Since there are two parts to your question. I will address the one directly dealing with Binomial.

For the purposes of discrete mathematics, the binomial is defined through its generating function: $$ (1+x)^{\alpha} = \sum_{m=0}^\infty \binom{\alpha}{m} x^m $$ It makes evaluations of sums using generating functions much easier if the sum were to run over all integers. So it is natural, in this context, to set $\binom{\alpha}{m} = 0$, $\forall m \in \mathbb{Z}_{< 0}$.

And this convention is indeed adopted in two books you link to, at the expense of breaking the symmetry $\binom{\alpha}{m} = \binom{\alpha}{\alpha-m}$, as explicitly emphasized in "Concrete Mathematics".

In Mathematica, Binomial[z,w] is a defined over $\mathbb{C} \times \mathbb{C}$, and the aforementioned symmetry holds almost everywhere, justifying automatic evaluation

In[62]:= Binomial[n, n - 3] Out[62]= 1/6 (-2 + n) (-1 + n) n 

But the symbolic polynomial above does not give zero for $n \in \mathbb{Z}_{\leqslant -1}$, so we have an inconsistency.

In order to fix all the sums for $c^r_n$ one may use Boole (also known as Iverson bracket). In fact it seems that this particular sum requires Boole even in version 7:

Since there are two parts to your question. I will address the one directly dealing with Binomial.

For the purposes of discrete mathematics, the binomial is defined through its generating function: $$ (1+x)^{\alpha} = \sum_{m=0}^\infty \binom{\alpha}{m} x^m $$ It makes evaluations of sums using generating functions much easier if the sum were to run over all integers. So it is natural, in this context, to set $\binom{\alpha}{m} = 0$, $\forall m \in \mathbb{Z}_{< 0}$.

And this convention is indeed adopted in two books you link to, at the expense of breaking the symmetry $\binom{\alpha}{m} = \binom{\alpha}{\alpha-m}$, as explicitly emphasized in "Concrete Mathematics".

In Mathematica, Binomial[z,w] is a defined over $\mathbb{C} \times \mathbb{C}$, and the aforementioned symmetry holds almost everywhere, justifying automatic evaluation

In[62]:= Binomial[n, n - 3] Out[62]= 1/6 (-2 + n) (-1 + n) n 

But the symbolic polynomial above does not give zero for $n \in \mathbb{Z}_{\leqslant -1}$, so we have an inconsistency.

In order to fix all the sums for $c^r_n$ one may use Boole (also known as Iverson bracket). In fact it seems that this particular sum requires Boole even in version 7: