smc

By using the properties of exponents, we can simplify the given expression as follows to obtain our answer: $$\begin{array}{rl}{2}^{-(2k+1)}-2^{-(2k-1)}+2^{-2k}& =2^{-2k-1}-2^{-2k+1}+2^{-2k}\\ & =\frac{2^{-2k}}{2}-2\cdot 2^{-2k}+2^{-2k}\\ & =2^{-2k}(\frac{1}{2}-2+1)\\ & =2^{-2k}(-\frac{1}{2})\\ & =-\frac{2^{-2k}}{2}\\ & =-2^{-2k-1}\\ & =\boxed{-2^{-(2k+1)}}.\end{array}$$