Simplifying Hypergeometric Function

Writing:

Sum[2^k n/(n - k) Binomial[n - k, 2 k] // FunctionExpand, {k, 0, n/3}] // ComplexExpand

I get:

2^(-1 + n) + Cos[(n \[Pi])/2]

which is what you want.

Just in case for the next time it pays to have several methods. Here is a different way:

Table[Sum[2^k n/(n - k) Binomial[n - k, 2 k], {k, 0, n/3}], {n, 3, 39,3}]
FindSequenceFunction[%, n] // ComplexExpand

{4, 31, 256, 2049, 16384, 131071, 1048576, 8388609, 67108864, \
536870911, 4294967296, 34359738369, 274877906944}

2^(-1 + 3 n) + Cos[(n \[Pi])/2]