Calculate the value of $$\displaystyle\sum_{k=0}^{n}\frac{1}{2^k}\binom{n}{k}$$
Setting $x=\frac{1}{2}$ in the binomial expansion $$(1+x)^n=\sum_{k=0}^{n}x^k\binom{n}{k}$$
gives the result immediately: $\boxed{\left(\frac{3}{2}\right)^n}$.