jmc

By the Binomial Theorem, $$(x+y{)}^{100}=(\genfrac{}{}{0ex}{}{100}{0})x^{100}+(\genfrac{}{}{0ex}{}{100}{1})x^{99}y+(\genfrac{}{}{0ex}{}{100}{2})x^{98}{y}^2+\cdots +(\genfrac{}{}{0ex}{}{100}{100})y^{100}.$$Setting $x=1$ and $y=-1,$ we get $$\left(\genfrac{}{}{0ex}{}{100}{0}\right)-\left(\genfrac{}{}{0ex}{}{100}{1}\right)+\left(\genfrac{}{}{0ex}{}{100}{2}\right)-\cdots +\left(\genfrac{}{}{0ex}{}{100}{100}\right)=\boxed{0}.$$