Is it allowed to write $\binom{n}{n+1}$ = 0?

Question

For ex. $\binom{n}{n}$ = $\binom{n-1}{n-1}$ + $\binom{n-1}{n}$ according to the rule $\binom{n}{i}$ = $\binom{n-1}{i-1}$ + $\binom{n-1}{i}$

Answer 1

Defining $\binom{n}{k}$ as the number of size-$k$ subsets of a size-$n$ set, your statement is indeed correct. You can even keep the usual formula in terms of factorials, viz. $\frac{1}{(-1)!}=\frac{1}{\infty}=0$.

Answer 2

Yes, and it's not an arbitrary "convention". By definition, $$\binom x{n+1}=\frac{x(x-1)(x-2)\cdots(x-n)}{(n+1)!},$$ a polynomial of degree $n+1$ with zeros at $x=0,1,2,\dots,n$.