# Generalizing (x_1+ x_n)^2

Hello, I am a recently graduated high school senior going off to college soon, and over the summer I have spend some of my free time experimenting with different problems in mathematics. This is one that I have spent many days on, and have come to a dead end.

Like the title says, I am attempting to generalize the squaring of any sum of variables

$$(x_1+ \dots + x_n)^2$$

So far I've just experimented with multiplying it out with different values of n and trying to spot a pattern.

For n = 3 it comes out to be
$$x^2_1 +x^2_2 +x^2_3 + 2(x_1 x_2 + x_1 x_3 + x_2 x_3)$$

The pattern seems to be the squares of the variables + 2 times all the combinations of 2 variables multiplied together.

I thought higher values of n would help me get a general pattern so I chose n = 6
$$(x_1 + \dots + x_6)^2 = x^2_1 + x^2_2 + x^2_3 + x^2_4 + x^2_5 + x^2_6 + 2(x_1 x_2 + x_1 x_3 + x_1 x_4 + x_1 x_5 + x_1 x_6 + x_2 x_3 + x_2 x_4 + x_2 x_5 + x_2 x_6 + x_3 x_4 + x_3 x_5 + x_3 x_6 + x_4 x_5 + x_4 x_6 + x_5 x_6)$$

Since there are so many terms to be added I thought using Sigma notation would help. The same formula above in Sigma notation is

$$( \sum_{i=1}^6 x_i )^2 = \sum_{i=1}^6 x^2_i + 2( \sum_{i=2}^6 x_1 x_i + \sum_{i=3}^6 x_2 x_i +\sum_{i=4}^6 x_3 x_i + \sum_{i=5}^6 x_4 x_i + x_5 x_6 )$$

This is obviously not a very compact formula, and I was wondering if there is a way to combine the sums of different indices. Or, if there is a better way to go about this.

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$$\left( \sum_{i=1}^n x_i \right)^{2} = \sum_{i=1}^{n}x_{i}^{2} + 2 \sum_{1 \leq i < j \leq n}x_{i} x_{j}$$

Does that help?

$$\left( \sum_{i=1}^n x_i \right)^{2} = \sum_{i=1}^{n}x_{i}^{2} + 2 \sum_{1 \leq i < j \leq n}x_{i} x_{j}$$

Does that help?
I've never used a summation with 2 indices without being a double sum. What exactly does the second sum mean?

It means the sum over all ordered pairs (i, j) where i<j. You could rewrite it using a double sum as follows:

$$\sum_{i=1}^{n} x_i^2 + 2 \sum_{i=1}^{n-1} \sum_{j=i+1}^{n} x_i x_j$$

Or more directly, without separating out the cross terms from the ones with i=j as:

$$\sum_{i=1}^{n} \sum_{j=1}^{n} x_i x_j$$

A quick way to realize this simplification might be:

$$(x_1+x_2 +\cdots + x_n)(x_1+x_2 +\cdots + x_n) = x_1(x_1+x_2 +\cdots + x_n) + x_2(x_1+x_2 +\cdots + x_n) + \cdots+ x_n(x_1+x_2 +\cdots + x_n)$$

The $k^{th}$ term above contains an $x_k^2$ and $x_ix_k$ for every $i \neq k$. So the $k^{th}$ and the $i^{th}$ are the only terms which contain $x_kx_i=x_ix_k$ giving you two copies of every non-square.

If your ring were not commutative, the general result would be

$$(x_1+x_2+\cdots + x_n)^2 = \sum_{i\le n} x_i^2 + \sum_{i \neq j} x_ix_j$$

Which looks (maybe deceptively) less complicated. Of course this applies to the commutative case as well.

disregardthat