A Dozen Proofs: Sum of Integers Formula (visual proofs) #SoME2

In this video, we explore the famous formula for the sum of the first n positive integers. In particular, we present twelve proofs of the sum formula using induction, area-based techniques, combinatorial techniques, physical techniques, and by using a couple of deep theorems. All of the proofs except the first are visually inspired or have a visual component. #SoME2 #manim #visualproof

Comment with your favorite of these twelve or let me know if you have a different favorite proof of this fact!

This video is my submission to the "Summer of Math Exposition 2" contest. The key takeaway is that we can gain exposure to many areas of mathematics by "thinking deeply of simple things" as suggested by mathematician Arnold Ross.

0:00 Introduction : Think Deeply About Simple Things
1:08 Proof by induction
2:49 Classic visual proof and "reverse and add"
4:45 Triangle area proof
5:30 Fundamental theorem of calculus proof
7:04 Trapezoid area proof
7:55 Double counting proof
9:15 Bijective proof
10:40 Linear recurrence proof
12:10 Pick's theorem proof
14:11 Euler's formula proof
16:10 Water flow diagram proof
17:49 Center of mass proof
19:19 Concluding remarks
20:38 Citations

#sumformula #sumintegers #integers​ #mathvideo​ #math​ #mtbos​ #animation​ #theorem​ #pww​ #proofwithoutwords​​ #proof​ #iteachmath #mathematics #3b1bsome2 #combinatorialproof #combinatorics #integralcalculus #area #areas #bijection #trapezoid #triangle #physics #moments #weight #centerofmass #waterflow #recurrences #linearrecurrence #gauss #doublecount

That paper includes many references, but here are a few more relevant sources for proofs from this video:

Joe DeMaio and Joey Tyson, Proof without words: A graph theoretic summation of the

Jaime Gaspar, Proof without words: using trapezoids to compute triangular numbers,

Tom Edgar, Proof without words: matchstick triangles, College Math. J. 47 (2016),

Tom Edgar, Proof without words: a recursion for triangular numbers and more, Math. Mag.

David Treeby, A moment’s thought: centers of mass and combinatorial identities, Math.

If you enjoyed this video, please like and subscribe. Also feel free to leave a comment noting your favorite of the 12 proofs!

To learn more about animating with manim, check out:
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Music in this video:

[MathVisualProofs]

Tell me which of the 12 is your favorite! Or maybe you have a favorite I didn't include?

[peytonritchie]

You have a great reading voice, I could easily see you having success in voice acting/audiobooks

[Mr_Happy_Face]

This is amazing. I’ve always wanted to find a YouTube video that proves the same thing in many many different ways.

I think in the future you could do a similar video on the connection between binomial expansion, Pascal’s triangle, and the number of ways to choose from a collection.

[johnchessant]

If f(x) is the generating function for a sequence {a_n}, then f(x)/(1-x) is the generating function for the partial sums of a_n. Using this fact, we see that the generating function for 1, 3, 6, 10, ..., is 1/(1-x)^3.
So the sum 1+2+...+n is the x^(n-1) coefficient in the series expansion of 1/(1-x)^3, which we can evaluate as 1/(n-1)! d^(n-1)/dx^(n-1) [1/(1-x)^3] = 1/(n-1)! 3*4*...*n*(n+1) / (1-0)^(n+2) = n(n+1)/2.

[buffalocat]

Here's one that came to me after I couldn't stop thinking about this video:
Consider the function f(x) = 1 + x + ... + x^n = (x^(n+1) - 1) / (x - 1). Evaluating f'(1) in the first expression gives the sum of the first n positive integers. Starting from the right expression and computing lim x->1 f'(x), we can use l'Hopital's rule to get the desired formula!
edit: I see I wasn't the first to come up with this!!

[monsieur]

The center of mass proof is amazing! All of them are, but that one is the best one

[edkhil]

I came here from YouTube shorts and I was wondering if you had a public repository with all the manim scripts. That would be very helpful!

[Mutual_Information]

I haven’t seen all your videos-you have many-but I’ve seen a lot.. and this is my favorite one. It offers a ton in a short amount of time. Excellent

[mostly_mental]

This is really well done. Lots of beautiful connections (some I'd never seen before), all explained clearly. I'm always a fan of recurrence relations (and the hidden generating functions), but I think Pick's Theorem is my favorite.

[mathflipped]

Great job, Tom! And good luck with this impressive #SoME2 submission. I'm still working on mine. It will be an opener for the Visual Group Theory series.

[keinKlarname]

9:45 This bijection has had an aha-effect on me! So easy - and so brilliant.

[behackl]

Great job with this collection, kudos! 👏

[samueldeandrade]

This guy right here is the real MVP of Mathematics on Youtube. 3blue1brown? Who is that?

[زكريا_حسناوي]

فيديو مذهل، أقدّر هذا المجهود الكبير وأحييك على إتقانك

أتساءل إن كان هناك برهان باستخدام علم المثلثات

[williamweatherall]

very cool. I started to get lost during the combinometric proofs, but the water one got me back. Maybe one day I'll return to this and see it more deeply.

[hamarana]

all I know about math is that 2+2=5. the rest of it I don´t know much.. great videos it gives insights to make us start loving what the universe of numbers is all about!

[neizod]

Reminds me of Philip Ording's 99 Variations on a Proof. Great job!

[andychen]

A few years ago, I used the formula “(n(n+1)) / 2” to make another formula, where instead of counting up by “1”, you count up by “x”.

(n(n+x)) / (2x)

[pengin]

Very nice video! I appreciate the work that went into this. I also didn't know some of the proofs, I really liked the one using Euler's formula! However, we used that the number of faces inside of the graph is n². This is equivalent to the summation formula we want to prove so it feels like we hide something at that step, right?

Here is one proof I just thought of: Consider the expression

f(x)=1+x+x²+...+xⁿ

If we calculate f'(1), we can either use the summation rule to obtain the sum 1+2+...+n. On the other hand we can first use the geometric formula for this expression which gives

f(x)=(xⁿ⁺¹ - 1)/(x-1) (for x≠1)

Now we use the quotient rule we obtain
f'(x)=((n+1)xⁿ *(x-1) - (xⁿ⁺¹ -1))/(x-1)²

We can't just plug in x=1 because this is undefined. But we can use L'Hôpital twice to obtain

f'(1)=((n+1)(n(n+1)-n(n-1)) - (n(n+1)))/2=n(n+1)/2

[markzuckerbread]

subscribed before i even started watching :D
great video :D