Sum of the first n natural numbers, 3 simples ways

1+2+3+...+n,

blackpenredpen,
math for fun,

[aymanabdellatief]

The story goes that Gauss was disrupting his elementary math class and his teacher sent him away and told him to come back when added the first hundred integers thinking it would take Gauss a while to figure this out. Gauss saw the pattern and within a short period of time he came back with the answer and showed the teacher the formula.

[karstenmeinders]

The third way we learned in school at an age of 14, 15. The teacher has us develop the formula on our own, so it is not impossible for kids. The Intuition to build a right-angled triangle helps. The second way surprised me - thanks for this! I am keen to see more ways to prove the sum formula.

[seleneteam]

A sum can always be considered as an integral. Here we have a step function. Fortunately it's obvious, that the linear function y(n)=n+1/2 represents the same area under the graph. Integrate y(n)=n+1/2 from 0 to n and you're done.

[akshat]

The first one is very intuitive. Love it ❤️

[timperry]

Dude, I always love your videos. If my math teachers had been nearly as excited about teaching math, I might have stuck at it.

[BigDBrian]

Here's a geometric way to approach the problem:
S0 = 0

S1: 1
x

S2: 3
x x
x

S3: 6
x x x
x x
x

S4: 10
x x x x
x x x
x x
x

S5: 15
x x x x x
x x x x
x x x
x x
x

etc
now for any of these you can complete the square by adding the previous term, e.g. with S5:
x x x x x
x x x x o
x x x o o
x x o o o
x o o o o

so you get S5+S4 = 5²
or in general, Sn + S(n-1) = n²
note that S(n-1) = Sn - n
Sn + Sn - n = n²
2Sn = n² + n
Sn = (n² + n)/2
QED

Alternatively for the final step you can be a bit quicker, adding one term to itself to get an n by (n+1) rectangle. again I'll use S5 as an example:
x x x x x o
x x x x o o
x x x o o o
x x o o o o
x o o o o o

so we get Sn + Sn = n(n+1)
Sn = n(n+1)/2
QED

[gsittly]

You can also solve it graphically. the sum is a triangle just like
#
##
###


so it is half a rectangle with width n and n+1 height (because of the fact that the diagonal is counted to the sum). So the area is n(n+1) / 2

[heliocentric]

A proof using right triangle:

*O* O O ... O
*O* *O* O ... O
. . .
. . . O
*O* *O* ... *O*

This is nxn square.
So there are (Sn) *O* 's [lower triangle, including the first diagonal]
Notice that 2(Sn) - n = n^2.
[Where n represents the number of *O* 's in the first diagonal, and n^2 is the total number of *O* 's and O 's]
Therefore (Sn)= (n^2+n)/2

[DarshanShah]

1st and 2nd ways even work for higher order sums, like 1^2 + 2^2 + 3^2 + ... and 1^3 + 2^3 + 3^3 + ... and so on! Thanks for this shortcut! 😄

[rohitg]

You are one of the only channels with so many subscribers who doesn't get any dislikes on your videos

[abraund]

Huge fan of the raw practicality of this, if you've not used mathematics for years labouring over some text book is an horrendous thought. It's great to be able to grok some methods and then just apply them where your problem lies. The binomial one in particular is fantastic, I vaguely remember that number pyramid and it's connection to probability, what a great link that you can use it to derive a summation formula.

[sabriath]

I've been just using this formula:

(A+B)*C/2

A = first number in sequence
B = last number in sequence
C = count of numbers in total sequence

I use it in stocks when dithering my money based on the buy orders I can make available. For example, if I want X shares of stock at each level starting at 44 cents and each 44 cents up until $7.48....well, that's 7.48/0.44 = 17 count, which comes to (0.44+7.48)*17/2 = 67.32. If I have $1000 to invest, then I can just do 1000/67.32 = 14 shares at each level (with 57.52 left over).

[hankinoco]

12:56 I am still fascinated by Gauss, such an elegant way to solve this problem.

[Anteater]

9:57 Thanks for the pick up equation. Them girls be loving me now.

[DjVortex-w]

The way I like to approach the problem is to think that the sum is equal to the _average_ of all the numbers multiplied by their amount. Thus I need to resolve what the average of the numbers is, and how many there are.

The nice thing about this approach is that it works the same with any linear series of numbers, not just 1+2+3+4+...+n. For example, you can easily use the same principle to come up with a formula for questions like "what is the sum of all even numbers between 2 and 100", or "what is the sum of all numbers divisible by 7 between 1 and 100" and so on.

[ylevision]

I found: n(n+1)

2

This is just the formula for a 90 degree triangle plus the completion of half squares at the hypotenuse to make the result in whole numbers. I discovered this after seeing that a triangle-like shape made out of cubes would have 1+2+3 + ... + n number of cubes for each diagonal row.

[manla]

Just a additional note for this sum: with this formula of sum of first n integers, you can use it to find out the sum of square integers and then cubic, and quads....

[زكريا_حسناوي]

شكراً جزيلاً لك، الطريقة الأولى كانت رائعة، وصلنا إلى الصيغة من البداية كأشخاص لا نعرف عن الصيغة شيئاً ولا تحتاج للكثير من المعرفة المسبقة، وبالنسبة لطريقة غاوس فهي جميلة ولكنها تصح مع مجموعة ضيقة من المتسلسلات أما هذه الطريقة عامة رائعة تصلح لمجموعة واسعة من المتتاليات

[ThePianofreaky]

The geometric way to prove it is the best way. N + (N-1) + ... + 3 + 2 + 1 is the area of a staircase with height N. To get the staircase, we first take the square of N and divide it by two. Geometrically, we cut it in half on the diagonal. This is already almost the staircase. What's missing is the zig zag triangles that make it a staircase and not a smooth slope. The area of each little triangle is 1/2. There are N small triangles in total. That means we add N/2 to the area. In the end we get 1/2 N^2 + 1/2 N.

[elnotacom]

And there is the geometric way. The square of size n has n x n elements and it is equal to n + 2 × S where n = elements in diagonal and S = = elements over/below the diagonal. So you add n to the equation and it is equivalent to method 3. Sorry for my bad english