Đáp án đúng: C
Giải chi tiết:\(\begin{array}{l}F = {2^{100}} - {2^{99}} - {2^{98}} - \ldots - {2^2} - 2 - 1\\ \Rightarrow 2F = 2.\left( {{2^{100}} - {2^{99}} - {2^{98}} - \ldots - {2^2} - 2 - 1} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {2^{101}} - {2^{100}} - {2^{99}} - \ldots - {2^3} - {2^2} - 2\\ \Rightarrow 2F - F = \left( {{2^{101}} - {2^{100}} - {2^{99}} - \ldots - {2^3} - {2^2} - 2} \right) - \left( {{2^{100}} - {2^{99}} - {2^{98}} - \ldots - {2^2} - 2 - 1} \right)\\ \Rightarrow F = {2^{101}} - {2^{100}} - {2^{99}} - {2^3} - {2^2} - 2 - {2^{100}} + {2^{99}} + {2^{98}} + \ldots + {2^2} + 2 + 1\\\,\,\,\,\,\,\,\,\,\,\, = {2^{101}} - \left( {{2^{100}} + {2^{100}}} \right) - \left( {{2^{99}} - {2^{99}}} \right) - \left( {{2^{98}} - {2^{98}}} \right) - \left( {2 - 2} \right) + 1\\\,\,\,\,\,\,\,\,\,\,\, = {2^{101}} - {2.2^{100}} + 1\\\,\,\,\,\,\,\,\,\,\,\, = {2^{101}} - {2^{101}} + 1\\\,\,\,\,\,\,\,\,\,\,\, = 1.\end{array}\)
Chọn C.
