Binomial formula for multiple summands.

Question

I'm sorry for this simple question and only asking because I can't find information on it, anywhere (maybe I'm looking for the wrong words).

But does

$(x_1+x_2...+x_n)^2=x_1^2+x_2^2+...+x_n^2+2x_1x_2+...+2x_{n-1}x_n$

hold true?

Thanks for any help.

Answer 1

Yes, it does.

Here's a proof by induction:

Let the base case be $n=2$.We'd have $(x_1+x_2)^2 = x_1^2 + x_2^2 + 2x_1x_2$ which is equal to what you have in your question. Now, suppose this is true for $n$ and we'd like to prove it for $n+1$. $(x_1+x_2+...+x_{n+1})^2 - (x_1+x_2+...+x_n)^2 = (x_1+x_2+...+x_{n+1} - x_1-x_2-...-x_n)(x_1+x_2+...+x_{n+1}+x_1+x_2+...+x_n = x_{n+1}(2(x_1 + x_2+...+x_n)+x_{n+1}) = 2x_1x_{n+1}+2x_2x_{n+1} + ... + 2x_nx_{n+1}+x_{n+1}^2$. To get the result you were looking for, we'd have to add $(x_1+x_2+...+x_n)^2 = x_1^2+x_2^2+...+2x_1x_2+...+2x_{n-1}x_n$ to $(x_1+x_2+...+x_{n+1})^2 - (x_1+x_2+...+x_n)^2 = 2x_1x_{n+1}+2x_2x_{n+1} + ... + 2x_nx_{n+1}+x_{n+1}^2 $, which would yield the result you're looking for.

Answer 2

Yes. You can use the Multinomial Theorem. Or you could prove it directly by induction.