State the binomial expansion of $(1+x)^n$
So I can do this $$(1+x)^n=\sum_{i=0}^{n} {n\choose i}x^i$$
Then given $n=2k$ is even. Derive an expression for
$$\sum_{i=0}^{2k} (-1)^i{2k\choose i}$$ this is where I am stuck cause this is the same as above with $x=-1$ so is the answer just $$(1+(-1))^{2k}=\sum_{i=0}^{2k} {2k\choose i}(-1)^i=0^{2k}=0 ~\forall k\in \mathbb{N}$$?
A combinatorial interpretation: you are correct in thinking that the sum is zero. One way to express this is that the number of subsets of odd cardinality of an $n$ element set is the same as the number of subsets of even cardinality of an $n$ element set. Indeed we can put these in bijection by taking the symmetric difference of a set with $\{1\}$, i.e. $$ A\mapsto A\triangle \{1\}. $$
