 * n/2 (n/2 + 1) + 2 = 2 * (n/2 + 1)
To get the evens of 2 to 50, you’d have to multiply 25 * (25 + 1) = 700
What about odd numbers like one + three + five + seven … and n?
It’s exactly the same similar formula to the even one, however, each number is one less than the other (we have 1 instead of 2, 3, instead of 4 etc.). The next largest odd number (n + 1)) and remove the additional (n + 1)/2 “-1 items: