A Floor-Sum Identity

Problem

Let $n \in \mathbb{N}^*$ . Prove that

\sum_{i=0}^{\left\lfloor n/3 \right\rfloor} \left\lfloor \frac{n - 3i}{2} \right\rfloor = \left\lfloor \frac{n^2 + 2n + 4}{12} \right\rfloor.

Answer

Solution

Difficulty	7/10
Topics	number theory, Induction, Casework

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