Determine the value of summation form.

Let's start with the famous identity known from Pascal's triangle, $$\binom{k}{n}=\binom{k-1}{n-1}+\binom{k-1}{n}.$$ It's valid for all $k$ and $n$, if we assume $\binom{k}{n}=0$ for $n>k$ or $n<0$. So we can define our sum as $$s_n=\sum_{k\le2n}\binom{k}{n}2^{-k}=\sum_{k\le2n}\binom{k-1}{n-1}2^{-k}+\sum_{k\le2n}\binom{k-1}{n}2^{-k}$$ Obviously, we can rewrite the RHS as $$\frac12\sum_{k\le2n}\binom{k-1}{n-1}2^{-k+1}+\frac12\sum_{k\le2n}\binom{k-1}{n}2^{-k+1},$$ and with a translation of the summation index, this becomes $$\frac12\sum_{k\le2n-1}\binom{k}{n-1}2^{-k}+\frac12\sum_{k\le2n-1}\binom{k}{n}2^{-k}.$$ The sums on the RHS can be expressed by $s_{n-1}$ and $s_n$, so we obtain $$\frac12\,s_{n-1}+\binom{2n-1}{n-1}2^{-2n}+\frac12\,s_n-\binom{2n}{n}2^{-2n-1}=\frac12\,s_{n-1}+\frac12\,s_n$$ because of $$\binom{2n}{n}=\binom{2n-1}{n-1}+\binom{2n-1}{n}=2\binom{2n-1}{n-1}.$$ But this implies $s_n=s_{n-1}$, and the final result is $$s_n=s_1=\binom{1}{1}2^{-1}+\binom{2}{1}2^{-2}=1.$$