Alternating Sum and Difference of Binomial Coefficients for Given n/Proof 2

Theorem

$\ds \forall n \in \Z: \sum_{i \mathop = 0}^n \paren {-1}^i \binom n i = \delta_{n 0}$

$\ds \sum_{i \mathop = 0}^0 \binom 0 0 = 1$

$\Box$

Let $n > 0$.

From the Binomial Theorem, we have that:

$\ds \forall n \in \Z_{\ge 0}: \paren {x + y}^n = \sum_{i \mathop = 0}^n \binom n i x^{n - i} y^i$

Putting $x = 1, y = -1$, we get:

$\ds 0$

$=$

$\ds \paren {1 - 1}^n$

which holds for all $n > 0$

$\ds $

$=$

$\ds \sum_{i \mathop = 0}^n \binom n i 1^{n - i} \paren {-1}^i$

$\ds $

$=$

$\ds \sum_{i \mathop = 0}^n \paren {-1}^i \binom n i$

$\blacksquare$

1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text I$. Algebra: The Binomial Theorem: Relations between coeffficients