A nice generalization

Probability Level 3

k=0nb3(n3k+b)\large \sum_{k=0}^{\left\lfloor \frac{n-b}{3} \right\rfloor} \binom{n}{3k+b}

Let n,bZ,n, b \in \mathbb{Z}, where n1n \geq 1 and 0b<30 \leq b < 3. Find the closed form of the sum above.

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