$n,m$ 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.

Using the rules: $F(n,1)=F(n,1)=1; F(n+1,m+1)=F(n,m+1)+F(n+1,m)$:
$$\begin{align}F(1,1)&=1;\\
F(1,2)&=F(2,1)=1;\\
F(2,2)&=F(1,2)+F(2,1)=2;\\
F(3,2)&=F(2,2)+F(3,1)=2+1=3;\\
F(2,3)&=F(1,3)+F(2,2)=1+2=3;\\
\vdots\end{align}$$
We get the Pascal triangle for $F(m,n)$:
$$\begin{array}{c|c|c|c|c|c|c}
m/n & 1 & 2 & 3 &4 &5&6&\cdots \\
\hline
1 & 1 & 1 & 1 &1&1&1 \\
\hline
2 & 1 & 2 & 3&4&5&6 \\
\hline
3 & 1 & 3 & 6&10&15&21\\
\hline
4 & 1 & 4 & 10&20&35&56\\
\hline
5 & 1 & 5 & 15&35&70&126\\
\hline
6 & 1 & 6 & 21&56&126&252\\
\hline
\vdots \\
\end{array}$$
Hence:
$$\begin{align}F(m,n)&= {m+n-2\choose n-1}=\frac{(m+n-2)!}{(n-1)!(m-1)!}.\\
F(2,n)&={n\choose n-1}=\frac{n!}{1!(n-1)!}=n.\\
F(3,n)&={n+1\choose n-1}=\frac{(n+1)!}{2!(n-1)!}=\frac{n(n+1)}{2}.\end{align}$$
Note: $F(m,n)=F(n,m)$.
