Evaluate $ \binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\cdots+\binom{n}{2k}+\cdots$ [duplicate]

Binomial expansion allows one to find both $\binom n0+\binom n1+\binom n2+\dots=(1+1)^n$ and $(\binom n0-\binom n1+\binom n2+\dots)=(1-1)^n$. Combining these two yields desired result: $\binom n0+\binom n2+\binom n4+\dots=\frac12\bigl((1+1)^n+(1-1)^n\bigr)=2^{n-1}$.

The given sums count the number of even-cardinality subsets of an $n$-element set. I'll provide a simple argument that shows that the value of that sum is $2^{n-1}$.

Fix $x \in X$ (where $|X|=n$). The map $f \colon \mathcal{P}(X) \to \mathcal{P}(X)$ (where $\mathcal{P}(X)$ denotes the power set of $X$) defined by $$ f(Y) = \begin{cases} Y \cup \{x\} &\text{if } x \notin Y \\ Y - \{x\} &\text{if } x \in Y \end{cases} $$

is a bijection of $\mathcal{P}(X)$ that takes even-cardinality subsets to odd-cardinality subsets and vice-versa. Thus there are the same number of even-cardinality subsets as odd-cardinality subsets. Since there are $2^{n}$ total subsets, there must be $2^{n}/2 = 2^{n-1}$ even-cardinality subsets.