<s>[INST] $n,m$ are natural positive integers. F$(n,1)$=F$(1,n)=1$ F$(n,m+1)$+F$(n+1,m)=$F$(n+1,m+1)$ Write F($2,n$) and F($3,n$) as a function of $n$ How do I solve this problem? I do not know where to begin nor does anyone close to me. I am lost and I need help. Thank you so much for whoever helps.
 [/INST] Picture a quarter-plane rectangular grid in $n$ and $m$.  Then the equations
$$
F(1,n) = F(n,1) = 1
$$
place the value $1$ along the edges of that grid.  Now look at the equation
$$
F(n+1,m+1) = F(n+1,m) + F(n, m+1)
$$
Whenever we have an empty square with filled in values below and to the immediate left, we can now fill in that square using said equation.  Fo for example
$$F(2,2) = F(2,1) + F(1,2) = 1+1 = 2\\
F(3,2) = F(3,1) + F(2,2) = 1+2 = 3 \\ \vdots \\
F(n,2) = F(n,1) + F(n-1,2) = 1 + n-1 = n
$$
And once you have filled in that row,
$$F(2,3) = F(2,2) + F(1,3) = 2+1 = 3\\
F(3,3) = F(3,2) + F(2,3) = 3+3 = 6 \\ 
F(4,3) = F(4,2) + F(3,3) = 6+6 = 10 \\ \vdots \\
F(n,3) = F(n,2) + F(n-1,3) = n + \frac{n(n-1)}{2} = \frac{(n+1)n}{2}
$$
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