Pythagoras and finite orthogonal sums

At $n=0$ the sum is $0$ and both sides vanish, and at $n=1$ the identity is $\mathrm{\parallel }{x}_1{\mathrm{\parallel }}^2=\mathrm{\parallel }{x}_1{\mathrm{\parallel }}^2$.

For two orthogonal vectors $x,y$, expansion gives $\mathrm{\parallel }x+y{\mathrm{\parallel }}^2=\mathrm{\parallel }x{\mathrm{\parallel }}^2+⟨x,y⟩+⟨y,x⟩+\mathrm{\parallel }y{\mathrm{\parallel }}^2=\mathrm{\parallel }x{\mathrm{\parallel }}^2+\mathrm{\parallel }y{\mathrm{\parallel }}^2$, and the same computation with $y$ replaced by $-y$ shows that the sum of a finite orthogonal family may be split off one term at a time.

Suppose the identity holds for orthogonal families of $n-1$ terms, $n\ge 2$, and let $x_1,\dots ,x_n$ be pairwise orthogonal; the partial sum $s={\sum }_{j=1}^{n-1}{x}_j$ satisfies $⟨s,x_n⟩={\sum }_{j=1}^{n-1}⟨x_j,x_n⟩=0$ by linearity in the first argument, and $\mathrm{\parallel }s{\mathrm{\parallel }}^2={\sum }_{j=1}^{n-1}\mathrm{\parallel }{x}_j{\mathrm{\parallel }}^2$ by the induction hypothesis, so $\mathrm{\parallel }{\sum }_{j=1}^n{x}_j{\mathrm{\parallel }}^2=\mathrm{\parallel }s+x_n{\mathrm{\parallel }}^2=\mathrm{\parallel }s{\mathrm{\parallel }}^2+\mathrm{\parallel }{x}_n{\mathrm{\parallel }}^2={\sum }_{j=1}^n\mathrm{\parallel }{x}_j{\mathrm{\parallel }}^2$.

Induction on $n$ from the cases $n=0,1$ of step 1.1 and the induction step of step 2.1 proves the identity for every $n$, so pairwise orthogonal vectors satisfy $\mathrm{\parallel }{\sum }_{j=1}^n{x}_j{\mathrm{\parallel }}^2={\sum }_{j=1}^n\mathrm{\parallel }{x}_j{\mathrm{\parallel }}^2$.