Answer
$(1-1)^n=0$
Work Step by Step
First we look for a pattern using Pascal's triangle.
For $n=1$ we have:
$$1-1=0$$
For $n=2$ we have:
$$1-2+1=0$$
For $n=3$ we have:
$$1-3+3-1=0$$
For $n=4$ we have:
$$1-4+6-4+1=0$$
For $n=5$ we have:
$$1-5+10-10+5-1=0$$
We notice that the sum of alternating binomial coefficients is zero.
We prove it by using the Binomial Theorem:
$$\begin{align*}
(1-1)^n&=0=\binom{n}{0}(1^n)+\binom{n}{1}1^{n-1}(-1)+\dots+\binom{n}{n}(-1)^n\\
&=\binom{n}{0}-\binom{n}{1}+\binom{n}{2}\dots+(-1)^n\binom{n}{n}.
\end{align*}$$
