Đáp án đúng: D
Giải chi tiết:\(\begin{array}{l}S = 1 + 11 + 111 + ... + \underbrace {11...111}_{n\,so\,1}\\S = 1 + \left( {{{10}^1} + 1} \right) + \left( {{{10}^2} + {{10}^1} + 1} \right) + ...\left( {{{10}^{n - 1}} + {{10}^{n - 2}} + ... + {{10}^1} + 1} \right)\\S = 1 + \left( {1 + {{10}^1}} \right) + \left( {1 + {{10}^1} + {{10}^2}} \right) + ... + \left( {1 + {{10}^1} + {{10}^2} + ... + {{10}^{n - 1}}} \right)\end{array}\)
Áp dụng công thức tổng của n số hạng đầu tiên của cấp số nhân \({S_n} = \frac{{{u_1}\left( {1 - {q^n}} \right)}}{{1 - q}}\) ta có :
\(\begin{array}{l}1 + {10^1} = \frac{{1 - {{10}^2}}}{{1 - 10}} = \frac{{1 - {{10}^2}}}{{ - 9}}\\1 + 10 + {10^2} = \frac{{1 - {{10}^3}}}{{ - 9}}\\1 + 10 + {10^2} + {10^3} = \frac{{1 - {{10}^4}}}{{ - 9}}\\...\\1 + 10 + ... + {10^{n - 1}} = \frac{{1 - {{10}^n}}}{{ - 9}}\\ \Rightarrow S = 1 + \left( {1 + {{10}^1}} \right) + \left( {1 + {{10}^1} + {{10}^2}} \right) + ... + \left( {1 + {{10}^1} + {{10}^2} + ... + {{10}^{n - 1}}} \right)\\\,\,\,\,\,\,\,\,\,\, = 1 + \frac{{1 - {{10}^2}}}{{ - 9}} + \frac{{1 - {{10}^3}}}{{ - 9}} + ... + \frac{{1 - {{10}^n}}}{{ - 9}}\\\,\,\,\,\,\,\,\,\,\, = 1 - \frac{1}{9}\left( {\underbrace {1 + 1 + ... + 1}_{n - 1\,so\,1} - \left( {{{10}^2} + {{10}^3} + ... + {{10}^n}} \right)} \right)\\\,\,\,\,\,\,\,\,\,\, = 1 - \frac{1}{9}\left( {n - 1 - \frac{{{{10}^2}\left( {1 - {{10}^{n - 1}}} \right)}}{{1 - 10}}} \right)\\\,\,\,\,\,\,\,\,\,\, = 1 - \frac{1}{9}n + \frac{1}{9} + \frac{{100\left( {{{10}^{n - 1}} - 1} \right)}}{{81}}\\\,\,\,\,\,\,\,\,\,\, = \frac{{100\left( {{{10}^{n - 1}} - 1} \right)}}{{81}} + \frac{{10}}{9} - \frac{n}{9}\\\,\,\,\,\,\,\,\,\,\, = \frac{{10\left( {{{10}^n} - 10} \right)}}{{81}} + \frac{{90}}{{81}} - \frac{n}{9}\\\,\,\,\,\,\,\,\,\,\, = \frac{{10\left( {{{10}^n} - 10 + 9} \right)}}{{81}} - \frac{n}{9}\\\,\,\,\,\,\,\,\,\,\, = \frac{{10\left( {{{10}^n} - 1} \right)}}{{81}} - \frac{n}{9}\end{array}\)
Chọn D.
