Toolkit 24

Square of a sum (n terms)

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

Proof

Using the distributive law,

\begin{aligned} (x_1+x_2+\cdots+x_n)^2 &= (x_1+x_2+\cdots+x_n)(x_1+x_2+\cdots+x_n). \end{aligned}

When the expression is expanded, each square

x_1^2,\ x_2^2,\ \ldots,\ x_n^2

appears once.

Every product of two different terms appears twice. For example, $x_1x_2$ appears once when $x_1$ is selected from the first factor and $x_2$ from the second, and once again as $x_2x_1$ .

Therefore,

\begin{aligned} (x_1+x_2+\cdots+x_n)^2 &= x_1^2+x_2^2+\cdots+x_n^2 \\ &\quad + 2x_1x_2+2x_1x_3+\cdots+2x_1x_n \\ &\quad + 2x_2x_3+\cdots+2x_2x_n \\ &\quad + \cdots \\ &\quad + 2x_{n-1}x_n. \quad\square \end{aligned}