If \({2}^{{2020}}-{2}^{{2019}}-{2}^{{2018}}+{2}^{{2017}} = \ {k}\times {2}^{{2017}}\), find k.
Notice that if we factor out \(2^{2017}\) from each term on the left side, we have:
\(2^{2017}(2^3-2^2-2^1+1)=2^{2017}(3)\)
Thus, the value of \(k\) is three(3).
