Biết 1 + i + i^2 + ...+ i^100 = a + bi (a, b thuộc R)

* Hướng dẫn giải

Đáp án B

$$\begin{gathered} \mathrm{Ta} \mathrm{có} {\mathrm{i}}^{2\mathrm{k}}+{\mathrm{i}}^{2\mathrm{k}+2}={\mathrm{i}}^{2\mathrm{k}}(1+{\mathrm{i}}^2)=0 \\ \mathrm{và} {\mathrm{i}}^{2\mathrm{k}+1}+{\mathrm{i}}^{2\mathrm{k}+3}={\mathrm{i}}^{2\mathrm{k}+1}(1+{\mathrm{i}}^2)=0 \\ \Rightarrow 1+\mathrm{i}+{\mathrm{i}}^2+...+{\mathrm{i}}^{100}=1+(\mathrm{i}+{\mathrm{i}}^3)+({\mathrm{i}}^2+{\mathrm{i}}^4)+...+({\mathrm{i}}^{97}+{\mathrm{i}}^{99})+({\mathrm{i}}^{98}+{\mathrm{i}}^{100}) \\ =1+0+0+...+0+0=1=\mathrm{a}+\mathrm{bi} \\ \Rightarrow \left\{\begin{array}{l}\mathrm{a}=1\\ \mathrm{b}=0\end{array}\right. \end{gathered}$$