Explain how $\binom{n}{0}-\binom{n}{1}+\binom{n}{2}+...+(-1)^{n}\binom{n}{n} = 0$

Question

I understand how the sum of the bottom row of Pascal's triangle is equal to $2^n$, but how is this formula equal to zero? Shouldn't it be 1?

$\binom{n}{0}-\binom{n}{1}+\binom{n}{2}+...+(-1)^{n}\binom{n}{n} = 0$

Answer 1

Reposing my comment as an answer, at callculus' request:

The above sum should equal zero, as it can be viewed as a binomial theorem expansion of $\big[1 + (-1)\big]^n$.

Answer 2

Consider the sequence of $n+1$ numbers,

$$\color{red}{1},0,0,0,0,0,...0$$

Taking forward differences, differences between each consecutive terms, gives,

$$\color{red}{-1},0,0,0,0,...0$$

Again,

$$\color{red}{1},0,0,0,...0$$

And so on. In total we do this "difference" process $n$ times. This gives the sequence of $n+1$ numbers $\color{red}{1,-1,1,....}$.

So the Umbral Taylor Series, if we take the first term to be $a_0$, for this sequence is given by:

$$a_x=\sum_{i=0}^{n} (-1)^{i}{x \choose i}$$

This works for $0 \leq x \leq n$ and $x$ and integer, as defined by our sequence. And obviously if $x>0$ we have $a_x=0$. So take $x=n$ and we get what we want.