# Sum of squares

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

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

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

$\sum {a}$ means cyclic Note by Mahdi Raza
3 months, 1 week ago

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

- 3 months, 1 week ago

What about $n > 1$?

- 3 months, 1 week ago

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

- 3 months, 1 week ago

My question was $a = b > 1$.

- 3 months, 1 week ago

Just try $a=b=2$. What do you get?

- 3 months, 1 week ago

$16 \neq 12$

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

- 3 months, 1 week ago

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

- 3 months, 1 week ago

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

- 3 months, 1 week ago

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

- 3 months, 1 week ago

- 3 months, 1 week ago

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

- 3 months, 1 week ago

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

- 3 months, 1 week ago

I said I'm out!

- 3 months, 1 week ago

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

- 3 months, 1 week ago

If you put $p=2$ you get

$\sum_{b_0=0}^{j}\sum_{b_1}^{j}a_{b_0}a_{b_1}=a_0^2+a_0a_1+a_0a_2...+a_0a_j+a_1a_0+a_1^2+a_1a_2+a_1a_3...+a_1a_j+a_2a_0+a_2a_1+a_2^2+a_2a_3...=(a_0^2+a_1^2+a_2^2...a_j^2)+(a_0a_1+a_0a_2...+a_0a_j+a_1a_0+a_1a_2+a_1a_3...+a_1a_j...)=\sum_{i=0}^{j}a_i^2+\sum_{k=0}^{j}\sum_{l=0}^{j}a_ka_l$

- 3 months, 1 week ago

@Mahdi Raza see here

- 3 months, 1 week ago

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

- 3 months, 1 week ago

Here's a visual that might help you see how the property works: The grid shows the multiplication of $(a + b + c + d)$ with $(a + b + c + d)$, 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 ($a^2$, $b^2$, $c^2$, and $d^2$) appears once, and each of the other terms ($ab$, $ac$, $ad$, $bc$, $bd$, and $cd$) appears twice because they have a reflection.

- 3 months, 1 week ago

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

- 3 months, 1 week ago

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

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

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

$\sum {a}$ means cyclic

- 3 months, 1 week ago

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

- 3 months, 1 week ago

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

$(a + b + c + d + ...)^3 = (a^3 + b^3 + c^3 + d^3 + ...) + 3(a^2b + a^2c + a^2d + ab^2 + b^2c + b^2d + ...) + 6(abc + abd + acd + bcd + ...)$

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

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

- 3 months, 1 week ago

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

- 3 months, 1 week ago

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 $(a + b + c + d + ...)^3$.

- 3 months, 1 week ago

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

- 3 months, 1 week ago

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

- 3 months, 1 week ago

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

- 3 months, 1 week ago

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

- 3 months, 1 week ago

@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 $\bigg(\sum {a}\bigg)^2$ and $\bigg(\sum {a}\bigg)^3$, in case i make one.

- 3 months, 1 week ago

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. :-)

- 3 months, 1 week ago

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

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

- 3 months, 1 week ago

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).

- 3 months, 1 week ago

Yeah

"Do you think this is right?"

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

- 3 months, 1 week ago

ha ha classic

- 3 months, 1 week ago

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!

- 3 months, 1 week ago

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!

- 3 months, 1 week ago