Alternating Sum and Difference of Binomial Coefficients for Given n/Proof 1

Theorem

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

Proof

Lemma

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

$\Box$

For $n > 0$:

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

\(=\)

\(\ds \binom n 0 + \sum_{i \mathop = 1}^{n - 1} \paren{-1}^i \binom n i + \paren{-1}^n \binom n n\)

\(\ds \)

\(=\)

\(\ds \binom n 0 + \sum_{i \mathop = 1}^{n - 1} \paren{-1}^i \paren{\binom {n - 1} {i - 1} + \binom {n - 1} i} + \paren{-1}^n \binom n n\)

Pascal's Rule

\(\ds \)

\(=\)

\(\ds \binom n 0 - \sum_{i \mathop = 1}^{n - 1} \paren{\paren{-1}^{i-1} \binom {n - 1} {i - 1} - \paren{-1}^i \binom {n - 1} i} + \paren{-1}^n \binom n n\)

\(\ds \)

\(=\)

\(\ds \binom n 0 - \paren{-1}^{1 - 1} \binom {n - 1} {1 - 1} + \paren{-1}^{n - 1} \binom {n - 1} {n - 1} + \paren{-1}^n \binom n n\)

Telescoping Series/Example 1

\(\ds \)

\(=\)

\(\ds 1 - 1 + \paren{-1}^{n - 1} - \paren{-1}^{n - 1}\)

Binomial Coefficient with Zero, Binomial Coefficient with Self

\(\ds \)

\(=\)

\(\ds 0\)

Hence the result.

$\blacksquare$