Tính tổng \[{S_n} = 1 + 11 + 111 + ... + 11...11\] (có 10 chữ số 1)

Cấp số nhân\[2;\,8;\,x;\,128\] theo thứ tự đó sẽ là\[{u_1};\,\,{u_2};\,\,{u_3};\,\,{u_4}\]ta có

\(\left\{ {\begin{array}{*{20}{c}}{\frac{{{u_2}}}{{{u_1}}} = \frac{{{u_3}}}{{{u_2}}}}\\{\frac{{{u_3}}}{{{u_2}}} = \frac{{{u_4}}}{{{u_3}}}}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{\frac{8}{2} = \frac{x}{8}}\\{\frac{{128}}{x} = \frac{x}{8}}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{x = 32}\\{{x^2} = 1024}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{x = 32}\\{\left[ {\begin{array}{*{20}{c}}{x = 32}\\{x = - 32}\end{array}} \right.}\end{array}} \right. \Leftrightarrow x = 32\)

Chọn B

Đáp án cần chọn là: B

Từ giả thiết ta có:

\(\left\{ {\begin{array}{*{20}{c}}{(x - y) + (3x - 3y) = 2(x + y)}\\{{{(y + 2)}^2} = (x - 2)(2x + 3y)}\end{array}} \right.\)

\(\begin{array}{l}\\ \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{x = 3y}\\{{{(y + 2)}^2} = (3y - 2)(9y)}\end{array}} \right.\\ \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{x = 3y}\\{13{y^2} - 11y - 2 = 0}\end{array}} \right.\\ \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{x = 3y}\\{\left[ {\begin{array}{*{20}{c}}{y = 1}\\{y = - \frac{2}{{13}}}\end{array}} \right.}\end{array}} \right.\end{array}\)

Vậy  \[x = 3;y = 1\] hoặc\[x = - \frac{6}{{13}};y = - \frac{2}{{13}}\]

Đáp án cần chọn là: C