Sum over k of -2 Choose k

Theorem

 * $\ds \sum_{k \mathop = 0}^n \binom {-2} k = \paren {-1}^n \ceiling {\dfrac {n + 1} 2}$

where:
 * $\dbinom {-2} k$ is a binomial coefficient
 * $\ceiling x$ denotes the ceiling of $x$.

Proof
When $n$ is even, we have:

As $n$ is even, $n + 1$ is odd, and so:


 * $\dfrac {n + 2} 2 = \paren {-1}^n \ceiling {\dfrac {n + 1} 2}$

When $n$ is odd, we have:

As $n$ is odd, $n + 1$ is even, and so $\dfrac {n + 1} 2$ is an integer.

Thus from Real Number is Integer iff equals Ceiling:
 * $\paren {-1} \dfrac {n + 1} 2 = \paren {-1}^n \ceiling {\dfrac {n + 1} 2}$

Thus:
 * $\ds \sum_{k \mathop = 0}^n \binom {-2} k = \paren {-1}^n \ceiling {\dfrac {n + 1} 2}$

whether $n$ is odd or even.

Hence the result.