How is $\sum_{r=0}^n C(n-1,r-1)$ possible $\sum_{r=0}^n C(n-1,r-1)$ = C(n-1,-1) + C(n-1,0) + C(n-1,1)+....+ C(n-1,n-1)
How is the 1st term valid i.e C(n-1,-1). If the 1st term is invalid, then $\sum_{r=0}^n C(n-1,r-1)$ is also invalid. Isn't it? Please explain.
But I know if I take C(n,r) = C(n-1,r) + C(n-1,r-1)
S0, C(n-1,r-1) = C(n,r) - C(n-1,r)
Hence $\sum_{r=0}^n C(n-1,r-1)$ = $\sum_{r=0}^n C(n,r)$ - $\sum_{r=0}^n C(n-1,r)$, which equals to $2^{n-1}$. How is this possible?
 A: For integral (even complex) $n$ and integral $r$ the following definition holds:
\begin{align*}
\binom{n}{r}=
\begin{cases}
\frac{n(n-1)\cdots (n-r+1)}{r!}&\quad r\geq 0\\
0&\quad r<0\tag{1}
\end{cases}
\end{align*}
See for instance formula (5.1) in chapter Binomial Coefficients of Concrete Mathematics
by D.E. Knuth, R.L. Graham and O. Patashnik.
From (1) we conclude $\binom{n-1}{-1}=0$.

We obtain
\begin{align*}
\color{blue}{\sum_{r=0}^n\binom{n-1}{r-1}}&=\sum_{r=1}^n\binom{n-1}{r-1}\tag{2}\\
&=\sum_{r=0}^{n-1}\binom{n-1}{r}\tag{3}\\
&\,\,\color{blue}{=2^{n-1}}\tag{4}
\end{align*}

Comment:

*

*In (2) we start with index $r=1$, since $\binom{n-1}{-1}=0$ according to (1).


*In (3) we shift the index $r$ by one to start with $r=0$.


*In (4) we apply the binomial theorem.
A: Recall that $$C(A, B)=\frac{A!}{B!(A-B)!}$$
What is often unconsciously over-applied is the recursive factorial definition $$(k-1)!=\frac{k!}{k}$$
Indeed, this is true wherever $k\in \Bbb N$, and the formula you state $$C(n,r)=C(n-1,r)+C(n-1,r-1)$$
flagrantly uses this very definition. However, when we apply it to $k=0$, we arrive at the massive problem that $(-1)!=\frac 10$. The formula still works under that condition, indeed, if we set $r=0$, we have:
$$\frac{n!}{0!n!}=\frac{(n-1)!}{0!(n-1)!}+\frac{(n-1)!}{(-1)!n!}\implies\frac{1}{(-1)!n!}=0$$ but clearly the use of an undefined number makes this impossible to use  in our sum.
