Combinatorial argument for solution of recursion behaving similarly as Pascals triangle?

Let $G(n,d)=\frac{F(n,d)+1}{2}$. Then $G$ is given by the recurrence $$G(n,d) = G(n-1,d) + G(n-1,d-1)$$ with initial conditions $G(0,d)=1$ and $G(n,1)=1$. We will show that $$H(n,d)=\sum_{k=0}^{d-1}\binom{n}{k}$$ satisfies the same recurrence and so is equal to $G(n,d)$, from which your identity for $F$ follows easily. The initial conditions are trivial to verify.

Combinatorially, $H(n,d)$ counts the number of subsets of $\{1,\dots,n\}$ with fewer than $d$ elements. Such a subset either contains $n$ as an element or doesn't. The subsets that do contain $n$ are in bijection with the subsets of $\{1,\dots,n-1\}$ with fewer than $d-1$ elements (by removing $n$) and the subsets that do not contain $n$ are in bijection with the subsets of $\{1,\dots,n-1\}$ with fewer than $d$ elements. That is, $$H(n,d)=H(n-1,d-1)+H(n,d-1),$$ as desired.