Choosing k positions among n with no more distance d each other

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Let $\binom{n}{k}_{<d}$ the number of combinations of k elements among [1,n] with constrained spacing :

no element can be at distance d or more from its successor.

$$\binom{n}{k}_{<d} = \sum_j (-1)^j \binom{k-1}{j}\binom{n-dj}{k} $$

How to prove this?

First, I simply thought the identity means that all $\binom{n}{k}$ - (1 case at distance d) - (2 cases) - (3 cases)- ...

But there is $(-1)^j$ term, so that it is not as intuitive as I expected.