Sum of squares

Can anyone share a proof or a reference for the following property

\[(a + b + c + d + \ldots)^2 = (a^2 + b^2 + c^2 + \ldots ) + 2(ab + bc + cd \ldots )\]

(a)2=a2+2ab\bigg(\sum {a}\bigg)^2 = \sum {a^2} + 2\sum {ab}

a\sum {a} means cyclic

Note by Mahdi Raza
3 months, 1 week ago

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This isn't true. Set a=b=1a=b=1. You get (1+1)2=(12+12)+(1×1) ( 1 + 1)^2 = (1^2 + 1^2) + (1 \times 1) , which is obviously false.

Pi Han Goh - 3 months, 1 week ago

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What about n>1n > 1?

A Former Brilliant Member - 3 months, 1 week ago

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What about it? I've shown a counterexample. What else is there?

Pi Han Goh - 3 months, 1 week ago

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@Pi Han Goh My question was a=b>1a = b > 1.

What's your answer to that?

A Former Brilliant Member - 3 months, 1 week ago

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@A Former Brilliant Member Just try a=b=2a=b=2. What do you get?

Pi Han Goh - 3 months, 1 week ago

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@Pi Han Goh 161216 \neq 12

I have now understood. I'll tell @Mahdi Raza

A Former Brilliant Member - 3 months, 1 week ago

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No no no.. I've added a clarification of what i meant @Pi Han Goh

Mahdi Raza - 3 months, 1 week ago

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Yup, that's what I assumed. My answer still holds.

Pi Han Goh - 3 months, 1 week ago

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Whoops.. Sorry @Pi Han Goh and @Yajat Shamji.. Forgot a 2 in there

Mahdi Raza - 3 months, 1 week ago

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@Mahdi Raza Read up multinomial theorem.

Pi Han Goh - 3 months, 1 week ago

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@Zakir Husain, @Mahdi Raza needs your help...

A Former Brilliant Member - 3 months, 1 week ago

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@Yajat Shamji, @Mahdi Raza - It is easy to prove, I'll write a note on it's proof. Just wait for sometime!

Zakir Husain - 3 months, 1 week ago

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I said I'm out!

A Former Brilliant Member - 3 months, 1 week ago

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@Mahdi Raza- I have proved the equation (in rough not on brilliant) : (i=0jai)p=b0=0jb1=0jb2=0j...bp1=0jab0ab1ab2...abp1(\sum_{i=0}^{j}a_i)^p=\sum_{b_0=0}^{j}\sum_{b_1=0}^{j}\sum_{b_2=0}^{j}...\sum_{b_{p-1}=0}^{j}a_{b_0}a_{b_1}a_{b_2}...a_{b_{p-1}} where pZp \in Z

Zakir Husain - 3 months, 1 week ago

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If you put p=2p=2 you get

b0=0jb1jab0ab1=a02+a0a1+a0a2...+a0aj+a1a0+a12+a1a2+a1a3...+a1aj+a2a0+a2a1+a22+a2a3...=(a02+a12+a22...aj2)+(a0a1+a0a2...+a0aj+a1a0+a1a2+a1a3...+a1aj...)=i=0jai2+k=0jl=0jakal\sum_{b_0=0}^{j}\sum_{b_1}^{j}a_{b_0}a_{b_1}=a_0^2+a_0a_1+a_0a_2...+a_0a_j+a_1a_0+a_1^2+a_1a_2+a_1a_3...+a_1a_j+a_2a_0+a_2a_1+a_2^2+a_2a_3...=(a_0^2+a_1^2+a_2^2...a_j^2)+(a_0a_1+a_0a_2...+a_0a_j+a_1a_0+a_1a_2+a_1a_3...+a_1a_j...)=\sum_{i=0}^{j}a_i^2+\sum_{k=0}^{j}\sum_{l=0}^{j}a_ka_l

Zakir Husain - 3 months, 1 week ago

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@Mahdi Raza see here

Zakir Husain - 3 months, 1 week ago

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Hmm.. it seems very elaborative but it's hard to see/understand.

Mahdi Raza - 3 months, 1 week ago

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Here's a visual that might help you see how the property works:

The grid shows the multiplication of (a+b+c+d)(a + b + c + d) with (a+b+c+d)(a + b + c + d), and each term is put as a row or column header. Each cell in the grid is then the product of its row and column header.

There is reflective symmetry along the diagonal of the grid, so each term along the diagonal (a2a^2, b2b^2, c2c^2, and d2d^2) appears once, and each of the other terms (abab, acac, adad, bcbc, bdbd, and cdcd) appears twice because they have a reflection.

David Vreken - 3 months, 1 week ago

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Lovely proof!! Thanks for sharing. This is by far the most easiest I’ve seen.

Mahdi Raza - 3 months, 1 week ago

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@David Vreken, i was thinking whether something similar can be created for:

(a+b+c+d+)3(a + b + c + d + \ldots)^3

(a)3\bigg(\sum {a}\bigg)^3

a\sum {a} means cyclic

Mahdi Raza - 3 months, 1 week ago

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@Mahdi Raza see my comments, there i proved an equation for any power pp

Zakir Husain - 3 months, 1 week ago

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You can show that using a cube, but it starts to get large and cumbersome. Here are the 44 different layers of the 4×4×44 \times 4 \times 4 cube showing (a+b+c+d)3(a + b + c + d)^3:

which shows that:

(a+b+c+d+...)3=(a3+b3+c3+d3+...)+3(a2b+a2c+a2d+ab2+b2c+b2d+...)+6(abc+abd+acd+bcd+...)(a + b + c + d + ...)^3 = (a^3 + b^3 + c^3 + d^3 + ...) + 3(a^2b + a^2c + a^2d + ab^2 + b^2c + b^2d + ...) + 6(abc + abd + acd + bcd + ...)

(a)3=a3+3a2b+6abc(\sum a)^3 = \sum a^3 + 3 \sum a^2b + 6 \sum abc

Unfortunately, trying to show a fourth power or higher using this method is extremely complicated because it needs more than three dimensions.

David Vreken - 3 months, 1 week ago

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@David Vreken You don't need to do it for more big numbers see my comments it is for general power

Zakir Husain - 3 months, 1 week ago

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@Zakir Husain Yes, I saw that, and it is very good, but I was answering @Mahdi Raza's question if a similar method (of using the grid) can be created for (a+b+c+d+...)3(a + b + c + d + ...)^3.

David Vreken - 3 months, 1 week ago

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@Zakir Husain @Zakir Husain, can you provide a proof for your general-power solution?

Mahdi Raza - 3 months, 1 week ago

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@Mahdi Raza Yes I'll but before that I'm more interested in areas of different regions remembered this question

Zakir Husain - 3 months, 1 week ago

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@Zakir Husain Actually, the preface of that daily challenge has the answer. Can you access the archive? If not I'll help you with a screenshot for that

Mahdi Raza - 3 months, 1 week ago

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@Mahdi Raza See my new note here also try the Bonus given in the last.

Zakir Husain - 3 months, 1 week ago

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@David Vreken @David Vreken, Superb!! Yeah you're right, we would need 3D visualization which is indeed cumbersome (and messy as well to show so many terms). Thanks for proving these!! Can I use this idea to make a video for (a)2\bigg(\sum {a}\bigg)^2 and (a)3\bigg(\sum {a}\bigg)^3, in case i make one.

P.S. I found your YouTube Channel as well, nice!

Mahdi Raza - 3 months, 1 week ago

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@Mahdi Raza I assume you mean the YouTube Channel with some of my math lectures? I made that for my students when we had to switch to online learning for the quarantine. There's nothing dazzling (like animations) on it, but just some good basic high school math lessons. :-)

David Vreken - 3 months, 1 week ago

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@David Vreken Yeah this one. It's good, i watched one or two.

And, Can I use this idea to make a video for (a)2\bigg(\sum {a}\bigg)^2 and (a)3\bigg(\sum {a}\bigg)^3, in case i make one.

Mahdi Raza - 3 months, 1 week ago

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@Mahdi Raza I hope you are enjoying the puns, too!

Yes, you can use that idea to make a video. It's not original to me, anyway, and at least the two-dimensional grid is a method I found in some textbook (I can't remember which one, though).

David Vreken - 3 months, 1 week ago

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@David Vreken Yeah

"Do you think this is right?"

"I don't know, is it 90 degrees" lol

Mahdi Raza - 3 months, 1 week ago

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@Mahdi Raza ha ha classic

David Vreken - 3 months, 1 week ago

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@Mahdi Raza By the way, I sometimes get a notification when you comment on an old daily challenge problem, but since I don't have a Premium account I can't read them if the daily challenge problem is expired (they usually expire after about a week). So I'm sorry that I'm not responding to those comments!

David Vreken - 3 months, 1 week ago

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@David Vreken Ohh, totally forgot about that. No problem at all, I was just commenting on how great (actually BRILLIANT) your solutions were. Especially those with the animations. I was not on brilliant for a year so i was exploring the archives section. Pretty great work done only on M.S. paint by you. Thanks for making those!

Mahdi Raza - 3 months, 1 week ago

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@Mahdi Raza I'm glad you enjoyed them!

David Vreken - 3 months, 1 week ago

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