I saw (in the book $A~ Walk~ Through~ Combinatorics$) that $\sum_{n \geq 0}{-3 \choose n} = \sum_{n \geq 0}{n+2 \choose 2}(-1)^n$, which confuses me. It seems that it can be derived directly from binomial thm, but is there any explicit formula about this?
Any help is appreciated!
Consider $P = \binom {-n}k$. This is defined as $$P = {-n\cdot (-n-1)\cdot (-n-2) \cdots (n-(k-2))\cdot(-n-(k-1))\over k\cdot (k-1)\cdots 2\cdot 1}$$ as an analog to the binomial coefficient for positive numbers*. We take $-1$ out of each of the k factors in the numerator to get $$= (-1)^k\cdot{(n+k-1)(n+k-2)\cdots(n+k-1-(k-2))(n+k-1-(k-1))\over k!}$$ $$P= (-1)^k \binom{n+k-1}k$$
*Note that this is merely an extension of the binomial coefficient/combination to negative numbers, since factorial does not make too much sense for negative numbers. We expand the binomial coefficient to achieve this:
$$n(n-1)(n-2)\cdots(n-(k-1))(n-k)(n-(k+1))(n-(k+2))\cdots 2\cdot 1\over (n-k)(n-k-1)(n-k-2)\cdots 2\cdot 1 \cdot k!$$ $$={n(n-1)(n-2)\cdots (n-k+1)\over k!}$$
This way the processes are equivalent. Indeed, the Maclaurin series of $(1+x)^{-n}, n\in \mathbf N$ would be represented as a polynomial with such coefficients, just like $(1+x)^n$, and in fact these binomial coefficients can be generalised for $n\in\mathbf Q$.
I believe there should be a $(-1)^n$ in the right sum?
In any case, for $n \geq k \geq 0$ we may write $$\binom{n}{k} = \frac{n!}{k!(n-k)!} = \frac{n! / (n-k)!}{k!} = \frac{n(n-1)\cdots(n-k+1)}{k!}$$ It is this last form that we use to define $\binom{n}{k}$ when $n$ is any real number and $k$ is a non-negative integer, or if you are comfortable with the product notation: $$\binom{n}{k} = \frac{\displaystyle\prod_{m=0}^{k-1} (n-m)}{k!}\quad , \qquad n\in \mathbb{R}, k \in \mathbb{N}$$
