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

Given the following recursion:

$$F(n,d) = F(n-1,d) + F(n-1,d-1) + 1$$

With initial conditions $$F(0,d)=1,F(n,1)=1$$ and $$n\in\mathbb N_0, d\in\mathbb N$$.

I noticed that it holds (By writing out the table for $$n,d$$ and doing some tweaking):

$$F(n,d)=2\sum_{k=1}^{d-1}\binom{n}{k}+1$$

Can this solution be justified (proven) by a combinatorial argument?

I want to avoid proving it by solving the above recursion in two vairables.

The recursion seems similar to the one for diagonals in Pascals triangle.