Xin chia sẻ cách tính dãy tổng quát:
$\begin{array}{l}
S = \overline k + \overline {kk} + \overline {kkk} + ... + \overline {\underbrace {kkk......k}_n} \\
\Leftrightarrow S = k\left( {\overline 1 + \overline {11} + \overline {111} + ... + \overline {\underbrace {111......1}_n} } \right) \\
\Leftrightarrow S = \dfrac{k}{9}\left( {\overline 9 + \overline {99} + \overline {999} + ... + \overline {\underbrace {999......9}_n} } \right) \\
\Leftrightarrow S = \dfrac{k}{9}.\left( {10 - 1 + {{10}^2} - 1 + {{10}^3} - 1 + ... + {{10}^n} - 1} \right) \\
\Leftrightarrow S = \dfrac{k}{9}.\left( {10 + {{10}^2} + {{10}^3} + ... + {{10}^n} - (\underbrace {1 + 1 + 1 + ... + 1}_n)} \right) \\
\Leftrightarrow S = \dfrac{k}{9}\left( {\dfrac{{{{10}^{n + 1}} - 10}}{9} - n} \right) \\
\end{array}$
Ở bài trên ứng với k = 3 và n = 33 thì:
$\begin{array}{l}
S = \dfrac{3}{9}\left( {\dfrac{{{{10}^{33 + 1}} - 10}}{9} - 33} \right) = \dfrac{{10.\left( {{{10}^{33}} - 1} \right)}}{{3.9}} - 11 \\
= \dfrac{{10.\underbrace {999...99}_{33}}}{{3.9}} - 11 = \dfrac{{\underbrace {111...11}_{33}0}}{3} - 11 \\
= \underbrace {370370...370}_{11\;so\;370} - 11 = \underbrace {370370...370}_{10\;so\;370}359 \\
\end{array}$
