Sum of the squares of 'n' Consecutive integers - Simple Proof

In this video I show the proof for determining the formula for the sum of the squares of "n" consecutive integers, i.e. 1^2 + 2^2 + 3^2 +.... + n^2. This is a pretty abstract proof and makes use of the useful "telescoping" or collapsing sum which I illustrate in the video. The formula for the sum of squares comes up very often in calculus so it's a good idea to understand the proof!

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[mes]

I don't always prove the sum of squares but when I do I usually first go over the amazing Telescoping Sum ;)

[mylesbenandjustin]

Thank you so much! I've been looking all over the internet for this!

[PancakeDoesGaming]

Ah, that's quite a clever way to go about it, and that telescoping sum you used (which I've never heard of at all before) is interesting! Thank you! If it weren't for you, I would've either had to use up some more of my time to find an understandable proof, or I would've been puzzled on how to prove such a thing... by the way, I've just used this to prove that the volume of a pyramid, any pyramid (no matter how curvy or whatever the base shape below is), is always Bh/3, where B is the base area and h is the "perpendicular" (as opposed to slant) height, so again, thank you for the video!

[leksikolanko]

Your style is perfect actually. Dab on these haters (it's called the pause button, duh). You are both hilariously hasty and extremely accurate. Informative & succinct. No BS. THANK you. Sincerely, A Girl Who Likes Math

[chessandmathguy]

Beautiful. Very easy to follow and was exactly what I was looking for. Nice job! Thanks!

[carlosnava]

only understood because my teacher explains it like this. Do you know if there is any way to find the formulas of of i^(whatever number)?

[apolllos]

How did someone create this proof. . . . I need to know!!!

[kcwilliamson]

Great videos! I have been looking for an easier way to derive these formulas than creating systems, which gets pretty tedious if you don't have a matrix calculator handy. This was just the ticket, and they're all derived the same way simply starting with the collapsing sum formula.

[mes]

thanks for watching! But what do you mean they run off the page? I can view them fine on my computer.

[GusTheWolfgang]

You explained this so badly, you're a mess. I love it <3

[xzanathar]

How did you choose the collapsing sum to start the proof?

[soulcutters]

Glad to find this video, other wise im going to stay curious until exam. My teacher didnt explain how the hell he gets this formula from...

[mes]

That's a great idea! I think I will start attaching links to download the PDF notes and within those notes there will be a link to the Word File notes for those that want to edit them if they would like. I have made this video's notes as PDF. Thanks for the advice :)

[benkao]

So there is a more simple way and intuitive to find proofs for sum. Which rely on the pascal triangle. But not the actual pascal triangle, the arbitrarily modified pascal triangle where you start with the diagonal row you want to summation (sequence or whatever) then perform addition to get only the next row (note: all numbers on one side are ones).
The next row represents the sum of the sequence above, as with all pascal diagonal rows that goes constant, 123, quadratic, cubic, etc.

The summed row should be one degree higher than the sequence you started so whatever degree parent function the sum row is take the number of points necessary (which is just x is one, y is one, x is two, y is the next term etc) to complete the system of equation necessary for the formula for sum of 1, 2, 3, ...x = (n(n+1)/2, or xˆ2, or x to the any power.

Since higher degree polynomial functions require a heck lot of points, I suggest the method of repeated adding of A(n)=A(n-1)+A(n-2), A(n)=A(n-2)+A(n-3)... and so on.

[animohene]

Can you please explain how the telescoping sum is equivalent to the original sum (sum of positive squares)? I understand the proof from the telescoping sum but what is the connection between that and the original problem. How does the sum of i^2 == the sum of (i +1)^3 - i^3, please?

[Kartik_]

Omg
Sought that
Finding from 2 days
Thanks 🙏🙇sir

[mes]

Yes that would have been faster. I just did it manually for those that aren't too familiar with Pascal's triangle.

[kenmawer]

I think you can also prove this by the way Pascal's triangle works. (i.e. nCr)

[shanthicomputercentrebyndo]

Sir DO you think that any derivations available for sum of squares formula?

[MoonLight]

Sir, you could've used Pascal's triangle to expand (n+1)^3.