Biết 1 + (1-i) +(1-i)^2+...+(1-i)^10 = a +bi(a,b thuộc R)

* Hướng dẫn giải

Đáp án D

$$\begin{gathered} \mathrm{Đặt} \mathrm{P}=\left(1-\mathrm{i}\right)+{\left(1-\mathrm{i}\right)}^2+...+{\left(1-\mathrm{i}\right)}^{10} \\ \Rightarrow \mathrm{S}+1+\mathrm{P}=\mathrm{a}+\mathrm{bi} \\ \mathrm{Ta} \mathrm{thấy} \mathrm{P} \mathrm{là} \mathrm{tổng} \mathrm{của} 1 \mathrm{cấp} \mathrm{số} \mathrm{nhân} \mathrm{với} {\mathrm{U}}_1=1-\mathrm{i}, \mathrm{q}=1-\mathrm{i}, \mathrm{n}=10 \\ \Rightarrow \mathrm{P}={\mathrm{U}}_1.\frac{1-{\mathrm{q}}^{\mathrm{n}}}{1-\mathrm{q}}=\left(1-\mathrm{i}\right).\frac{1-{\left(1-\mathrm{i}\right)}^{10}}{1-\left(1-\mathrm{i}\right)} \\ \mathrm{Mặt} \mathrm{khác}: {\left(1-\mathrm{i}\right)}^{10}={\left[{\left(1-\mathrm{i}\right)}^2\right]}^5={\left(-2\mathrm{i}\right)}^5=-32{\mathrm{i}}^5=-32\mathrm{i} \\ \Rightarrow \mathrm{P}=\frac{\left(1-\mathrm{i}\right)\left(1+32\mathrm{i}\right)}{\mathrm{i}}=31-33\mathrm{i} \\ \Rightarrow \mathrm{S}=1+31-33\mathrm{i}=32-33\mathrm{i}=\mathrm{a}+\mathrm{bi} \\ \Rightarrow \left\{\begin{array}{l}\mathrm{a}=32\\ \mathrm{b}=-33\end{array}\right. \end{gathered}$$