There’s a popular story that Gauss, mathematician extraordinaire, had a lazy teacher. The so-called educator wanted to keep the kids busy so he could take a nap; he asked the class to add the numbers 1 to 100.

Gauss approached with his answer: 5050. So soon? The teacher suspected a cheat, but no. Manual addition was for suckers, and Gauss found a formula to sidestep the problem:

Let’s share a few explanations of this result and really understand it intuitively. For these examples we’ll add 1 to 10, and then see how it applies for 1 to 100 (or 1 to any number).

Table of Contents

## Technique 1: Pair Numbers

Pairing numbers is a common approach to this problem. Instead of writing all the numbers in a single column, let’s wrap the numbers around, like this:

```
1 2 3 4 5
10 9 8 7 6
```

An interesting pattern emerges: **the sum of each column is 11**. As the top row increases, the bottom row decreases, so the sum stays the same.

Because 1 is paired with 10 (our n), we can say that each column has (n+1). And how many pairs do we have? Well, we have 2 equal rows, we must have n/2 pairs.

which is the formula above.

## Wait — what about an odd number of items?

Ah, I’m glad you brought it up. What if we are adding up the numbers 1 to 9? We don’t have an even number of items to pair up. Many explanations will just give the explanation above and leave it at that. I won’t.

Let’s add the numbers 1 to 9, but instead of starting from 1, let’s count from 0 instead:

```
0 1 2 3 4
9 8 7 6 5
```

By counting from 0, we get an “extra item” (10 in total) so we can have an even number of rows. However, our formula will look a bit different.

Notice that each column has a sum of n (not n+1, like before), since 0 and 9 are grouped. And instead of having exactly n items in 2 rows (for n/2 pairs total), we have n + 1 items in 2 rows (for (n + 1)/2 pairs total). If you plug these numbers in you get:

which is the same formula as before. It always bugged me that the same formula worked for both odd and even numbers – won’t you get a fraction? Yep, you get the same formula, but for different reasons.

## Technique 2: Use Two Rows

The above method works, but you handle odd and even numbers differently. Isn’t there a better way? Yes.

Instead of looping the numbers around, let’s write them in two rows:

```
1 2 3 4 5 6 7 8 9 10
10 9 8 7 6 5 4 3 2 1
```

Notice that we have 10 pairs, and each pair adds up to 10+1.

The total of all the numbers above is

But we only want the sum of one row, not both. So we divide the formula above by 2 and get:

Now this is cool (as cool as rows of numbers can be). It works for an odd or even number of items the same!

## Technique 3: Make a Rectangle

I recently stumbled upon another explanation, a fresh approach to the old pairing explanation. Different explanations work better for different people, and I tend to like this one better.

Instead of writing out numbers, pretend we have beans. We want to add 1 bean to 2 beans to 3 beans… all the way up to 5 beans.

```
x
x x
x x x
x x x x
x x x x x
```

Sure, we could go to 10 or 100 beans, but with 5 you get the idea. How do we count the number of beans in our pyramid?

Well, the sum is clearly 1 + 2 + 3 + 4 + 5. But let’s look at it a different way. Let’s say we mirror our pyramid (I’ll use “o” for the mirrored beans), and then topple it over:

```
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? In case you’re wondering whether it “really” lines up, it does. Take a look at the bottom row of the regular pyramid, with 5′x (and 1 o). The next row of the pyramid has 1 less x (4 total) and 1 more o (2 total) to fill the gap. Just like the pairing, one side is increasing, and the other is decreasing.

Now for the explanation: How many beans do we have total? Well, that’s just the area of the rectangle.

We have n rows (we didn’t change the number of rows in the pyramid), and our collection is (n + 1) units wide, since 1 “o” is paired up with all the “x”s.

Notice that this time, we don’t care about n being odd or even – the total area formula works out just fine. If n is odd, we’ll have an even number of items (n+1) in each row.

But of course, we don’t want the total area (the number of x’s and o’s), we just want the number of x’s. Since we doubled the x’s to get the o’s, the x’s by themselves are just half of the total area:

And we’re back to our original formula. Again, the number of x’s in the pyramid = 1 + 2 + 3 + 4 + 5, or the sum from 1 to n.

## Technique 4: Average it out

We all know that

`average = sum / number of items`

which we can rewrite to

`sum = average * number of items`

So let’s figure out the sum. If we have 100 numbers (1…100), then we clearly have 100 items. That was easy.

To get the average, notice that the numbers are all equally distributed. For every big number, there’s a small number on the other end. Let’s look at a small set:

```
1 2 3
```

The average is 2. 2 is already in the middle, and 1 and 3 “cancel out” so their average is 2.

For an even number of items

```
1 2 3 4
```

the average is between 2 and 3 – it’s 2.5. Even though we have a fractional average, this is ok — since we have an **even** number of items, when we multiply the average by the count that ugly fraction will disappear.

Notice in both cases, 1 is on one side of the average and N is equally far away on the other. So, we can say the average of the entire set is actually just the average of 1 and n: (1 + n)/2.

Putting this into our formula

And voila! We have a fourth way of thinking about our formula.

## So why is this useful?

Three reasons:

1) Adding up numbers quickly can be useful for estimation. Notice that the formula expands to this:

Let’s say you want to add the numbers from 1 to 1000: suppose you get 1 additional visitor to your site each day – how many total visitors will you have after 1000 days? Since thousand squared = 1 million, we get `million / 2 + 1000/2 = 500,500`

.

2) This concept of adding numbers 1 to N shows up in other places, like figuring out the probability for the birthday paradox. Having a firm grasp of this formula will help your understanding in many areas.

3) Most importantly, this example shows there are many ways to understand a formula. Maybe you like the pairing method, maybe you prefer the rectangle technique, or maybe there’s another explanation that works for you. **Don’t give up** when you don’t understand — try to find another explanation that works. Happy math.

By the way, there are more details about the history of this story and the technique Gauss may have used.

## Variations

**Instead of 1 to n, how about 5 to n?**

Start with the regular formula (1 + 2 + 3 + … + n = n * (n + 1) / 2) and subtract off the part you don’t want (1 + 2 + 3 + 4 = 4 * (4 + 1) / 2 = 10).

```
Sum for 5 + 6 + 7 + 8 + … n = [n * (n + 1) / 2] – 10
```

And for any starting number a:

```
Sum from a to n = [n * (n + 1) / 2] – [(a - 1) * a / 2]
```

We want to get rid of every number from 1 up to a – 1.

**How about even numbers, like 2 + 4 + 6 + 8 + … + n?**

Just double the regular formula. To add evens from 2 to 50, find 1 + 2 + 3 + 4 … + 25 and double it:

```
Sum of 2 + 4 + 6 + … + n = 2 * (1 + 2 + 3 + … + n/2) = 2 * n/2 * (n/2 + 1) / 2 = n/2 * (n/2 + 1)
```

So, to get the evens from 2 to 50 you’d do 25 * (25 + 1) = 650

**How about odd numbers, like 1 + 3 + 5 + 7 + … + n?**

That’s the same as the even formula, except each number is 1 less than its counterpart (we have 1 instead of 2, 3 instead of 4, and so on). We get the next biggest even number (n + 1) and take off the extra (n + 1)/2 “-1″ items:

```
Sum of 1 + 3 + 5 + 7 + … + n = [(n + 1)/2 * ((n + 1)/2 + 1)] – [(n + 1) / 2]
```

To add 1 + 3 + 5 + … 13, get the next biggest even (n + 1 = 14) and do

```
[14/2 * (14/2 + 1)] – 7 = 7 * 8 – 7 = 56 – 7 = 49
```

**Combinations: evens and offset**

Let’s say you want the evens from 50 + 52 + 54 + 56 + … 100. Find all the evens

```
2 + 4 + 6 + … + 100 = 50 * 51
```

and subtract off the ones you don’t want

```
2 + 4 + 6 + … 48 = 24 * 25
```

So, the sum from 50 + 52 + … 100 = (50 * 51) – (24 * 25) = 1950

Phew! Hope this helps.

Ruby nerds: you can check this using

```
(50..100).select {|x| x % 2 == 0 }.inject(:+)
1950
```

## Leave a Reply

276 Comments on "Techniques for Adding the Numbers 1 to 100"

Thanks alot

You’re welcome Sandra — I’m always interested in finding several explanations for an idea and seeing which ones work for different people.

This is great. It is exactly what I wanted. A big thank you

wow this has been buggin me for quit some while, Thank you for the answer

[…] But we remember that adding the numbers 1 to n = n(n + 1)/2. Don’t confuse this with n(n-1)/2, which is C(n,2) or the number of pairs of n items. They look almost the same! […]

hi our maths teacher mrs donally taught us this using step 2 it was very intresting

hi i like this website i’ve never thought about his before

how do you send this to a friend by mail?

Hi Suzie, I’m glad you liked it!

If you’d like to share by email, you can copy and paste this link:

http://betterexplained.com/articles/techniques-for-adding-the-numbers-1-to-100/

But appreciate the suggestion, I may need to add an “email this article” button. Thanks!

Here is a more generic way to think about this that lets you calculate any equally spaced series of numbers:

1+2+3+4+5 = 15

n = number of digits in the set

a = the first digit in the set

b = the last digit in the set

(n(a+b))/2 = total

(5(1+6))/2 = 15

1 3 5 7 9 = 25

(5(1+9))/2 = 25

The explanation is really the same as the first explanation above. The difference is that we don’t automatically add 1, we add the first number in the set. Adding the first number to the last number is how we “pair” each number together. Pairing means to group each number in twos such that all pairs sum to the same number (which is why this only works on equally spaced digits).

3 5 7 9 11 = 35

(5(3+11))/2 = 35

5 9 13 = 27

(3(5+13))/2 = 27

Hi xilplaxim, thanks for the insight! Yep, you can extend the explanations above to almost any sequence. Here’s something interesting as well that you made me think of — let’s say your pattern is

1 2 _ 4 5 _ 7 8 _ 10 11 _ 13 14 …

(multiples of 3 are intentionally removed). That seems like a tricky pattern to add up, but you can realize it’s the same as the full pattern

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

minus the “holes”

3 6 9 12 15

You can use the formula to find the full pattern (1 to 15 = 15*16/2 = 120) and subtract off the holes using your formula (3+6+9+12+15) = 5*18/2 = 45). So we do 120 – 45 and get 75, which is the sum (details here).

Your pattern formula can help subtract away any “gap” you want, which is pretty interesting! Thanks again for the note.

can we use the following method to sum when there are odds?

Instead of 0 can we assume n+1 as our last no i.e.n and afterwards subtract tht no. from the sum.

@Archana: That’s interesting way of looking at it, you can do that too. Let’s say you’re adding 1-5. Rather than n=5, have m=6. Then you can do

Sum from 1-6 = m * (m+1)/2 = 6*7/2 = 21. And you subtract m to get

21 – 6 = 15. [which is the sum of 1-5].

Of course, the formula n(n+1)/2 works no matter if n is even or odd. But that’s a good way to think about the odd case.

Thanks, this helped me a lot

So how would I figure out the ammount of odd numbers from 1-100?

Hi PJ, good question. To find the odd numbers from 1-100 (1 to 99, really), I’d use technique #2 (Use two rows):

Notice how each row ads to 100 (n). How many rows do we have? 50. [1-4 is 2 odd numbers, 1-6 has 3 odd numbers… 1-n will have n/2 odd numbers). Lastly, we have twice the amount of numbers we need, so we must divide by 2 again.

So the sum of odd numbers 1 to n [where n is even] would be n * (n/2) * 1/2. In this case, 100 * 50 * 1/2 = 2500.

When I was 12 or 13 (about 3 years ago), I found this way to derive the formula, which is quite similar but not the same as your triangle method.

x

xx

xxx

xxxx

xxxxx

The number of Xs is the sum of 1, 2, 3, 4, and 5. Here’s how I derived the formula (Xs are the important elements of the triangle, Os are used to show what I don’t count, parentheses are used to show what I eliminate, and N is the number you want the sum up to).

xxxxx N

This is the number you want the sum of.

xoooo

xxooo

xxxoo N^2

xxxxo

xxxxx

Squaring N, you get the triangle and some extra elements (which need to be eliminated)

xoooo

xxooo

xxxoo N^2-N

xxxxo

(xxxxx)

There are no Os you need to eliminate in the bottom row, so you can remove it and add it back later.

x(oooo)

xx(ooo)

xxx(oo) (N^2-N)/2

xxxx(o)

(xxxxx)

Half of the elements above the bottom row are Os, so dividing by two gives you the number of Xs.

x(oooo)

xx(ooo)

xxx(oo) (N^2-N)/2+N

xxxx(o)

xxxxx

Add the original bottom row. You now have the triangle.

Time for my favorite part: changing how an equation looks. (N^2-N)/2+N is too sloppy.

(N^2-N)/2+N Original Equation

N^2/2 -N/2+N Distribute

N^2/2+N/2 Add -N/2 to N

(N^2+N)/2 Factor out 1/2

(N)(N+1)/2 Factor out N

That’s how I thought about it at the time, when I was bored one night. Yes, I developed theorems when I was 13. I still haven’t stopped. I like that way of looking at it, and when I showed that to my math class, most everyone understood what I was doing. Just goes to show you that there are no end to the ways that you can derive a formula.

Hi Zac, thanks for the awesome comment! That’s a cool way to look at it, I like the approach of making a full box (n^2) and then taking pieces away.

Yep, it goes to show that there are so many ways of looking at a single formula :).

[…] FYI: The trick is to think – hmm, take one number from the top and one number from the bottom – ie, 99+1=100. Then take the next one from top and bottom because 98+2=100 also. 97+3=100, 96+4=100, etc. So now you’ve gotten rid of 49 of them (all the way to 49+51=100), so you have 4900, then you just have the 50 and the 100 left over, so the answer is 5050. I even looked online to see if this shortcut was there, but this one guy went all way too complex with his “possible solutions.” […]

[…] I just read an entry, which explains the following formula, not in 2 but 4 different ways. Sum from 1 to n = […]

Hi. What if I wanted to find the sum of all even numbers between 80 to 560?

I like your ‘Many explanations will just give the explanation above and leave it at that. I won’t.’

Kudos on your dedication. I have bookmarked your page for future reference.

Best regards and good health,

Leonard Juska

Costa Mesa, CA USA

P.S. If there ever is time, I want to revisit a book I read as a child called

The Trachtenburg Speed System of Basic Mathematics by Ann Cutler and Rudolph McShane of the work by Jakow Trachtenburg. Bantam Books 553 07020 150 First printing 1960; I have a 1973 copy. Intuitive adventure.

@Jenn: Hi Jenn, you could try matching up the numbers like this (using technique #2):

80 82 84 86 …

560 558 556 554 …

and then dividing the sum by 2.

@Leonard: Thanks for the comment! Glad you enjoyed it; that looks like an interesting book :)

Thanks alot :)

Thank you so much for that explanation. I had learned this same formula in my Maths class without knowing how it had been derived. I had to use this formula to find the sum of series in arithmetic progression, and now, I wonder why I hadn’t seen this site before.

Keep up the good work!!