Alternating Sum and Difference of r Choose k up to n

Theorem

Let $r \in \R, n \in \Z$.

Then:

$\displaystyle \sum_{k \le n} \left({-1}\right)^k \binom r k = \left({-1}\right)^n \binom {r - 1} n$

Proof

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sum_{k \le n} \left({-1}\right)^k \binom r k\)	\(=\)	\(\displaystyle \sum_{k \le n} \binom {k - r - 1} k\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Negated Upper Index of Binomial Coefficient
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \binom {-r + n} n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Sum of r+k Choose k up to n
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({-1}\right)^n \binom {r - 1} n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Negated Upper Index of Binomial Coefficient

$\blacksquare$