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Algebra Difficulty 6.2 National olympiad Prove it

9. 证明：
[a1,a2,a3,⋯ ,an]=[[a1,a2],a3,⋯ ,an]=[[a1,⋯ ,ar],[ar+1,⋯ ,an]].\begin{array}{l} \quad\left[a_{1}, a_{2}, a_{3}, \cdots, a_{n}\right]=\left[\left[a_{1}, a_{2}\right], a_{3}, \cdots, a_{n}\right] \\ =\left[\left[a_{1}, \cdots, a_{r}\right],\left[a_{r+1}, \cdots, a_{n}\right]\right] . \end{array}

Solution

9. L(a1,a2,⋯ ,an)=L([a1,a2],a3,⋯ ,an)=L([a1,⋯ ,ar],[ar+1,⋯ ,an])\mathscr{L}\left(a_{1}, a_{2}, \cdots, a_{n}\right)=\mathscr{L}\left(\left[a_{1}, a_{2}\right], a_{3}, \cdots, a_{n}\right)=\mathscr{L}\left(\left[a_{1}, \cdots, a_{r}\right],\left[a_{r+1}, \cdots, a_{n}\right]\right).

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.