Sum over k of -2 Choose k

Theorem

 * $\displaystyle \sum_{k \mathop = 0}^n \binom {-2} k = \left({-1}\right)^n \left\lceil {\dfrac {n + 1} 2}\right\rceil$

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

Proof
When $n$ is even, we have:

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


 * $\dfrac {n + 2} 2 = \left({-1}\right)^n \left\lceil {\dfrac {n + 1} 2}\right\rceil$

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:
 * $\left({-1}\right) \dfrac {n + 1} 2 = \left({-1}\right)^n \left\lceil {\dfrac {n + 1} 2}\right\rceil$

Thus:
 * $\displaystyle \sum_{k \mathop = 0}^n \binom {-2} k = \left({-1}\right)^n \left\lceil {\dfrac {n + 1} 2}\right\rceil$

whether $n$ is odd or even.

Hence the result.